chapter 10.pdf

60 2 Kinetic Theory of Gases 11.1 Introduction. (i) E3 In gases the intermolecular forces are very weak and its molecule may fly apart in all directions. So the gas is characterised by the following properties. It has no shape and size and can be obtained in a vessel of any shape or size. ID (ii) It expands indefinitely and uniformly to fill the available space. (iii) It exerts pressure on its surroundings. 11.2 Assumption of Kinetic Theory of Gases. U Kinetic theory of gases relates the macroscopic properties of gases (such as pressure, D YG temperature etc.) to the microscopic properties of the gas molecules (such as speed, momentum, kinetic energy of molecule etc.) Actually it attempts to develop a model of the molecular behaviour which should result in the observed behaviour of an ideal gas. It is based on following assumptions : (1) Every gas consists of extremely small particles known as molecules. The molecules of a given gas are all identical but are different than those of another gas. (2) The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses. (3) Their size is negligible in comparison to intermolecular distance (10–9 m) U (4) The volume of molecules is negligible in comparison to the volume of gas. (The volume of molecules is only 0.014% of the volume of the gas). ST (5) Molecules of a gas keep on moving randomly in all possible direction with all possible velocities. (6) The speed of gas molecules lie between zero and infinity (very high speed). (7) The number of molecules moving with most probable speed is maximum. (8) The gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic. (i.e. the total energy before collision = total energy after the collision). (9) Molecules move in a straight line with constant speeds during successive collisions. (10) The distance covered by the molecules between two successive collisions is known as free path and mean of all free paths is known as mean free path. Kinetic Theory of Gases 3 (11) The time spent M a collision between two molecules is negligible in comparison to time between two successive collisions. (12) The number of collisions per unit volume in a gas remains constant. (13) No attractive or repulsive force acts between gas molecules. 60 (14) Gravitational attraction among the molecules is ineffective due to extremely small masses and very high speed of molecules. (15) Molecules constantly collide with the walls of container due to which their momentum E3 changes. The change in momentum is transferred to the walls of the container. Consequently pressure is exerted by gas molecules on the walls of container. (16) The density of gas is constant at all points of the container. 11.3 Pressure of an Ideal Gas. ID Consider an ideal gas (consisting of N molecules each of mass m) enclosed in a cubical box of side L.   It’s any molecule moves with velocity v in any direction where v  v x ˆi  v y ˆj  v z kˆ v x and rebounds with velocity  v x. U This molecule collides with the shaded wall ( A1 ) with velocity Y D YG The change in momentum of the molecule P  (mv x )  (mv x )  2mv x vx – vx X L As the momentum remains conserved in a collision, the change in momentum of the wall A1 is P  2mv x Z After rebound this molecule travel toward opposite wall A2 with velocity  v x , collide to it and again rebound with velocity v x towards wall A1. U (1) Time between two successive collision with the wall A1. Distance travelled by molecule between two successive collision 2L  Velocity of molecule vx ST t   Number of collision per second n  v 1  x t 2 L (2) The momentum imparted per unit time to the wall by this molecule nP  vx m 2mv x  v x2 L 2L This is also equal to the force exerted on the wall A 1 due to this molecule  F  (3) The total force on the wall A 1 due to all the molecules Fx  (4) Now pressure is defined as force per unit area m L v 2 x m 2 vx L 4 Kinetic Theory of Gases Fx m  A AL  Px  So Px  Py  Pz  m V 2 x m V  (v 2 x v m V 3P  m 2 (v1  v 22  v33 ......) V or 3P  mN V or 3P  mN 2 v rms V 2 y and Pz  m V v 2 z E3  v 12  v 22  v 32  v 42 ....      N     As root mean square velocity of the gas molecule  ID 1mN 2 v rms 3 V D YG Important points (i) P  v U P m V [As Px  Py  Pz  P and v 2  v x2  v y2  v z2 ] 2 v 12  v 22  v 32  v 42 .....   N  or Py  Similarly  v y2  v z2 ) 3P  v rms  2 x 60 v v 1mN 2 v rms 3 V or P (m N )T V 2 T] [As v rms (a) If volume and temperature of a gas are constant P  mN i.e. Pressure  (Mass of gas). i.e. if mass of gas is increased, number of molecules and hence number of collision per second increases i.e. pressure will increase. (b) If mass and temperature of a gas are constant. P  (1/V), i.e., if volume decreases, U number of collisions per second will increase due to lesser effective distance between the walls resulting in greater pressure. ST (c) If mass and volume of gas are constant, P  (v rms )2  T i.e., if temperature increases, the mean square speed of gas molecules will increase and as gas molecules are moving faster, they will collide with the walls more often with greater momentum resulting in greater pressure. (ii) P   P 1mN 2 1M 2 v rms  v rms 3 V 3 V 1 2  v rms 3 [As M = mN = Total mass of the gas] M   As   V    (iii) Relation between pressure and kinetic energy Kinetic Theory of Gases 5 1M 2 1 1 2 2  Kinetic energy per unit volume (E)    v rms   v rms M v rms 2 V  2 2 and we know P  From (i) and (ii), we get P  1 2  v rms 3 …..(i) …..(ii) 2 E 3 60 Kinetic energy  i.e. the pressure exerted by an ideal gas is numerically equal to the two third of the mean kinetic energy of translation per unit volume of the gas. Problem 1. E3 Sample Problems based on Pressure The root mean square speed of hydrogen molecules of an ideal hydrogen gas kept in a gas chamber at 0°C is 3180 m/s. The pressure on the hydrogen gas is (Density of hydrogen gas is 8.99  10 2 kg / m 3 , 1 atmosphere  1.01  10 5 N / m 2 ) Solution : (d) As P  (c) 2.0 atm (d) 3.0 atm 1 1 2  v rms  (8.99  10  2 )  (3180 ) 2  3.03  10 5 N /m 2  3.0 atm 3 3 The temperature of a gas is raised while its volume remains constant, the pressure exerted by a gas on the walls of the container increases because its molecules U Problem 2. (b) 1.5 atm ID (a) 0.1 atm D YG (a) Lose more kinetic energy to the wall (b) Are in contact with the wall for a shorter time (c) Strike the wall more often with higher velocities (d) Collide with each other less frequency Solution : (c) Due to increase in temperature root mean square velocity of gas molecules increases. So they strike the wall more often with higher velocity. Hence the pressure exerted by a gas on the walls of the container increases. Problem 3. A cylinder of capacity 20 litres is filled with H 2 gas. The total average kinetic energy of U translatory motion of its molecules is 1.5  10 5 J. The pressure of hydrogen in the cylinder is ST (a) 2  10 6 N / m 2 [MP PET 1993] (b) 3  10 6 N / m 2 (c) 4  10 6 N / m 2 (d) 5  10 6 N / m 2 Solution : (d) Kinetic energy E = 1.5  10 5 J , volume V = 20 litre = 20  10 3 m 3 Pressure  Problem 4. 2 E 2  1.5  10 5   3V 3  20  10 3    5  10 6 N /m 2.   N molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel at temperature T. The mean square of the velocity of molecules of gas B is v2 and the mean square of x component of the velocity of molecules of w2 gas A is w2. The ratio 2 is [NCERT 1984; MP PMT 1990] v (a) 1 (b) 2 (c) 1 3 (d) 2 3 6 Kinetic Theory of Gases Solution : (d) Mean square velocity of molecule  3 kT m For gas A, x component of mean square velocity of molecule  w 2 3kT m For B gas mean square velocity  v 2  Problem 5. 3w 2 v2  3kT …..(ii) 2m w2 2 2 so 2 . 3 1 v E3 From (i) and (ii) …..(i) 60  Mean square velocity  3w 2  A flask contains 10 3 m 3 gas. At a temperature, the number of molecules of oxygen are 3.0  10 22. The mass of an oxygen molecule is 5.3  10 26 kg and at that temperature the rms velocity of molecules is 400 m/s. The pressure in N / m 2 of the gas in the flask is (b) 2.87  10 4 (c) 25.44  10 4 (d) 12.72  10 4 ID (a) 8.48  10 4 Solution : (a) V  10 3 m 3 , N  3.0  10 22 , m  5. 3  10 26 kg , v rms  400 m /s Problem 6. 1 mN 2 1 5. 3  10 26  3.0  10 22 v rms   (400 ) 2  8. 48  10 4 N /m 2. 3 V 3 10  3 U P A gas at a certain volume and temperature has pressure 75 cm. If the mass of the gas is doubled at the same volume and temperature, its new pressure is (b) 75 cm D YG (a) 37.5 cm Solution : (c) P  (c) 150 cm (d) 300 cm 1 M 2 MT v rms  P  3 V V At constant volume and temperature, if the mass of the gas is doubled then pressure will become twice. 11.4 Ideal Gas Equation. ST U A gas which strictly obeys the gas laws is called as perfect or an ideal gas. The size of the molecule of an ideal gas is zero i.e. each molecule is a point mass with no dimension. There is no force of attraction or repulsion amongst the molecule of the gas. All real gases are not perfect gases. However at extremely low pressure and high temperature, the gases like hydrogen, nitrogen, helium etc. are nearly perfect gases. The equation which relates the pressure (P), volume (V) and temperature (T) of the given state of an ideal gas is known as gas equation. Ideal gas equations For 1 mole or NA molecule or M gram or 22.4 litres of gas PV = RT For  mole of gas PV = RT For 1 molecule of gas  R PV    NA  T  kT  Kinetic Theory of Gases 7 For N molecules of gas PV = NkT For 1 gm of gas R PV   T  rT M for n gm of gas PV = nrT R 60 (1) Universal gas constant (R) : Dimension [ML2 T 2 1 ] PV Pressure  Volume Work done   T No. of moles  Temperatu re No. of moles  Temperatu re S.T.P. value : 8.31 E3 Thus universal gas constant signifies the work done by (or on) a gas per mole per kelvin. litre  atm Joule cal  1.98  0.8221 mole  kelvin mole  kelvin mole  kelvin k ID (2) Boltzman's constant (k) : Dimension [ML2 T 2 1 ] R 8.31   1.38  10 23 Joule/kelvin 23 N 6.023  10 R ; M Unit : Joule gm  k elvin D YG r U (3) Specific gas constant (r) : Dimension [L2 T 2 1 ] Since the value of M is different for different gases. Hence the value of r is different for different gases. Sample Problems based on Ideal gas equation Problem 7. A gas at 27°C has a volume V and pressure P. On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be (a) 1800°C (b) 162°C U Solution : (c) From ideal gas equation PV  RT we get (c) 1527°C T2  P2   V2   T1  P1   V1 (d) 600°C   2 P1   3 V1        P1   V1    6  ST  T2  6 T1  6  300  1800 K  1527 C. Problem 8. A balloon contains 500 m 3 of helium at 27°C and 1 atmosphere pressure. The volume of the helium at – 3°C temperature and 0.5 atmosphere pressure will be (a) 500 m 3 Solution : (c) From PV  RT we get Problem 9. (b) 700 m 3 V2  T2   P1    V1  T1   P2 (c) 900 m 3 (d) 1000 m 3   270   1  9 9     V2  500   900 m 3   5   300   0.5  5 When volume of system is increased two times and temperature is decreased half of its initial temperature, then pressure becomes (a) 2 times Solution : (c) From PV  RT we get (b) 4 times P2  T2   V1    P1  T1   V2 (c) 1 / 4 times   T1 / 2   V1        T1   2 V1  1 P    P2  1 4  4 (d) 1 / 2 times 8 Kinetic Theory of Gases Problem 10. The equation of state corresponding to 8g of (a) PV  8 RT O 2 is (c) PV  RT (b) PV  RT / 4 (d) PV  RT / 2 So from PV  RT we get PV  RT 1 RT or PV  4 4 1 4 60 Solution : (b) As 32 gm O 2 means 1 mole therefore 8 gm O 2 means 1 / 4mole i.e.   Problem 11. A flask is filled with 13 gm of an ideal gas at 27°C and its temperature is raised to 52°C. The mass of the gas that has to be released to maintain the temperature of the gas in the flask at 52°C and the pressure remaining the same is (b) 2.0 g (c) 1.5 g (d) 1.0 g E3 (a) 2.5 g Solution : (d) PV  Mass of gas  Temperature In this problem pressure and volume remains constant so M 1 T1 = M2T2 = constant M2 T 12 12 (27  273 ) 300 12  1     13  gm  12 gm  M 2  M1  M1 T2 (52  273 ) 325 13 13 13 ID  i.e. the mass of gas released from the flask = 13 gm – 12 gm = 1 gm. Problem 12. Air is filled at 60°C in a vessel of open mouth. The vessel is heated to a temperature T so U that 1 / 4th part of air escapes. Assuming the volume of vessel remaining constant, the value of T is [MP PET 1996, 99] (a) 80°C (b) 444°C (c) 333°C D YG Solution : (d) M 1  M , T1  60  273  333 K , M 2  M  (d) 171°C M 3M  [As 1 / 4th part of air escapes] 4 4 If pressure and volume of gas remains constant then MT = constant  T2 M 4 4  M  4  1   T2   T1   333  444 K  171 C  3 3 T1 M 2  3M / 4  3 Problem 13. If the intermolecular forces vanish away, the volume occupied by the molecules contained in 4.5 kg water at standard temperature and pressure will be given by (b) 4. 5 m 3 (a) 5. 6 m 3 (d) 11.2 m 3 4. 5 kg Mass of water   250 , T = 273 K and P  10 5 N /m 2 (STP) Molecular wt. of water 18  10  3 kg U Solution : (a)   (c) 11.2 litre ST From PV  RT  V  RT P  250  8.3  273 10 5  5.66 m 3. Problem 14. The pressure P, volume V and temperature T of a gas in the jar A and the other gas in the jar B at pressure 2P, volume V/4 and temperature 2T, then the ratio of the number of molecules in the jar A and B will be [AIIMS 1982] (a) 1 : 1 (b) 1 : 2  N Solution : (d) Ideal gas equation PV  RT    NA number  N 1  P1   V1   N 2  P2   V2   T2     T1 (c) 2 : 1   RT  (d) 4 : 1 where N = Number of molecule, NA = Avogadro   P   V   2T  4        .   2 P   V /4   T  1 Kinetic Theory of Gases 9 Problem 15. The expansion of an ideal gas of mass m at a constant pressure P is given by the straight line D. Then the expansion of the same ideal gas of mass 2m at a pressure P/ 2 is given by the straight line Volume 8 (a) E 6 4 B C 60 (b) C A 2 1 (c) B D E Temperatur e (d) A M M T ; Here   represents the slope of curve drawn on volume and P P E3 Solution : (d) From PV  MT or V  temperature axis. For second condition slope ID M For first condition slope   graph is D (given in the problem) P 2M M  4   i.e. slope becomes four time so graph A is correct in P /2 P U this condition. Problem 16. If the value of molar gas constant is 8.3 J/mole-K, the n specific gas constant for hydrogen in J/mole-K will be (b) 8.3 (c) 16.6 D YG (a) 4.15 Solution : (a) Specific gas constant r  (d) None of these Universal gas constant (R) 8.3   4.15 Joule/mole-K. Molecular weight of gas (M ) 2 Problem 17. A gas in container A is in thermal equilibrium with another gas in container B. both contain equal masses of the two gases in the respective containers. Which of the following can be true (a) PA VA  PB VB (c) PA  PB , VA  VB (d) PA P  B VA VB According to problem mass of gases are equal so number of moles will not be equal U Solution : (b, c) i.e.  A   B (b) PA  PB , VA  VB ST From ideal gas equation PV  RT  PA V A A  PB VB B [As temperature of the container are equal] From this relation it is clear that if PA  PB then Similarly if V A  VB then  VA  A  1 i.e. V A  VB B VB  PA  A  1 i.e. PA  PB. B PB Problem 18. Two identical glass bulbs are interconnected by a thin glass tube. A gas is filled in these bulbs at N.T.P. If one bulb is placed in ice and another bulb is placed in hot bath, then the pressure of the gas becomes 1.5 times. The temperature of hot bath will be (a) 100°C Ice Hot bath 10 Kinetic Theory of Gases (b) 182°C (c) 256°C (d) 546°C Solution : (d) Quantity of gas in these bulbs is constant i.e. Initial No. of moles in both bulb = final number of moles 1.5 PV 1.5 PV PV PV    R(273 ) R(273 ) R(273 ) R(T ) 60 1   2  1'   2' 2 1.5 1.5   273 273 T  T  819 K  546 C.  E3 Problem 19. Two containers of equal volume contain the same gas at pressures P1 and P2 and absolute temperatures T1 and T2 respectively. On joining the vessels, the gas reaches a common pressure P and common temperature T. The ratio P/T is equal to P1 P2  T1 T2 (b) P1T1  P2T2 (T1  T2 )2 (c) P1T2  P2T1 (T1  T2 )2 (d) ID (a) Solution : (d) Number of moles in first vessel  1  P V P1 V and number of moles in second vessel  2  2 RT 2 RT1 If both vessels are joined together then quantity of gas D YG P(2 V ) P1 V P2 V   RT RT1 RT 2 U remains same i.e   1   2 P1 P  2 2T1 2T2 P P P  1  2 T 2T1 2T2 Initiall y P1 T 1 V P2 T 2 V Finally PT V P T V Problem 20. An ideal monoatomic gas is confined in a cylinder by a spring-loaded piston if cross-section ST U 8  10 3 m 2. Initially the gas is at 300K and occupies a volume of 2.4  10 3 m 3 and the spring is in a relaxed state. The gas is heated by a small heater coil H. The force constant of the spring is 8000 N/m, and the atmospheric pressure is 1.0  10 5 Pa. The cylinder and piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and cylinder. There is no heat loss through heater coil wire leads and thermal capacity of the heater coil is negligible. With all the above assumptions, if the gas is heated by the heater until the piston moves out slowly by 0.1m, then the final temperature is (a) 400 K Gas (b) 800 K H Spring (c) 1200 K (d) 300 K Solution : (b) V1  2.4  10 3 m 3 , P1  P0  10 5 N m2 and T1 = 300 K (given) If area of cross-section of piston is A and it moves through distance x then increment in volume of the gas = Ax Kinetic Theory of Gases 11 and if force constant of a spring is k then force F = kx and pressure = V2  V1  Ax  2.4  10 3  8  10 3  0.1  3.2  10 3 and P2  P0  kx  10 5   2  10 5 A 8  10 3 P1 V1 P V 10 5  2.4  10 3 2  10 5  3.2  10 3  2 2   T2  800 K  T1 T2 300 T2 60 From ideal gas equation F kx  A A 8000  0.1 Problem 21. Two identical containers each of volume V0 are joined by a small pipe. The containers contain identical gases at temperature T0 and pressure P0. One container is heated to temperature 2T0 while maintaining the other at the same temperature. The common (a) P  2P0 Solution : (b, c) (b) P  E3 pressure of the gas is P and n is the number of moles of gas in container at temperature 2T0 4 P0 3 Initially for container A (c) n  (d) n  3 P0 V0 2 RT0 P0 V0  n 0 RT 0 P0 V0  n 0 RT 0  n 0  P0 V0 RT 0 Initiall y n0, V0 P0, T0 ID For container B 2 P0 V0 3 RT0 Total number of moles  n 0  n 0  2n 0 (A) n0, V0 P0, T0 (B) U Since even on heating the total number of moles is conserved Hence......(i) n1  n 2  2n 0 If P be the common pressure then D YG For container A PV0  n1 R 2T0 For container B PV0  n 2 RT 0 (A) PV0  n2  RT 0 Substituting the value of n 0 , n1 and n 2 in equation (i) we get P 4 P0 3 No. of moles in container A (at temperature 2T0 ) = n1  U Finally n1, V0 P, 2T0 PV0  n1  2RT 0 n2, V0 P, T0 (B) PV0 PV0 2.P0 V0   2 RT 0 RT 0 RT 0  PV0 2 P0 V0 4  V   P0  0  3 RT 0 2RT 0 3  2 RT 0 ST 4    As P  3 P0    Problem 22. At the top of a mountain a thermometer reads 7°C and a barometer reads 70 cm of Hg. At the bottom of the mountain these read 27°C and 76 cm of Hg respectively. Comparison of density of air at the top with that of bottom is o 7 C, 70 cm of Hg (a) 75/76 (b) 70/76 (c) 76/75 (d) 76/70 Solution : (a) Ideal gas equation, in terms of density 27oC, 76 cm of Hg P1 P2  P T   constant  1  1  2  1 T1  2 T2 2 P2 T1 12 Kinetic Theory of Gases   Top  Bottom  PTop PBottom  TBottom 70 300 75    TTop 76 280 76 11.5 Vander Waal's Gas Equation. 60 All real gases do not obey the ideal gas equation. In order to explain the behaviour of real gases following two modification are considered in ideal gas equation. (i) Non-zero size of molecule : A certain portion of volume of a gas is covered by the Hence the effective volume becomes (V – b) E3 molecules themselves. Therefore the space available for the free motion of molecules of gas will be slightly less than the volume V of a gas. (ii) Force of attraction between gas molecules : Due to this, molecule do not exert that force on the wall which they would have exerted in the absence of intermolecular force. Therefore the observed pressure P of the gas will be less than that present in the absence of ID a   intermolecular force. Hence the effective pressure becomes  P  2  V   Waal’s equation or real gas equation. U The equation obtained by using above modifications in ideal gas equation is called Vander Vander Waal's gas equations a    P  2  (V  b)  RT V   For  moles of gas 2   P  a  V2  D YG For 1 mole of gas   (V  b)   RT   Here a and b are constant called Vander Waal’s constant. U Dimension : [a] = [ML5 T 2 ] and [b] = [L3] Units : a = N  m4 and b = m3. ST 11.6 Andrews Curves. The pressure (P) versus volume (V) curves for actual gases are called Andrews curves. (1) At 350°C, part AB represents vapour phase of water, 1  in this part Boyle’s law is obeyed  P  . Part BC represents V  the co-existence of vapour and liquid phases. At point C, vapours completely change to liquid phase. Part CD is parallel to pressure axis which shows that compressibility of the water is negligible. D G Gas 380°C P H Liquid vapour region 374.1°C Liquid 370°C F C 360°C E 350°C Andrews curve for water B Vapou rA V (2) At 360°C portion representing the co-existence of liquid vapour phase is shorter. Kinetic Theory of Gases 13 (3) At 370°C this portion is further decreased. (4) At 374.1°C, it reduces to point (H) called critical point and the temperature 374.1°C is called critical temperature (Tc) of water. (5) The phase of water (at 380°C) above the critical temperature is called gaseous phase. 60 Critical temperature, pressure and volume The point on the P-V curve at which the matter gets converted from gaseous state to liquid state is known as critical point. At this point the difference between the liquid and vapour E3 vanishes i.e. the densities of liquid and vapour become equal. (i) Critical temperature (Tc) : The maximum temperature below which a gas can be liquefied by pressure alone is called critical temperature and is characteristic of the gas. A gas cannot be liquefied if its temperature is more than critical temperature. and H2O (374.1°C) ID CO2 (304.3 K), O2 (–118°C), N2 (–147.1°C) (ii) Critical pressure (Pc) : The minimum pressure necessary to liquify a gas at critical temperature is defined as critical pressure. O2 (49.7atm) U CO2 (73.87 bar) and D YG (iii) Critical volume (Vc) : The volume of 1 mole of gas at critical pressure and critical temperature is defined as critical volume. CO2 (95 10–6 m3) (iv) Relation between Vander Waal’s constants and Tc, Pc, Vc : Tc  a 8a , Pc  , Vc  3 b , 27 Rb 27 b 2 a 27 R 2 Tc2 RT , b   c 64 Pc 8  Pc    and Pc Vc 3  R Tc 8 U Sample problems based on Vander Waal gas equation Problem 23. Under which of the following conditions is the law PV = RT obeyed most closely by a real ST gas [NCERT 1974; MP PMT 1994, 97; MP PET 1999; AMU 2001] (a) High pressure and high temperature (b) Low pressure and low temperature (c) Low pressure and high temperature (d) High pressure and low temperature Solution : (c) At low pressure and high temperature real gas obey PV = RT i.e. they behave as ideal gas because at high temperature we can assume that there is no force of attraction or repulsion works among the molecules and the volume occupied by the molecules is negligible in comparison to the volume occupied by the gas. 14 Kinetic Theory of Gases  aT 2  V c  (RT  b) , where a, b, c and R are Problem 24. The equation of state of a gas is given by  P   V   constants. The isotherms can be represented by P  AV m  BV n , where A and B depend only on temperature then [CBSE PMT 1995] (b) m  c and n  1 (c) m  c and n  1 (d) m  c and n  1 60 (a) m  c and n  1  aT 2  c Solution : (a)  P  V  RT  b  P  aT 2 V 1  RTV c  bV c  P  (RT  b)V c  (aT 2 )V 1   V   E3 By comparing this equation with given equation P  AV m  BV n we get m   c and n  1. Problem 25. An experiment is carried on a fixed amount of gas at different temperatures and at high pressure such that it deviates from the ideal gas behaviour. The variation of ID shown in the diagram. The correct variation will correspond to PV/RT 2.0 (a) Curve A (b) Curve B PV with P is RT A B 1.0 U (c) Curve C (d) Curve D 0, 0 C D 20 40 60 80 100 P (atm) PV  constant but RT when pressure increase, the decrease in volume will not take place in same proportion so PV will increases. RT D YG Solution : (b) At lower pressure we can assume that given gas behaves as ideal gas so Problem 26. The conversion of ideal gas into solids is (a) Possible only at low pressure (b) Possible only at low temperature (c) Possible only at low volume (d) Impossible U Solution : (d) Because there is zero attraction between the molecules of ideal gas. 11.7 Various Speeds of Gas Molecules. ST The motion of molecules in a gas is characterised by any of the following three speeds. (1) Root mean square speed : It is defined as the square root of mean of squares of the speed of different molecules i.e. v rms  v 12  v 22  v 32  v 42 .... N (i) From the expression for pressure of ideal gas P  v rms  (ii) v rms  3 PV  mN 3 PV  Mass of gas 3PV  Mass of gas Mass of gas    As    V   3P  3 RT  M 1mN 2 v rms 3 V 3 RT M [As if M is the molecular weight of gas PV = RT and Mass of gas =  M ] Kinetic Theory of Gases 15 (iii) v rms  3 RT  M 3 N A kT  N AM 3kT m [As M = NAm and R = NAk] 3P  Root mean square velocity v rms    3 RT  M 3kT m 60 Important points (i) With rise in temperature rms speed of gas molecules increases as v rms  T. E3 (ii) With increase in molecular weight rms speed of gas molecule decreases as v rms  1. M e.g., rms speed of hydrogen molecules is four times that of oxygen molecules at the same temperature. 3 RT M and vs  times that of speed of sound in gas RT U v rms  3 M    v rms  D YG As 3  8.31  273  1840 m / s. 2  10 3 3 RT  M e.g., At NTP for hydrogen gas (v rms )  (iv) rms speed of gas molecules is ID (iii) rms speed of gas molecules is of the order of km/s 3 vs (v) rms speed of gas molecules does not depends on the pressure of gas (if temperature remains constant) because P   (Boyle’s law) if pressure is increased n times then density will also increases by n times but vrms remains constant. (vi) Moon has no atmosphere because vrms of gas molecules is more than escape velocity (ve). U A planet or satellite will have atmosphere only and only if v rms  v e ST (vii) At T = 0; vrms = 0 i.e. the rms speed of molecules of a gas is zero at 0 K. This temperature is called absolute zero. (2) Most probable speed : The particles of a gas have a range of speeds. This is defined as the speed which is possessed by maximum fraction of total number of molecules of the gas. e.g., if speeds of 10 molecules of a gas are 1, 2, 2, 3, 3, 3, 4, 5, 6, 6 km/s, then the most probable speed is 3 km/s, as maximum fraction of total molecules possess this speed. Most probable speed v mp  2P   2 RT  M 2kT m (3) Average speed : It is the arithmetic mean of the speeds of molecules in a gas at given temperature. 16 Kinetic Theory of Gases v av  v1  v 2  v 3  v 4 ..... N and according to kinetic theory of gases Note :   8 RT 8 kT   M  m  60 8P Average speed v av  vrms > vav > vmp (order remembering trick) (RAM) 3: 8  : 2  3 : 2.5 : 2 E3  vrms : vav : vmp =  For oxygen gas molecules vrms = 461 m/s, vav = 424.7 m/s and vrms = 376.4 m/s Sample Problems based on Various speeds ID Problem 27. At room temperature, the rms speed of the molecules of certain diatomic gas is found to be 1930 m/s. The gas is (a) H 2 (b) F2 (c) O 2 3 RT  1930 m /s M 3 RT  M   3  8.31  300  2  10  3 kg  2 gm i.e. the gas is hydrogen. 1930  1930 D YG (1930 ) 2 (given) U Solution : (a) Root means square velocity v rms  Problem 28. Let A and B the two gases and given : TA T  4. B ; where T is the temperature and M is the MA MB molecular mass. If C A and CB are the rms speed, then the ratio (a) 2 (b) 4 3 RT M  CA  CB U Solution : (a) As v rms  (d) Cl 2 T A / TB  MA / MB (c) 1 4 2 CA will be equal to CB (d) 0.5  TA  M  4 A given   As MB  TB  Problem 29. The rms speed of the molecules of a gas in a vessel is 400 ms–1. If half of the gas leaks out at constant temperature, the rms speed of the remaining molecules will be (b) 400 2 ms 1 ST (a) 800 ms–1 (c) 400 ms–1 (d) 200 ms–1 Solution : (c) Root mean square velocity does not depends upon the quantity of gas. For a given gas and at constant temperature it always remains same. Problem 30. The root mean square speed of hydrogen molecules at 300 K is 1930 m/s. Then the root mean square speed of oxygen molecules at 900 K will be (a) 1930 (b) 836 m/s 3m/s (c) 643 m/s (d) 1930 m/s 3 Solution : (b) v rms  3 RT M  v H2 v O2  TH 2 M H2  M O2 TO 2  1930  v O2 300 32  2 900  v O2  1930  3  836 m/s. 4 Kinetic Theory of Gases 17 Problem 31. At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at 47°C Solution : (a) For oxygen v O 2  3 RT O 2 M O2 According to problem   TO 2  M O2 TH 2 M H2  (c) – 73 K For hydrogen v H 2  3 R and 3 RT O 2 M O2  3R TH 2 M H2 TH 2 320 47  273  2  20 K.  TH 2   32 32 2 (d) 3 K TH 2 M H2 60 (b) 80 K E3 (a) 20 K Problem 32. Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will (b) Decrease ID (a) Increase (c) Remain same others (d) Decrease for some, while increase for Solution : (c) If a lorry is moving with constant velocity then the v rms of gas molecule inside the U 2 container will not change and we know that T  v rms. So temperature remains same. Problem 33. The speeds of 5 molecules of a gas (in arbitrary units) are as follows : 2, 3, 4, 5, 6. The root D YG mean square speed for these molecules is (a) 2.91 Solution : (d) v rms  (b) 3.52 v 12  v 22  v 32  v 42  v 52  5 Problem 34. Gas at a pressure (c) 4.00 22  32  4 2  52  62  5 (d) 4.24 100  20  4.24 5 P0 in contained as a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to [NCERT 1984; MNR 1995; MP PET 1997; MP PMT 1997; RPET 1999; UPSEAT 1999, 2000] U (a) 4 P0 1 mN 2 v rms 3 V ST Solution : (b) P  (b) 2P0 2  P  m v rms so (c) P0 2 v  P2 m m / 2  2v  2   2   1  1 P1 m 1  v1  m 1  v1 (d) P0 2 2    2  P2  2 P1  2 P0  Problem 35. Let v , v rms and v mp respectively denote the mean speed, root mean square speed and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. Then (a) No molecule can have speed greater than [IIT-JEE 1998] 2 v rms (b) No molecule can have speed less than v mp / 2 (c) v mp  v  v rms (d) The average kinetic energy of a molecule is 3 2 mv mp 4 18 Kinetic Theory of Gases 3 RT 8 RT , v av  and  M M We know that v rms  v rms : v av : v mp   v rms  v mp and v mp  2 RT M 3 : 2.5 : 2 so v mp  v av  v rms 3 2 3 2  v mp or v rms 2 2  Average kinetic energy  1 1 3 2 3 2 2 m v rms  m v mp.  mv mp 2 2 2 4 60 Solution : (c, d) Problem 36. The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is [Roorkee 1996] (b) v 2v Solution : (c) vrms  (c) 2v (d) 4v E3 (a) 3 RT. According to problem T will becomes T/2 and M will becomes M/2 so the value M of vrms will increase by 4  2 times i.e. new root mean square velocity will be 2v. Problem 37. The molecules of a given mass of a gas have a rms velocity of 200 m/sec at 27°C and 100 2 3 (b) 100 2 (c) U (a) ID 1.0  10 5 N / m 2 pressure. When the temperature is 127°C and pressure is 0.5  10 5 N / m 2 , the rms velocity in m/sec will be 400 [AIIMS 1985; MP PET 1992] (d) None of these 3 As D YG Solution : (c) Change in pressure will not affect the rms velocity of molecules. So we will calculate only the effect of temperature. vrms  T  v 300 o 300  400  v 400 o 3 4  200  v 400 3 4  v 400  200  2 3  400 3 Problem 38. Which of the following statement is true m /s. [IIT-JEE 1981] (a) Absolute zero degree temperature is not zero energy temperature (b) Two different gases at the same temperature pressure have equal root mean square velocities U (c) The rms speed of the molecules of different ideal gases, maintained at the same temperature are the same ST (d) Given sample of 1cc of hydrogen and 1cc of oxygen both at N.T.P.; oxygen sample has a large number of molecules Solution : (a) At absolute temperature kinetic energy of gas molecules becomes zero but they possess potential energy so we can say that absolute zero degree temperature is not zero energy temperature. Problem 39. The ratio of rms speeds of the gases in the mixture of nitrogen oxygen will be (a) 1 : 1 Solution : (c) vrms  v N2 3 RT   v O2 M (b) M O2 M N2 (c) 3 :1  32  28 8: 7 (d) 6: 7 8 7 Problem 40. A vessel is partitioned in two equal halves by a fixed diathermic separator. Two different ideal gases are filled in left (L) and right (R) halves. The rms speed of the molecules in L Kinetic Theory of Gases 19 part is equal to the mean speed of molecules in the R part. Then the ratio of the mass of a molecule in L part to that of a molecule in R part is 3 2 (b)  /4 (c) 2/3 R L 60 (a) (d) 3 / 8 v rms  Mean or average speed of molecule in right part v av  3 KT  mL 8 KT  mR  3 8  mL  mR 8 KT  mR  mL 3.  mR 8 ID According to problem 3 KT mL E3 Solution : (d) Root means square velocity of molecule in left part Problem 41. An ideal gas ( = 1.5) is expanded adiabatically. How many times has the gas to be U expanded to reduce the root mean square velocity of molecules 2 times (a) 4 times (b) 16 times (c) 8 times (d) 2 times v rms  T ) D YG Solution : (b) To reduce the rms velocity two times, temperature should be reduced by four times (As  T1  T T2  T , V1  V 4 V From adiabatic law TV  1  constant we get  2  V1 3/2 given] 1 U  V2  V1 (4 ) 3 / 2 1  V1 (4 ) 2  16 V1      1  T1 4 T2 1  V2  (4 ) 1 V1 [ = V2  16 V1 ST 11.8 Kinetic Energy of Ideal Gas. Molecules of ideal gases possess only translational motion. So they possess only translational kinetic energy. Quantity of gas Kinetic energy of a gas molecule (Emolecule) Kinetic energy of 1 mole (M gram) gas (Emole) Kinetic energy  1  3 kT  3 1 2  m m v rms   kT 2  m  2 2  3kT   As v rms  m    1 1 3 RT 3 2 M v rms  M  RT 2 2 M 2   As v rms   3 RT   M  20 Kinetic Theory of Gases Kinetic energy of 1 gm gas (Egram)  3 R 3 k NA 3 k 3 T  T  T  rT 2 M 2 m NA 2m 2 Here m = mass of each molecule, M = Molecular weight of gas and NA = Avogadro number = 6.023  1023 60 Important points (1) Kinetic energy per molecule of gas does not depends upon the mass of the molecule but only depends upon the temperature of the gas. 3 kT or E  T i.e. molecules of different gases say He, H2 and O2 etc. at same 2 temperature will have same translational kinetic energy though their rms speed are different.  3k T  v rms   m   E3 E ID As (2) Kinetic energy per mole of gas depends only upon the temperature of gas. 3 k T 2m  Egram  T m D YG Egram  U (3) Kinetic energy per gram of gas depend upon the temperature as well as molecular weight (or mass of one molecule) of the gas. From the above expressions it is clear that higher the temperature of the gas, more will be the average kinetic energy possessed by the gas molecules at T = 0, E = 0 i.e. at absolute zero the molecular motion stops. Sample Problems based on Kinetic energy Problem 42. Read the given statements and decide which is/are correct on the basis of kinetic theory of gases [MP PMT 2003] U (I) Energy of one molecule at absolute temperature is zero (II) rms speeds of different gases are same at same temperature For one gram of all ideal gas kinetic energy is same at same (IV) temperature For one mole of all ideal gases mean kinetic energy is same at same (a) All are correct (b) I and IV are correct (c) IV is correct ST (III) temperature (d) None of these Solution : (c) If the gas is not ideal then its molecule will possess potential energy. Hence statement (I) is wrong. rms speed of different gases at same temperature depends on its molecular weight  1   v rms  . Hence statement (II) also wrong.   M  1   Kinetic energy of one gram gas depends on the molecular weight  E gm  . Hence statement M  (III) also wrong. Kinetic Theory of Gases 21 3   But K.E. of one mole of ideal gas does not depends on the molecular weight  E  RT . 2   Hence (IV) is correct. Problem 43. At which of the following temperature would the molecules of a gas have twice the average kinetic energy they have at 20°C Solution : (c) E  T  (b) 80°C E2 T  2 E1 T1  (c) 313°C 2 E1 T2  E1 (20  273 ) (d) 586°C 60 (a) 40°C  T2  293  2  586 K  313 C. Problem 44. A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The E3 ratio of the average rotational kinetic energy per O 2 molecule to that per N 2 molecule is (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 ID (d) Depends on the moments of inertia of the two molecules Solution : (a) Kinetic energy per degree of freedom  1 kT 2 1  K.E.  2 kT   kT 2  U As diatomic gas possess two degree of freedom for rotational motion therefore rotational D YG In the problem both gases (oxygen and nitrogen) are diatomic and have same temperature (300 K) therefore ratio of average rotational kinetic energy will be equal to one. Problem 45. A gas mixture consists of molecules of type 1, 2 and 3 with molar masses m1  m 2  m 3. v rms and K are the rms speed and average kinetic energy of the gases. Which of the following is true [AMU (Engg.) 2000] (b) (vrms )1  (vrms )2  (vrms )3 and (K )1  (K )2  (K )3 (c) (vrms )1  (vrms )2  (vrms )3 and (K )1  (K )2  (K )3 (d) (vrms )1  (vrms )2  (vrms )3 and (K )1  (K )2  (K )3 U (a) (vrms )1  (vrms )2  (vrms )3 and (K )1  (K )2  (K )3 Solution : (a) The rms speed depends upon the molecular mass v rms  1 but kinetic energy does not depends M ST on it E  M 0 In the problem m 1  m 2  m 3  (v rms )1  (v rms ) 2  (v rms )3 but (K1 )  (K 2 )  (K 3 ) Problem 46. The kinetic energy of one gram mole of a gas at normal temperature and pressure is (R = 8.31 J/mole-K) (a) 0.56  10 4 J Solution : (d) E  [AFMC 1998; MH CET 1999; Pb. PMT 2000] (b) 1.3  10 2 J (c) 2.7  10 2 J (d) 3.4  10 3 J 3 3 RT   8.31  273  3.4  10 3 Joule 2 2 Problem 47. The average translational kinetic energy of O 2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N 2 (molar mass 28) molecules in eV at the same temperature is 22 Kinetic Theory of Gases [IIT-JEE 1997 Re-Exam] (a) 0.0015 (b) 0.003 (c) 0.048 (d) 0.768 Solution : (c) Average translational kinetic energy does not depends upon the molar mass of the gas. Different gases will possess same average translational kinetic energy at same temperature. Problem 48. The average translational energy and the rms speed of molecules in a sample of oxygen gas 60 at 300 K are 6.21  10 21 J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) (b) 8.78  10 21 J, 684 m / s (c) 6.21  10 21 J, 968 m / s (d) 12.42  10 21 J, 684 m / s E3 (a) 12.42  10 21 J, 968 m / s Solution : (d) E  T but v rms  T i.e. if temperature becomes twice then energy will becomes two time i.e. 2  6.21  10–21 = ID 12.42  10–21 J 2 times i.e. 484  2  684 m/s. But rms speed will become Problem 49. A box containing N molecules of a perfect gas at temperature T1 and pressure P1. The (b) P2  P1 , T2  T1 2 D YG (a) P2  P1 , T2  T1 U number of molecules in the box is doubled keeping the total kinetic energy of the gas same as before. If the new pressure is P2 and temperature T2 , then Solution : (b) Kinetic energy of N molecule of gas E  Initially E1  (c) P2  2P1 , T2  T1 (d) P2  2P1 , T2  T1 2 3 NkT 2 3 3 N 1 kT1 and finally E 2  N 2 kT2 2 2 U But according to problem E1  E 2 and N 2  2N 1  Since the kinetic energy constant T 3 3 N 1 kT1  (2 N 1 )kT2  T2  1 2 2 2 3 3 N 1 kT1  N 2 kT2  N 1 T1  N 2 T2  NT = constant 2 2 ST From ideal gas equation of N molecule PV  NkT  P1 V1  P2 V2  P1  P2 [As V1  V2 and NT = constant] Problem 50. Three closed vessels A, B and C are at the same temperature T and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only O2 , B only N 2 and C a mixture of equal quantities of O2 and N 2. If the average speed of the O2 molecules in vessel A is V1 , that of the N 2 molecules in vessel B is V2 , the average speed of the O2 molecules in vessel C is (where M is the mass of an oxygen molecule) [IIT-JEE 1992] (a) (V1  V2 ) / 2 (d) (b) V1 (c) (V1 V2 )1 / 2 3 kT / M Kinetic Theory of Gases 23 Solution : (b) Average speed of gas molecule v av  8 kT. It depends on temperature and molecular mass. m So the average speed of oxygen will be same in vessel A and vessel C and that is equal to V1. Problem 51. The graph which represent the variation of mean kinetic energy of molecules with E E E (b) t t Solution : (c) Mean K.E. of gas molecule E  E 3 3 k t   273 k 2 2 (d) t t 3 3 kT  k (t  273 ) where T = temperature is in kelvin and t = is 2 2 ID in centigrade  E (c) E3 (a) 60 temperature t°C is k = Boltzmann's constant U By comparing this equation with standard equation of straight line y  mx  c 3 3 k and c  273 k. So the graph between E and t will be straight line with 2 2 positive intercept on E-axis and positive slope with t-axis. D YG We get m  11.9 Gas Laws. (1) Boyle’s law : For a given mass of an ideal gas at constant temperature, the volume of a gas is inversely proportional to its pressure. i.e. V  1 P or PV = constant ST U m  (i) PV = P    constant   P   constant or P1 1  P2 2 N (ii) PV = P   constant n  P P P  constant or 1  2 n n1 n 2 or P1 V1  P2 V2 [As volume  m  [If m and T are constant] ] [As m = constant] [As number of molecules per unit volume n  [As N = constant] (iii) According to kinetic theory of gases P  1 mN 2 v rms 3 V N N  V ] V n 24 Kinetic Theory of Gases  P mass of gas T V [As v rms  T and mN = Mass of gas] If mass and temperature of gas remain constant then P  1. This is in accordance with V (iv) Graphical representation : If m and T are constant PV P V V 1/V 1/P ID V P E3 PV P 60 Boyle’s law. Sample Problems based on Boyle's law by [MP PET 2002] (b) 5.26% (c) 4.26% D YG (a) 5% U Problem 52. At constant temperature on increasing the pressure of a gas by 5% will decrease its volume (d) 4.76% Solution : (d) If P1  P then P2  P  5% of P = 1.05 P  From Boyle’s law PV = constant Fractional change in volume  100  105 5 V V2  V1    105 105 V V1 V 5  100 %    100 %   4.76 % i.e. volume decrease by V 105 U  Percentage change in volume 4.76%. V2 P 100 P  1   V1 P2 1.05 P 105 Problem 53. A cylinder contained 10 kg of gas at pressure 10 7 N / m 2. The quantity of gas taken out of ST cylinder if final pressure is 2.5  10 6 N / m is (assume the temperature of gas is constant) (a) Zero (b) 7.5 kg (c) 2.5 kg Solution : (b) At constant temperature for the given volume of gas  10 7 2.5  10 6  10 m2  m2  2. 5  10 6  10 10 7 (d) 5 kg P1 m  1 P2 m2  2. 5 kg  The quantity of gas taken out of the cylinder = 10 – 2.5 = 7.5 kg. Problem 54. If a given mass of gas occupies a volume of 10 cc at 1 atmospheric pressure and temperature of 100°C (373.15 K). What will be its volume at 4 atmospheric pressure; the temperature being the same [NCERT 1977] Kinetic Theory of Gases 25 (a) 100 cc Solution : (c) P  (b) 400 cc (c) 2.5 cc (d) 104 cc V P 1 1  2  1  V2  10     2. 5 cc V V1 P2 4 Problem 55. An air bubble of volume V0 is released by a fish at a depth h in a lake. The bubble rises to (b) V0 ( gh / P) (c) V0  gh   1   P    gh   (d) V0  1   P   E3 (a) V0 60 the surface. Assume constant temperature and standard atmospheric pressure P above the lake. The volume of the bubble just before touching the surface will be (density of water is ) Solution : (d) According to Boyle’s law multiplication of pressure and volume will remains constant at the bottom and top. ID If P is the atmospheric pressure at the top of the lake and the volume of bubble is V then from P1 V1  P2 V2 D YG  gh    V  V0 1  P    h (P1 V1) U  P  h g  (P  h g)V0  PV  V    V0 P   P2V2 Problem 56. The adjoining figure shows graph of pressure and volume of a gas at two temperatures T1 and T2. Which of the following interferences is correct P (a) T1  T2 p (b) T1  T2 T2 T1 (c) T1  T2 V1 V2 V U (d) No interference can be drawn Solution : (c) For a given pressure, volume will be more if temperature is more (Charle’s law) From the graph it is clear that V2 > V1 ST  T2 > T1 (2) Charle's law (i) If the pressure remains constant, the volume of the given mass of a gas increases or 1 decreases by of its volume at 0°C for each 1°C rise or fall in 273.15 Vt temperature. 1   Vt  V0 1  t. 273. 15   centigrade scale. This is Charle’s law V0 for – 273.15 O t(°C) 26 Kinetic Theory of Gases (ii) If the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature. V  T or V1 V2  T1 T2 or [If m and P are constant] V m   constant T T T  constant or [As volume V  or  1 T1   2 T2  ] [As m = constant] 1 mN 2 v rms 3 V E3 (iv) According to kinetic theory of gases P  or P  m 60 (iii) V  constant T Mass of gas T V ID If mass and pressure of the gas remains constant then V  T. This is in accordance with Charles law. V U (v) Graphical representation : If m and P are constant 1/V D YG V T V/ T T V/ T T or 1/T 1/T V or 1/V [All temperature T are in kelvin] Sample problems based on Charle's law Problem 57. A perfect gas at 27°C is heated at constant pressure to 327°C. If original volume of gas at 27°C is V then volume at 327°C is U (a) V (b) 3V ST Solution : (c) From Charle’s law V  T  (c) 2V (d) V/2 V2 T 327  273 600   2  V2  2 V.  2  27  273 300 V1 T1 Problem 58. Hydrogen gas is filled in a balloon at 20°C. If temperature is made 40°C, pressure remaining same, what fraction of hydrogen will come out (a) 0.07 Solution : (a) As V  T  (b) 0.25 V2 T  2 V1 T1 (c) 0.5  313   V2    V1  293   313    V1  V1 V2  V1 20  293     0.07. Fraction of gas comes out  V1 V1 293 (d) 0.75 Kinetic Theory of Gases 27 Problem 59. The expansion of unit mass of a perfect gas at constant pressure is shown in the diagram. Here a (a) a = volume, b = °C temperature O b 60 (b) a = volume, b = K temperature (c) a = °C temperature, b = volume (d) a = K temperature, b = volume Solution : (c) In the given graph line have a positive slop with X-axis and negative intercept on Y-axis. According to Charle’s law Vt ...... (i) E3 So we can write the equation of line y = mx – c V0 t  V0 , by rewriting this equation we get 273......(ii) ID  273   Vt  273 t    V0  By comparing (i) and (ii) we can say that time is represented on Y-axis and volume in Xaxis. U Problem 60. A gas is filled in the cylinder shown in the figure. The two pistons are joined by a string. If the gas is heated, the pistons will Gas D YG (a) Move towards left (b) Move towards right (c) Remain stationary (d) None of these U Solution : (b) When temperature of gas increases it expands. As the cross-sectional area of right piston is more, therefore greater force will work on it (because F = PA). So piston will move towards right. Problem 61. An ideal gas is initially at a temperature T and volume V. Its volume is increased by V due ST to an increase in temperature T, pressure remaining constant. The quantity   varies with temperature as   (a) T (b) T + T (Temp. K)  (c) T Solution : (c) From ideal gas equation or  T + T (Temp. K) T PV = RT PV  RT Dividing equation (ii) by (i) we get (d) T + T (Temp. K) T …..(i) …..(ii) V T V 1     V T VT T (given) T + T (Temp. K) V VT 28 Kinetic Theory of Gases    1. So the graph between  and T will be rectangular hyperbola. T (3) Gay-Lussac’s law or pressure law (i) The volume remaining constant, the pressure of a given mass of a gas increases or 1 of its pressure at 0°C for each 1°C rise or 273.15 fall in temperature.  t  Pt P0 E3 1  Pt  P0 1  273.15  60 decreases by This is pressure law for centigrade scale. – 273.15 t(°C) O (ii) The volume remaining constant, the pressure of a given mass of a gas is directly PT ID proportional to its absolute temperature. P P P  constant or 1  2 T T1 T2 or mass of gas T V D YG P or 1mN 2 v rms 3 V 2 [As v rms T] U (iii) According to kinetic theory of gases P  [If m and V are constant] If mass and volume of gas remains constant then P  T. This is in accordance with Gay Lussac’s law. (4) Graphical representation : If m and V are constants P/ T U P ST T 1/P P/ T T or 1/T P or 1/P P T 1/T [All temperature T are in kelvin] Sample problems based on Gay Lussac's law Problem 62. On 0°C pressure measured by barometer is 760 mm. What will be pressure on 100°C[AFMC 2002] (a) 760 mm Solution : (d) From Gay Lussac’s law (b) 730 mm (c) 780 mm P2 T  100  273  373  2   P1 T1  0  273  273 (d) None of these  373   P2     760  1038 mm.  273  Problem 63. If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1°C, the initial temperature must be (a) 250 K (b) 250°C (c) 2500 K (d) 25°C Kinetic Theory of Gases 29 Solution : (a) P1  P , T1 = T , P2  P  (0.4% of P)  P  P1 T  1 P2 T2 From Gay Lussac's law 0.4 P PP 100 250 P  P P 250  T2  T  1 T T 1 [As V = constant for closed 60 vessel] By solving we get T = 250 K. Problem 64. Pressure versus temperature graph of an ideal gas of equal number of moles of different volumes are plotted as shown in figure. Choose the correct alternative 4 E3 P (a) V1  V2 , V3  V4 and V2  V3 1 (c) V1  V2  V3  V4 T (d) V4  V3  V2  V1 R ID Solution : (a) From ideal gas equation PV  RT  P V T P V= constant Comparing this equation with y  mx R V i.e. V  1 tan  U Slope of line tan   m  2 3 (b) V1  V2 , V3  V4 and V2  V3  T D YG It means line of smaller slope represent greater volume of gas. For the given problem figure Point 1 and 2 are on the same line so they will represent same volume i.e. V1  V2 Similarly point 3 and 4 are on the same line so they will represent same volume i.e. V3  V4 But V1  V3 (= V 4 ) or V2  V3 (= V 4 ) as slope of line 1-2 is less than 3-4. U (5) Avogadro’s law : Equal volume of all the gases under similar conditions of temperature and pressure contain equal number of molecules. ST According to kinetic theory of gases PV  For first gas, PV  1 2 m N v rms 3 1 2 m 1 N 1 v rms (1) 3 …..(i) For second gas, PV  From (i) and (ii) 1 2 m 2 N 2 v rms (2) 3 …..(ii) 2 2 m 1 N 1 v rms 1  m 2 N 2 v rms 2 …..(iii) As the two gases are at the same temperature …..(iv) 1 1 3 2 2 2 2 m 1 v rms m 2 v rms kT  m 1 v rms 1  m 2 v rms 2 1  2  2 2 2 30 Kinetic Theory of Gases So from equation (iii) we can say that N 1  N 2. This is Avogadro’s law. (i) Avogadro’s number (NA) : The number of molecules present in 1 gm mole of a gas is defined as Avogadro number. 60 N A  6.023  10 23 per gm mole  6.023  10 26 per kg mole. (ii) At S.T.P. or N.T.P. (T = 273 K and P = 1 atm) 22.4 litre of each gas has 6.023  10 23 molecule. E3 (iii) One mole of any gas at S.T.P. occupy 22.4 litre of volume Example : 32 gm oxygen, 28 gm nitrogen and 2gm hydrogen occupy the same volume at S.T.P. (iv) For any gas 1 mole = M gram = 22.4 litre = 6.023  1023 molecule. ID Sample problems based on Avogadro's Law Problem 65. Temperature of an ideal gas is T K and average kinetic energy is E  2.07  10 23 T (a) 2.68  10 22 U Joule/molecule. Number of molecules in 1 litre gas at S.T.P. will be (b) 2.68  10 25 (d) 1.68  10 22 (c) 2.68  10 28  D YG Solution : (a) As we know that at S.T.P. 22.4 litre of gas contains 6.023  10 23 molecules 1 litre of gas contain 6.023  10 23  2.68  10 22 molecules. 22.4 Problem 66. The average kinetic energy per molecule of helium gas at temperature T is E and the molar gas constant is R, then Avogadro’s number is (a) RT 2E (b) 3 RT E U Solution : (d) Average kinetic energy per unit molecule E  ST But Avagadro number  N A  (c) E 2 RT (d) 3 RT 2E 3 2E kT  k  2 3T 3 RT R R  NA .  2E k (2 E / 3T ) Problem 67. One mole of a gas filled in a container at N.T.P., the number of molecules in 1 cm3 of volume will be (a) 6.02  10 23 / 22400 (b) 6.02  10 23 (d) 6.02  10 23 / 76 (c) 1/22400 Solution : (a) Number of molecule in 22.4 litre gas at N.T.P.  6.023  10 23 or number of molecule in 22.4  10 3 cm 3  6.023  10 23  22.4  10 3 cm 3 ]  Number of molecules in 1 cm 3  6.023  10 23. 22400 [As 22.4 litre Kinetic Theory of Gases 31 (6) Grahm’s law of diffusion : When two gases at the same pressure and temperature are allowed to diffuse into each other, the rate of diffusion of each gas is inversely proportional to the square root of the density of the gas. or v rms   1  60 3P We know v rms  and rate of diffusion of a gas is proportional to its rms velocity i.e., r  v rms 1  or r1  r2 2 1 E3  r (7) Dalton’s law of partial pressure : The total pressure exerted by a mixture of non- ID reacting gases occupying a vessel is equal to the sum of the individual pressures which each gases exert if it alone occupied the same volume at a given temperature. For n gases P  P1  P2  P3 ..... Pn where P = Pressure exerted by mixture and P1 , P2 , P3 ,...... Pn  Partial pressure of component U gases. Sample problems based on Dalton's law D YG Problem 68. The capacity of a vessel is 3 litres. It contains 6 gm oxygen, 8 gm nitrogen and 5 gm CO 2 mixture at 27°C. If R = 8.31 J/mole  kelvin, then the pressure in the vessel in N / m 2 will be (approx.) (a) 5  10 5 (b) 5  10 4 Solution : (a) Dalton’s law P  P1  P2  P3  V   2 RT V   3 RT V  m m  RT RT  m 1 [ 1   2   3 ]   2  3  V V  M1 M 2 M 3  8. 31  300  6 8 5     498  10 3 ~ 500  10 3 ~ 5  10 5 N /m 2.  3  32 28 44  3  10  U   1 RT (d) 10 5 (c) 10 6 Problem 69. Two gases occupy two containers A and B the gas in A, of volume 0.10 m 3 , exerts a pressure ST of 1.40 MPa and that in B of volume 0.15 m 3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then it the temperature remains constant, the final pressure in the container will be (in MPa) (a) 0.70 (b) 0.98 (c) 1.40 (d) 2.10 Solution : (b) As the quantity of gas remains constant  A   B   PA V A PB VB P(V A  VB ) P V  PB VB 1.4  0.1  0.7  0.15     P A A  P  0.98 MPa. V A  VB 0.1  0.15 RT RT RT Problem 70. The temperature, pressure and volume of two gases X and Y are T, P and V respectively. When the gases are mixed then the volume and temperature of mixture become V and T respectively. The pressure and mass of the mixture will be (a) 2P and 2M (b) P and M (c) P and 2M Solution : (a) From Dalton’s law, Pressure of mixture  P1  P2  P  P  2 P (d) 2P and M 32 Kinetic Theory of Gases Similarly mass also will become double i.e. 2M. Problem 71. A closed vessel contains 8g of oxygen and 7g of nitrogen. The total pressure is 10 atm at a given temperature. If now oxygen is absorbed by introducing a suitable absorbent the pressure of the remaining gas in atm will be (a) 2 (b) 10 (c) 4 (d) 5 P  P1  P2   1 RT V   2 RT V  60 Solution : (d) From Dalton’s law final pressure of the mixture of nitrogen and oxygen m 1 RT m 2 RT 8 RT 7 RT RT     M1 V M2 V 32 V 28 V 2V RT 2V …..(i) 7 RT 28 V  P E3 When oxygen is absorbed then for nitrogen let pressure is P   10  …..(ii) RT 4V From equation (i) and (ii) we get pressure of the nitrogen P  5 atm. (mass of gas)T V 2 [As v rms T] U P If mass of gas is constant then PV  T or PV = RT. This is ideal gas equation. D YG 11.10 Degree of Freedom. 1 mN 2 v rms 3 V ID (8) Ideal gas equation : From kinetic theory of gases P  The term degree of freedom of a system refers to the possible independent motions, systems can have. or The total number of independent modes (ways) in which a system can possess energy is called the degree of freedom (f). The independent motions can be translational, rotational or vibrational or any combination of these. (ii) Rotational degree of freedom (iii) Vibrational degree of freedom ST U So the degree of freedom are of three types : (i) Translational degree of freedom General expression for degree of freedom f = 3A – B ; where A = Number of independent particles, B = Number of independent restriction (1) Monoatomic gas : Molecule of monoatomic gas can move in any y direction in space so it can have three independent motions and hence 3 degrees of freedom (all translational) vy v vz z vx x Kinetic Theory of Gases 33 (2) Diatomic gas : Molecules of diatomic gas are made up of two y x z 60 atoms joined rigidly to one another through a bond. This cannot only move bodily, but also rotate about one of the three co-ordinate axes. However its moment of inertia about the axis joining the two atoms is negligible compared to that about the other two axes. Hence it can have only two rotational motion. Thus a diatomic molecule has 5 degree of freedom : 3 translational and 2 rotational. (3) Triatomic gas (Non-linear) : A non-linear molecule can rotate about any of three co-ordinate axes. Hence it has 6 degrees of freedom : 3 translational and 3 rotational. E3 y x z Example Monoatomic He, Ne, Ar Diatomic H2, O2 non Triatomic linear f = 3A – B 1 0 f=3 2 H2 O CO2, BeCl2 3 3 1 3 Figure A f=5 B A A A B B f=6 A B A A 2 f=7 A B B A  The above degrees of freedom are shown at room temperature. Further at high temperature, in case of diatomic or polyatomic molecules, the atoms with in the molecule may also vibrate with respect to each other. In such cases, the molecule will have an additional degrees of freedom, due to vibrational motion. U Note : B D YG Triatomic linear A U Atomicity of gas ID (4) Tabular display of degree of freedom of different gases ST  An object which vibrates in one dimension has two additional degree of freedom. One for the potential energy and one for the kinetic energy of vibration.  A diatomic molecule that is free to vibrate (in addition to translation and rotation) will have 7 (2 + 3 + 2) degrees of freedom.  An atom in a solid though has no degree of freedom for translational and rotational motion, due to vibration along 3 axes has 3  2 = 6 degrees of freedom (and not like an ideal gas molecule). When a diatomic or polyatomic gas dissociates into atoms it behaves as monoatomic gas whose degree of freedom are changed accordingly. 11.11 Law of Equipartition of Energy. 34 Kinetic Theory of Gases For any system in thermal equilibrium, the total energy is equally distributed among its various degree of freedom. And the energy associated with each molecule of the system per 1 degree of freedom of the system is kT. 2 where k  1.38  10 23 J / K , T = absolute temperature of the system. 60 If the system possess degree of freedom f then f kT 2 Total energy associated with each molecule Total energy associated with N molecules f RT 2 Total energy associated with each mole f 2 ID Total energy associated with  mole f kT 2 E3 N Total energy associated with each gram Total energy associated with M0 gram f rT 2 M0 U RT f rT 2 D YG Sample problems based on Law of equipartition of energy Problem 72. Energy of all molecules of a monoatomic gas having a volume V and pressure P is 3 PV. The 2 total translational kinetic energy of all molecules of a diatomic gas as the same volume and pressure is (a) 1 PV 2 [UPSEAT 2002] (b) f f RT  PV 2 2 U Solution : (b) Energy of 1 mole of gas  3 PV 2 (c) 5 PV 2 (d) 3 PV where f = Degree of freedom Monoatomic or diatomic both gases posses equal degree of freedom for translational ST motion and that is equal to 3 i.e. f = 3 Although total energy will be different,  E 3 PV 2 For monoatomic gas E total  For diatomic gas E total  5 PV 2 3 PV 2 [As f = 3] [As f = 5] Problem 73. The temperature of argon, kept in a vessel is raised by 1°C at a constant volume. The total heat supplied to the gas is a combination of translational and rotational energies. Their respective shares are [BHU 2000] (a) 60% and 40% (b) 40% and 60% (c) 50% and 50% (d) 100% and 0% Kinetic Theory of Gases 35 Solution : (d) As argon is a monoatomic gas therefore its molecule will possess only translatory kinetic energy i.e. the share of translational and rotational energies will be 100% and 0% respectively. Problem 74. CO 2 (O  C  O) is a triatomic gas. Mean kinetic energy of one gram gas will be (If N- (a) 3 / 88 NkT (b) 5 / 88 NkT (c) 6 / 88 NkT  E f RT 2 (d) 7 / 88 NkT E3 Solution : (d) Mean kinetic energy for  mole gas  . 60 Avogadro's number, k-Boltzmann's constant and molecular weight of CO 2  44 ) 1 7  7 7 m 7 NkT RT    NkT    NkT  44  2  88 2 M2 [As f = 7 and M = 44 for CO 2 ] Problem 75. At standard temperature and pressure the density of a gas is 1.3 gm/ m3 and the speed of (a) 3 (b) 4 m , sec N m2 D YG P  1.01  10 5 (c) 5 Substituting these value in v sound  Now from   1  P  (d) 6 Density of gas   1.3 U Solution : (c) Given velocity of sound v s  330 ID the sound in gas is 330 m/sec. Then the degree of freedom of the gas will be kg m3 , Atomic pressure we get   1.41 2 2 2   5. we get f  f   1 1.4  1 11.12 Mean Free Path. ST U The molecules of a gas move with high speeds at a given temperature but even then a molecule of the gas takes a very long time to go from one point to another point in the container of the gas. This is due to the fact that a gas molecule suffers a number of collisions with other gas molecules surrounding it. As a result of these collisions, the path followed by a gas molecule in the container of the gas is zig-zag as shown in the figure. During two successive collisions, a molecule of a gas moves in a straight line with constant velocity and the distance travelled by a gas molecule between two successive collisions is known as free path. The distance travelled by a gas molecule between two successive collisions is not constant and hence the average distance travelled by a molecule during all collisions is to be calculated. This average distance travelled by a gas molecule is known as mean free path. 36 Kinetic Theory of Gases Let 1 ,  2 ,  3 ,...n be the distance travelled by a gas molecule during n collisions respectively, then the mean free path of a gas molecule is given by   1 2nd 2 n ; where d = Diameter of the molecule, n = Number of molecules per unit 60 (1)   1   2  3 ....  n volume PV =  RT =  NkT So   N P   n  Number of molecule per unit volume V kT 1 kT 2 2 d P 2nd  2 m 2 (mn )d 2  m 2d 2  [As mn = Mass per unit volume = Density = ID 1 (3) From   E3 (2) As ] (4) If average speed of molecule is v then t  vT N U  v [As N = Number of collision in time t, T = time interval between D YG two collisions] Important points (i) As   a gas. 2d  2   1  i.e. the mean free path is inversely proportional to the density of 1 kT. For constant volume and hence constant number density n of gas 2 2 d P U (ii) As   m P is constant so that  will not depend on P and T. But if volume of given mass of a T 1 gas is allowed to change with P or T then   T at constant pressure and   at constant P temperature. ST molecules, Sample Problems based on Mean free path Problem 76. If the mean free path of atoms is doubled then the pressure of gas will become (a) P / 4 Solution : (b) As   (b) P / 2 1 kT 2 d P 2  P 1  (c) P / 8 (d) P i.e. by increasing  two times pressure will become half. Kinetic Theory of Gases 37 Problem 77. The mean free path of nitrogen molecules at a pressure of 1.0 atm and temperature 0°C is 0.8  10 7 m. If the number of density of molecules is 2.7  10 25 perm 3 , then the molecular diameter is (a) 3.2 nm (c) 3.2 m (b) 3.2 Å (d) 2.3mm 1 2nd 2 we get d  1.04  10 19  3.2  10 10 m  3.2 Å 11.13 Specific heat or Specific Heat Capacity. E3 Substituting these value in   60 Solution : (b) Mean free path   0.8  10–7 m number of molecules per unit volume n  2.7  10 25 per m3 ID It characterises the nature of the substance in response to the heat supplied to the substance. Specific heat can be defined by two following ways : Gram specific heat and Molar specific heat. Gram specific heat c  Joule cal cal , , gm  k elvin k g  k elvin gm  C D YG Units : Q m T U (1) Gram specific heat : Gram specific heat of a substance may be defined as the amount of heat required to raise the temperature of unit mass of the substance by unit degree. Dimension : [L2 T 2 1 ] (2) Molar specific heat : Molar specific heat of a substance may be defined as the amount of heat required to raise the temperature of one gram mole of the substance by a unit degree, it is represented by capital (C) Q T U C calorie calorie Joule , or mole  kelvin mole  C mole  kelvin ST Units : Important points (1) C  Mc  M Q 1 Q  m T  T m   As   M    i.e. molar specific heat of the substance is M times the gram specific heat, where M is the molecular weight of that substance. (2) Specific heat for hydrogen is maximum c  3.5 cal. gm  C 38 Kinetic Theory of Gases (3) In liquids, water has maximum specific heat c  1 cal. gm  C (4) Specific heat of a substance also depends on the state of substance i.e. solid, liquid or Example : c ice  0. 5 cal cal cal , c water  1 , c steam  0. 47 gm  C gm  C gm  C 60 gas. (5) Specific heat also depends on the conditions of the experiment i.e. the way in which E3 heat is supplied to the body. In general, experiments are made either at constant volume or at constant pressure. 11.14 Specific Heat of Gases. ID In case of solids and liquids, due to small thermal expansion, the difference in measured values of specific heats is very small and is usually neglected. However, in case of gases, specific heat at constant volume is quite different from that at constant pressure. In case of gases, heat energy supplied to a gas is spent not only in raising the temperature U of the gas but also in expansion of gas against atmospheric pressure. Hence specific heat of a gas, which is the amount of heat energy required to raise the temperature of one gram of gas through a unit degree shall not have a single or unique value.  D YG (i) If the gas is compressed suddenly and no heat is supplied from outside i.e. Q = 0, but the temperature of the gas raises on the account of compression. C Q 0 m (T ) i.e. C = 0 (ii) If the gas is heated and allowed to expand at such a rate that rise in temperature due to heat supplied is exactly equal to fall in temperature due to expansion of the gas. i.e. T = 0 C Q Q   m (T ) 0 U  i.e. C =  ST (iii) If rate of expansion of the gas were slow, the fall in temperature of the gas due to expansion would be smaller than the rise in temperature of the gas due to heat supplied. Therefore, there will be some net rise in temperature of the gas i.e. T will be positive.  C Q  positive m (T ) i.e. C = positive (iv) If the gas were to expand very fast, fall of temperature of gas due to expansion would be greater than rise in temperature due to heat supplied. Therefore, there will be some net fall in temperature of the gas i.e. T will be negative. C Q  negative m ( T ) i.e. C = negative Kinetic Theory of Gases 39 Hence the specific heat of gas can have any positive value ranging from zero to infinity. Further it can even be negative. The exact value depends upon the mode of heating the gas. Out of many values of specific heat of a gas, two are of special significance. 60 (1) Specific heat of a gas at constant volume (cv) : The specific heat of a gas at constant volume is defined as the quantity of heat required to raise the temperature of unit mass of gas (Q)v through 1 K when its volume is kept constant, i.e., c v  m T C v  Mcv  E3 If instead of unit mass, 1 mole of gas is considered, the specific heat is called molar specific heat at constant volume and is represented by capital Cv. M (Q)v 1 (Q)v  m T  T m   As   M    ID (2) Specific heat of a gas at constant pressure (cp) : The specific heat of a gas at constant pressure is defined as the quantity of heat required to raise the temperature of unit mass of gas (Q) p through 1 K when its pressure is kept constant, i.e., c P  m T 1 (Q) p  T m   As   M    D YG M (Q) p U If instead of unit mass, 1 mole of gas is considered, the specific heat is called molar specific heat at constant pressure and is represented by Cp. C p  MC p  m T  11.15 Mayer's Formula. Out of two principle specific heats of a gas, Cp is more than Cv because in case of Cv, volume of gas is kept constant and heat is required only for raising the temperature of one gram mole of the gas through 1°C or 1 K. U No heat, what so ever, is spent in expansion of the gas. It means that heat supplied to the gas increases its internal energy only i.e. (Q)v  U  C v T ST …..(i) while in case of Cp the heat is used in two ways (i) In increasing the temperature of the gas by T (ii) In doing work, due to expansion at constant pressure (W) So (Q) p  U  W   C p T From equation (i) and (ii)  …..(ii)  C p T   Cv T  W  T (C p  Cv )  PV [For constant P, W = PV] 40 Kinetic Theory of Gases  C p  Cv  PV  T [From PV = RT, At constant pressure PV = RT]  C p  Cv  R 60 This relation is called Mayer’s formula and shows that C p  C v i.e. molar specific heat at constant pressure is greater than that at constant volume. 11.16 Specific Heat in Terms of Degree of Freedom. E3 We know that kinetic energy of one mole of the gas, having f degrees of freedom can be given by f RT 2 …..(i) ID E U where T is the temperature of the gas but from the definition of Cv , if dE is a small amount of heat energy required to raise the temperature of 1 gm mole of the gas at constant volume, through a temperature dT then dE dT D YG dE  C v dT  C v dT or C v  [As  = 1] Putting the value of E from equation (i) we get C v   Cv  d f  f  RT   R dT  2  2 f R 2 f f  R  R    1 R 2 2   U From the Mayer’s formula C p  C v  R  C p  Cv  R  f  C p    1 R 2  ST  …..(ii) f    1 R 2  1 2  Ratio of Cp and Cv :   f Cv f R 2   1 Cp 2 f Important points (i) Value of  is always more than 1. So we can say that always Cp > Cv. (ii) Value of  is different for monoatomic, diatomic and triatomic gases. Kinetic Theory of Gases 41 (iii) As   1   Cv  2 f  2 f 1   1   f 2  1 f R R 2  1 Specific heat and kinetic energy for different gases E3 60     1  f   R and C p    1  R    1  R   2   1  1  A Restriction B Degree f = 3A – B of freedom Molar specific heat at constant volume f R R 2  1 Molar specific heat at constant pressure    f   R C p    1  R   2   1 Ratio of Cp and Cv 1 mole Cp Cv 1 E mole  U Kinetic energy of   ST Kinetic energy of 1 molecule Kinetic energy of 1 gm f RT 2 E molecule  E gram  2 f f kT 2 f rT 2 Triatomic linear 2 3 3 0 1 3 2 3 5 6 7 3 R 2 5 R 2 3R 7 R 2 5 R 2 7 R 2 4R 9 R 2 5~  1. 66 3 7~  1.4 5 4~  1. 33 3 9~  1. 28 7 3 RT 2 5 RT 2 3RT 7 RT 2 3 kT 2 5 kT 2 3kT 7 kT 2 3 rT 2 5 rT 2 3rT 7 rT 2 D YG Cv  Triatomic nonlinear 1 U Atomicity Diatomic ID Monoatomi c Sample Problems based on Specific heat Problem 78. Find the ratio of specific heat at constant pressure to the specific heat constant volume for NH 3 (a) 1.33 [RPMT 2003] (b) 1.44 (c) 1.28 (d) 1.67 Solution : (c) For polyatomic gas ratio of specific heat  < 1.33 Because we know that as the atomicity of gas increases its value of  decreases. 42 Kinetic Theory of Gases Problem 79. For a gas R  0.67. This gas is made up of molecules which are [CBSE PMT 1992; JIPMER 2001, 2002] Cv (a) Diatomic molecules (b) Mixture (c) Monoatomic (d) Polyatomic R  1 we get   1  0.67 diatomic and polyatomic or  = 1.67 i.e. the gas is 60 Solution : (c) By comparing with relation Cv  of monoatomic. Problem 80. 40 calories of heat is needed to raise the temperature of 1 mole of an ideal monoatomic gas (a) 20 calorie (b) 40 calorie E3 from 20°C to 30°C at a constant pressure. The amount of heat required to raise its temperature over the same interval at a constant volume (R  2 caloriemole 1 K 1 ) is (c) 60 calorie (d) 80 calorie Solution : (a) At constant pressure (Q) p   C p T  1  C p  (30  20 )  40  C p  4 calorie mole  kelvin Now (Q)v   C v T  1  2  (30  20 )  20 calorie ID 42  2  Cv  C p  R 3 2 5 2 (c) D YG (b) Solution : (c) Specific heat at constant volume C v    1  2 3  3R , then the value of  will be 2 U Problem 81. At constant volume the specific heat of a gas is (a) calorie mole kelvin   R 3R   1 2 5 3 (d) None of the above (given) 5. 3 Problem 82. For a gas the difference between the two specific heats is 4150 J/kg K. What is the specific (a) 8475 J/kg - K (b) 5186 J/kg - K U heats at constant volume of gas if the ratio of specific heat is 1.4 and Solution : (d) Given c p  c v  4150 …..(i) (c) 1660 J/kg - K cp cv  1.4  c p  1.4 c v (d) 10375 J/kg - K …..(ii) ST By substituting the value of c p in equation (i) we get 1.4 c v  c v  4150  0.4 c v  4150  cv  4150  10375 J /kg - K. 0.4 Problem 83. Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K. The piston of A is free to move while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is [IIT-JEE 1998] (a) 30 K (b) 18 K (c) 50 K (d) 42 K Solution : (d) In both cylinders A and B the gases are diatomic ( = 1.4). Piston A is free to move i.e. it is isobaric process. Piston B is fixed i.e. it is isochoric process. If same amount of heat Q is given to both then (Q)isobaric  (Q)isochoric Kinetic Theory of Gases 43  C p (T ) A   C v (T ) B  (T ) B  Cp Cv (T ) A   (T ) A  1.4  30  42 K. Problem 84. The specific heat of a gas (b) Has a unique value at a given temperature (c) Can have any value between 0 and  (d) Depends upon the mass of the gas 60 (a) Has only two values of C p and Cv Solution : (c) Range of specific heat varies from positive to negative and from zero to infinite. It depends upon the nature of process. Problem 85. The specific heat at constant volume for the monoatomic argon is 0.075 kcal/kg-K E3 whereas its gram molecular specific heat C v  2.98 cal/mole/K. The mass of the argon atom is (Avogadro’s number  6.02  10 23 molecules/mole) (a) 6.60  10 23 gm (b) 3.30  10 23 gm (c) 2.20  10 23 gm (d) 13.20  10 23 gm Cv  M  cv  2.98 calorie 0.075  10 3 calorie kcal M  M  0.075 3 gm  kelvin mole  kelvin kg - kelvin 10 39.7 6.023  10 23 of  6.60  10  23 gm. 6.023  10 23 atom = D YG  mass 2. 98  39. 7 gm 0. 075 U  molecular weight of argon M  i.e. ID Solution : (a) Molar specific heat = Molecular weight  Gram specific heat 39.7 gm  mass of single atom Problem 86. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is (a) 2/5 (b) 3/5 (c) 3/7 (d) 5/7 Solution : (d) When a gas is heated at constant pressure then its one part goes to increase the internal energy and another part for work done against external pressure i.e. (Q) p  U  W  C p T   C v T  PV U  ST So fraction of energy that goes to increase the internal energy C 7 U 1 5  v   [As   for Cp  7 (Q) p 5 diatomic gas] Problem 87. The temperature of 5 mole of a gas which was held at constant volume was changed from 100oC to 120oC. The change in internal energy was found to be 80 J. The total heat capacity of the gas at constant volume will be equal to (a) 8 J K 1 (b) 0.8 J K 1 (c) 4 J K 1 (d) 0.4 J K 1 Solution : (c) At constant volume total energy will be utilised in increasing the temperature of gas i.e. (Q)v   C v T   C v (120  100 )  80   Cv  80  4 Joule/kelvin. This is the heat capacity of 5 mole gas. 20 44 Kinetic Theory of Gases Problem 88. A gas, is heated at constant pressure. The fraction of heat supplied used for external work is (a)  1 (b)  1     1   1  (d)  1  2     (c)   1 1  60 Solution : (b) We know fraction of given energy that goes to increase the internal energy  So we can say the fraction of given energy that supplied for external work  1  1 . E3 Problem 89. A monoatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the expansion is Solution : (c) Fraction of (b) 25%, 75% energy supplied 5    As   3 for monoatomic gas    30  60 % 5 for increment in (d) 40%, 60% internal U  Percentage energy  (c) 60%, 40% ID (a) 75%, 25% 5 1 D YG  1 3 Fraction of energy supplied for external work done  1    5    Percentage energy  1 energy   1   3 5 2 5 3 2  100 %  40 %. 5 Problem 90. The average degrees of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by gas is U (a) 75 J (b) 100 J (d) 125 J 2 2 4 1  f 6 3 ST Solution : (b) As f = 6 (given)    1  (c) 150 J Fraction of energy given for external work W  1   1   Q   25 1    1   Q  4 /3  1 3 1  4 4  Q  25  4  100 Joule Problem 91. Certain amount of an ideal gas are contained in a closed vessel. The vessel is moving with a constant velocity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is (  C P / C V ) (a) Mv 2 2 R(  1) (b) Mv 2 (  1) 2R (c) Mv 2 2 R(  1) Solution : (b) If m is the total mass of the gas then its kinetic energy  1 mv 2 2 (d) Mv 2 2 R(  1) Kinetic Theory of Gases 45 When the vessel is suddenly stopped then total kinetic energy will increase the temperature 1 m mv 2   C v T  C v T of the gas (because process will be adiabatic) i.e. [As 2 M R Cv  ]  1  T  2 Mv 2 (  1). 2R 60 m R 1 T  mv M  1 2  Problem 92. The density of a polyatomic gas is standard conditions is 0.795 kgm 3. The specific heat of the gas at constant volume is (b) 1400 J- kg1 K 1 (c) 1120 J- kg1 K 1 Solution : (b) Ideal gas equation for m gram gas PV  mrT Specific m rT   rT V heat 4     3 for polyat omic gas     r at P 1.013  10 5   466.7  T 0.795  273 ID P [where r = Specific gas constant] constant volume cv  r 466.7   1 4 1 3  1400 J kg.kelvin U or (d) 925 J- kg1 K 1 E3 (a) 930 J-kg1 K 1 D YG Problem 93. The value of C p  Cv  1.00 R for a gas in state A and C p  C v  1.06 R in another state. If PA and PB denote the pressure and T A and T B denote the temperatures in the two states, then (a) PA  PB , TA  TB (b) PA  PB , TA  TB (c) PA  PB , TA  TB (d) PA  PB , TA  TB Solution : (c) For state A, C p  C v  R i.e. the gas behaves as ideal gas. For state B, C p  C v  1. 06 R ( R) i.e. the gas does not behave like ideal gas. and we know that at high temperature and at low pressure nature of gas may be ideal. U So we can say that PA  PB and T A  TB 11.17 Gaseous Mixture. ST If two non-reactive gases are enclosed in a vessel of volume V. In the mixture 1 moles of one gas are mixed with 2 moles of another gas. If NA is Avogadro’s number then Number of molecules of first gas N 1   1 N A and number of molecules of second gas N 2   2 N A (i) Total mole fraction   ( 1   2 ). (ii) If M 1 is the molecular weight of first gas and M 2 that of second gas. Then molecular weight of mixture will be M  1 M1  2 M 2 1  2 46 Kinetic Theory of Gases (iii) Specific heat of the mixture at constant volume will be    1 C V1   2 C V2  1   2 R C Vmix  m1 m 2  M1 M 2 1   2   1 2  R    1   2   1  1  2  1  m1 / M1 m 2 / M 2      2 1   1 1 60 C Vmix   R  R     2   1 1  2 1  1   C Pmix  1   2 R m1 m  2 M1 M 2  m1   M 1 R 1   2 D YG (v)  mixture  C Pmix C Vmix    1        2  2    1     1  1    2  1    1  m 2   2          1  1  M 2   2  1   ( 1 C P1   2 C P2 )  1 C P1   2 C P2 1   2  ( 1 C V1   2 C V2 )  1 C V   2 C V 1 2 1   2    1       R   2  2  R   1   1 1   2  1      R   R     2    1     1  1    2  1   1 1    2 2   1  2  1  1 1 ( 2  1)   2  2 ( 1  1)  1  1 2  1 ( 2  1)   2 ( 1  1)  1 1  2 1 ST U   mixture  ID C Pmix  1   2 U   1      R   2  2  R 1 1  2 1  1   1 C P1   2 C P2 E3 (iv) Specific heat of the mixture at constant pressure will be C Pmix  Sample problems based on Mixture Problem 94. If two moles of diatomic gas and one mole of monoatomic gas are mixed with then the ratio of specific heats is (a) 7 3 Solution : (c) 1  1 ,  1  [MP PMT 2003] (b) 5 4 (c) 19 13 (d) 15 19 5 7 (for monoatomic gas) and  2  2 ,  2  (for diatomic gas) 3 5 Kinetic Theory of Gases 47 5 7 2 3  5  1 1  2 2 5 7  1 1 1 1  2 1 5 / 2  7 19 5   3   1 2 1 2 3 / 2  5 13   5 7 1 1  2 1 1 1 3 5 From formula  mixture 60 1 Problem 95. 22 gm of CO 2 at 27°C is mixed with 16 gm of O 2 at 37°C. The temperature of the mixture is[CBSE PM (b) 27°C (c) 37°C Solution : (a) Let t is the temperature of mixture Heat gained by CO 2 = Heat lost by O 2 1 Cv1 T1   2 Cv2 T2  22 16  5  (3 R)(t  27 )   R  (37  t) 44 32  2   3(t  27 )  ID  U 5 (37  t) 2 By solving we get t  32 C. (d) 30.5°C E3 (a) 32°C Problem 96. A gas mixture consists of 2 mole of oxygen and 4 mole of argon at temperature T. D YG Neglecting all vibrational modes, the total internal energy of the system is (a) 4 RT (b) 15 RT (c) 9 RT Solution : (d) Total internal energy of system  U oxygen  U argon   1 2 ST U argon)] 5 3 RT  4 RT  5 RT  6 RT  11 RT 2 2 (d) 11 RT f1 f RT   2 2 RT 2 2 [As f1 = 5 (for oxygen) and f2 = 3 (for

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