Chemistry Chapter 4: Gases PDF

DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS CHEMISTRY CHAPTER # 4 GASES outline (full chapter handouts)  Definitions and reasons  theories and numericals  extra solved numericals  all important content Class: XI-E1 Su...

DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS CHEMISTRY CHAPTER # 4 GASES outline (full chapter handouts)  Definitions and reasons  theories and numericals  extra solved numericals  all important content Class: XI-E1 Subject Teacher: MUHAMMAD AJMAL Page 1 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Matter: Anything that has mass and occupies volume (takes up space) is reffered as matter. Fundamemtal states of matter: There are 4 fundamental states of matter. 1. Solid 2. Liquid 3. Gas 4. Plasma Gases: 1. Definition: Gases molecules remain far from each other and have very weak attractions in comparision with the molecules of other states. 2. Greek word: “Khaos” is a greek word from which the word “gas” was derived. the meaning of this word is emptiness or gap. 3. Movement: Molecules of gases can move in all directions. they follow straight path and collides with each other and with the wall of container. 4. Classification: On the basis of elemental compositions they can be classified in to mono atomic, di atomic and poly atomic gases. All noble gases are mono atomic in nature. Classification Examples Monoatomic Noble gases (Helium, Neon, Argon) etc. Diatomic Oxygen, Hydrogen, halogen gases (fluorine, chlorine) etc. Polyatomic Methane, Ethane, Carbondioxide. 5. Uses: CO2 helps in photosynthesis, Methane (natural gas) used in gasoline vehicles, poer station and in kitchen. Kinetic Molecular Theory (KMT) Of Gases: The theory was presented by Swiss mathematician Daniel Bernoulli (1738) and the modifications in the theory was done by Maxwell and Boltzmann. Page 2 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Postulates: Postulates of kinetic molecular theory are as follows: 1. Molecules All gases consists of tiny particles known as molecules, e.g: Monoatomic (noble gases), Diatomic (O2 and N2), Polyatomic (metane, ethane). 2. Volume: Actual volume of the gas is negligible as compared to its occupies volume. 3. Movement: Molecules of gases are in continuous random and haphazard motion. they move straightly and collide with other molecules and containers wall. 4. Mean free path: Average distance covered by the gas molecules before successive collision (no loss and gain of kinetic energy) is known as mean free path. 5. Pressure: They exert pressure after collision. 6. Forces: There is no force of attraction and repulsion in between molcules of ideal gases. 7. Energy: Average kinetic energy (K.E) of molecules is directly proportion to absolute temperature (An increase in temperature increases the speed in which the gas molecules move. All gases at a given temperature have the same average kinetic energy). K.E α Absolute temperature Pressure and Units: At 0 oC on sea level, a body bears 760torr pressue. Barometer: it used to measure pressure. it consists of a long glass tube filled with mercury. Manometer: It’s a type of barometer whivh measures pressure of a gas enclosed in container. Page 3 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Pressure: Force applied on an object per unit area. Its S.I unit is N/m2 (Pascal or Pa). Pressure = Force / Area Pascal: 1 Newton force distributed on unit area (square meter). 101325 Pa = 1atm = 760mmHg = 14.7psi Example#1: Convert 20psi in to (a) atm (b) KPa (c) torr Data: psi = 20 atm = ? KPa = ? torr = ? Solution: (a) For atm: 14.7 psi = 1atm 20 psi = 20/14.7 atm 20 psi = 1.360 atm (b) and, for Kpa, 14.7 psi = 101.325Kpa 20 psi = 101.325 x 20 / 14.7 20 psi = 137.857142 Kpa (c) Now, for torr, 14.7 psi = 760torr 20 psi = 760 x 20/14.7 Page 4 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS 20 psi = 1034.0136 torr Result The calculated pressures are (a) 1.360atm (b) 137.857142 Kpa (c) 1034.0136 torr. Example#2: Convert 30 torr in to (a) atm, (b) Kpa (c) psi Data: torr = 30 atm = ? Kpa = ? psi = ? Solution: (a) For atm, 760torr = 1atm 30torr = 30/760atm 30torr = 0.0394atm (b) For Kpa, 760torr = 101.325Kpa 30torr = 101.325 x 30 / 760 30torr = 3.9996Kpa (c) For psi, 760torr = 14.7psi 30torr = 14.7 x 30 / 760 30torr = 0.5802psi Result The calculated pressures are (a) 0.0394atm (b) 3.9996Kpa (c) 0.5802psi. Page 5 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Atmospheric Pressure And Its Effect On Weather (Short Note): 1. Weather Conditions: Cloud formation, M.P and B.P, vaporization etc are directly related with weather. 2. Atmospheric pressure: “Pressure exerted by air on earths surface is reffered as atmospheric pressure”, so it will increase as the number of particles that are making the air increases. 3. Reason of suffocating weather: Area having low atmospheric pressure becomes warmer and suffocated, these areas attracts clouds, winds and precipitation. 5. Reason of calm weather: Area having high air pressure becomes cold, dry, calm and possess clear sky. 6. Uses of gases: Gases Uses Oxygen Breathing purpose Nitrogen Controls oxidation Ozone Saves earth from dangerous radiations of sun Absolute Temperature Scale On The Basis Of Charles Law: Temperature: “The degree of hotness and coldness of a body is called temperature”. It can be measured in Celsius, Fahrenheit and Kelvin or Absolute scale (introduced by Lord Kelvin). Charles Law: Charles provided a relationship between volume and absolute temperature at constant pressue. Statement: “Volume of a given mass of gas is directly proportional to its Absolute temperature, at constant pressure”. Mathematically: V α T (constant pressure) Page 6 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Explanation: Volume of gas increses or decreases, by 1/273 times of its original volume at 0 oC by every degree rise or fall of temperature (pressure constant). Reason: Increasing temperature also increases kinetic energy of the molecules, by which molecules starts to move more freely, as a result of which gas expands and volume of the gas increses. Boyle’s Law: Robert boyle provided a relationship in between volume and pressure at conatant tempertrure. Statement: “Volume of given mass of gas is inversely proportional to pressure at constant temperature” Mathematically: V α 1/P (constant temperature) Explanation (Reason): Application of pressure on a gas present in any container decreses its volume because gases contracts by applying pressure. By removing or lowering pressure gas will expand and hence volume will increase. This contraction and repulsion is because of the presence of large intermolecular spaces in between gases. Graphical explanation of absolute zero: Absolute zero (Definition): “Temperature at which volume of an ideal gas become zero”. Graphical representation: Explanation: This is the straight line graph in between temperature (x-axis) and volume (y-axis). Straight line shows that both parameters have direct relation with each other. By extending this line, - Page 7 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS 273.15oC temperature will obtain. This is zero “kelvin” or “absolute zero”, at this temperature volume of ideal gas becomes zero. But this is not for real gases, because before this temperature, gases condensed in to liquid. Relation between Kelvin and Celsius scale: K = oC + 273 Example#3 Convert following temperature in to Kelvin. Data: (i) Temperature in oC = -12 oC (ii) Temperature in oC = 27 oC (iii) Temperature in oC = 43 oC (iv) Temperature in oC = 110oC (v) Temperature in oC = -786 oC Temperatures in Kelvin = ? Solution: (i) K = oC + 273 K = -12 + 273 K = 261 (ii) K = 27 + 273 K = 300 (iii) K = oC + 273 K = 43 + 273 K = 316 (iv) K = oC + 273 K = 110 + 273 K = 383 (v) K = oC + 273 K = -786 + 273 Page 8 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS K = 513 Result: The temperatures in kelvins are (i) 261K (ii) 300K (iii) 316K (iv) 383K (v) 513K Avgadro’s Law: He (in 1811) provided a relation in between volume and no. of moles of a gas at constant temperature and pressure. Statement: “At constant temperature and pressure conditions volume of a given mass of a gas is directly proportional to its number of moles”. Mathematically: V α n (constant temp and pressure) Or V = Kn V/n = K Above relation shows that ratio of volume with its number of moles will be constant (at constant temp and pressure. Suppose initial number of moles of a gas is “n1” having volume “V1”, by inserting more gases the number of moles will become “n2” having volume “V2”. So, V1/n1 = K and V2/n2 =K Now, V1/n1 = V2/n2 Example#4 A cylinder contains 1 mole of N2 gas at STP. When more gas is pumped in to the cylinder, the volume of gas becomes 4dm3 from 2dm3. Calculate how many moles of nitrogen were added? Data: V1 = 2dm3 V2 = 4dm3 Page 9 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS n1 = 1mol ∆n = ? Solution: V1/n1 = V2/n2 2/1 = 4/n2 n2 = 4/2 n2 = 2 ∆n = n2 – n1 ∆n = 2–1 ∆n = 1mol Result: The change in number of moles or the moles added will be 1mol. Example#5 At standard temperature and pressure 26.4dm3 of a gas contains 1.26moles. If 0.25moles are added to the gas, what will be the new volume of the gas? Data: V1 = 26.4dm3 n1 = 1.26mol n2 = 0.25mol V2 = ? Solution: We have, V1/n1 = V2/n2 26.4/1.26 = V2/0.25 V2 = 5.238dm3 Result: The volume occupied will be 5.238dm3 Page 10 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Example#6 Laughing gas (N2O) at 20oC and 800torr pressure occupies a volume of 10dm3. Calculate the volume that it will occupy at STP. Data: T1 = 20oC = 20+273 = 293K P1 = 800torr V1 = 10dm3 V2 = ? T2 = 273K P2 = 760torr Solution: P1V1/T1 = P2V2/T2 V2 = 800 X 10 X 273 / 293 X 760 V2 = 9.80dm3 Result: The calculated volume will be 9.80dm3 Example#7 A steel gas cylinder has a capacity of 15.8dm3 and filled with 785g of helium gas at a temperature of 20oC. Calculate the pressure of helium in the cylinder (mol.mass of helium = 4g/mol). Data: V = 15.8 dm3 Mass = 785g T = 20oC = 20+273 = 293K P = ? Solution: PV = nRT Since n = mass/molar mass Page 11 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS PV = (m/molar mass) RT P = (m/molar mass) RT / V P = (785/4) 0.0821 X 293 15.8 P = 298.7881atm Result: The calculated pressure is 298.7881atm. Gas Constants And Its Units: (a) value of R, when pressure is in atm and volume in dm3 At STP, P = 1atm moles = 1mol T = 273K V = 22.4dm3 By using formula: PV = nRT R = PV/nT R = 1 X 22.4 / 1 X 273 R = 0.0821atm dm3mole-1K-1 (a) value of R, when pressure is in N/m2 and volume in m3 (S.I unit): At STP, since 1atm = 101325 N/m2 P = 101325 N/m2 moles = 1mol T = 273K V = 0.0224m3 By using formula: PV = nRT Page 12 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS R = PV/nT R = 101325 X 0.024 / 1 X 273 R = 8.314 Nm mole-1K-1 R = 8.314 J mole-1K-1 (because 1N.m = 1J) Example#8 The density of a gas is 1.88g/dm3 at 27oC and 1atm pressure. Calculate the molar mass Data: d = 1.88g/dm3 T = 27oC = 27+273 = 300K P = 1atm M = ? Solution: Since d = PM/RT 1.88 = 1 x M / 0.0821 x 300 M = 46.32g/mol Result: The calculated molar mass is 46.32g/mol. Deviation From Ideal Gas Behaviour: Ideal gas: “Gases which obey all gas laws (bolyle’s, Charle’s and avagadro’s law), at all the conditions of temperature and pressure are known as ideal gases” Real gas: “In reality no gas obeys all gas laws at all conditions, such all gases are real gases”. Reason of deviation: At high pressure and low temperature, attractive forces become active as a result of whicjh gas liquefies and deviates from its idea behavior. Graphical Explanation Of Deviation Of real Gases: Page 13 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS For real gases general gas equation can be given as: PV = Z(nRT) Or Z = PV/nRT (Z = compresibilty factor) Positive and negative deviation: a) Value of Z = 1 (for ideal gas). b) Less than 1 shows negative deviation (for real gas). c) Greater than 1 shows positive deviation (for real gas). Plot btween Z and P: Graph between Z and P provides following information: a) At low pressure: At low pressures gases behave idealy and at this Z = 1. gases deviates from ideal behavior at high pressures. b) At high temperature: At high temperatures gases again behave ideally. Causes Of Deviation Of Real Gases From Ideal Behaviour: Faulty assumptions of kinetic molecular theory (KMT): 1. Volume: Actual volume of the gas is negligible as compared to the occupied volume. 2. Forces: There is no force of attraction and repulsion in between gaseous molecules. Deviations: 1. Deviations at high pressure: By increasing volume, the empty spaces between molecules remove and hence the actual volume becomes significant. 2. Deviations at low temperature: At low temperatures, the kinetic energy and velocity of molecules decreases by which molecules come closer and attractive forces become active. Example#9 5 moles of nitrogen gas is placed in 20dm3 vessel at 40oC. Calculate the pressure. Page 14 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS (a) gas behave idealy (b) gas behave as non idealy a = 1.30 dm atmmol and b = 0.030 dm mol-1 6 -2 3 Data: n = 5mol V = 20dm3 T = 40oC+273 = 313K P = ? Solution: (a) Since at ideal behaviour of gas, PV = nRT P = 5 x 0.0821 x 313 / 20 P = 6.42atm (b) Since at non ideal behaviour of gas, (P + an2/V2) (V - nb) = nRT (P + 1.30 x 52/ 202) (20-5x0.030) = 5 x 0.0821 x 313 (P + 1.30 x 25/ 400) (20-0.15) = 128.486 (P + 0.08125) (19.85) = 128.486 (P + 0.08125) = 128.486/19.85 (P + 0.08125) = 6.4728 P = 6.4728 - 0.08125 P = 6.3915atm Result: The calculated pressures are (a) 6.42atm (b) 6.3915atm Dalton’s Law Of Partial Pressure: He (1801) studied the effect of pressure given by non-reacting gases. Statement: “Total pressure exerted by the mixture of non-reacting gases (at constant volume and temperature) is equal to the sum of individual pressures of each gas”. Page 15 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Explanation: If PA, PB, PC etc are the pressures experted by individual gases, the the total pressure will be given as: PT = PA + PB + P C + … Pressure exerted by each gas is known as partial pressure of that gas. Applications Of Dalton’s Law Of Partial Pressure: 1. Calculation of pressure of dry gas: When the gas is collected over water by its downward displacement, the gas becomes moist. The measured pressure will be the sum of partial pressure of the gas and water (according to Dalton’s law). Mathematically: PTotal = Pdry gas + Pwater vapours Pdry gas = PTotal - Pwater vapours 2. Oxygen maintainance at high altitude: At high altitudes, because of low partial pressure of gas, cockpit of an airplane is pressurized to avoid uncomfortable breathing (pressure of air at high altitude is lower than sea level). 3. Maintainance of oxygen pressure under sea: Under sea, partial pressure of oxygen is high. it causes problem in respiration. Sea divers use “SCUBA (self contained under water breathing apparatus).” It contains 96% helium and 4% oxygen gas. Example#10 30dm3 cylinder is filled with 5moles of oxygen gas and 14 moles of nitrogen gas at 25oC. Calculate total pressure of gaseous mixture and partial pressure of individual gases. Data: V = 30dm3 moles of oxygen = 5mol moles of nitrogen = 14mol Total number of moles (nt) = moles of oxygen + moles of nitrogen 5 +14 = 19mol T = 25 + 273 = 298K Page 16 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS total pressure (Pt) = ? Poxygen = ? Pnitrogen = ? Solution: Since PtV = ntRT Pt = 19 x 0.0821 x 298 / 30 Pt = 15.49atm similarly, PoxygenV = n(oxygen) RT Poxygen = 5 x 0.0821 x 298 / 30 Poxygen = 4.077atm PnitrogenV = n(nitrogen) RT Pnitrogen = 14 x 0.0821 x 298 / 30 Pnitrogen = 11.41atm Result: The calculated pressures are Pt = 15.49atm, Poxygen = 4.077atm, Pnitrogen = 11.41atm Example#11 20cm3 of gas is collected over water at 25oC and 760torr pressure, what will be the mass of gas collected? (pressure of water at 25oC is 23.8torr, molar mass of gas =32) Data: V = 20cm3/ 1000 = 0.02 dm3 Ptotal = 760torr/760 = 1atm mass = ? Pwater = 23.8torr /760 = 0.0313atm T = 25oC + 273 = 298K Solution: Page 17 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Since, Pgas = Ptotal – Pwater Pgas = 1 – 0.0313 Pgas = 0.9687atm now, PV = nRT 0.9687 x 0.02 = n x 0.0821 x 298 n = 7.9x10-4mol mass = 7.9x10-4 x 32 mass = 0.0253g Result: The calculated mass is 0.0253g. Example#12 The mole fraction of oxygen in air is 0.2093, determine the partial pressure of oxygen in air if atmospheric pressure is 760torr. Data: mole fraction of oxygen (Xoxygen) = 0.2093mol Poxygen = ? Patmosphere = 760torr = 1atm Solution: Since, Poxygen = Xoxygen x Patmosphere Poxygen = 0.2093 x 1 Poxygen = 0.2093 atm Result: The calculated pressure of oxygen is 0.2093atm. Graham’s Law Of Diffusion And Effusion: Page 18 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Effusion: “The escaping of gas molecules from tiny hole in to the region of low pressure is known as effusion”. Diffusion: “The homogenously intermixing of different gases with each other is known as diffusion”. Graham’s Law: He (1831) provided relation in between rate od diffusion or effusion of gases and molar masses or densities (constant temperature and pressure). Statement: “At constant temperature and pressure, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its density or molar mass”. Mathematically: r α 1/√d here, d=density OR r α 1/√M here, M=molar mass Explanation: Gas will diffuse or effuse by difficulty, if it will have higher density or molar mass. Gas will diffuse or effuse easily, if it will have low density or molar mass. Applications: 1. Molar masses and densities of different gases can be determined. 2. Effect of hazardous gases can be reduced by diffusion in air. Example#13 Compare the rates of diffusion of methane and ethane gases. Data: molar mass of methane = 16 molar mass of ethane = 30 Solution: Since, by Graham’s law: Page 19 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS r(methane) / r(ethane) = √M(ethane) / M(methane) r(methane) / r(ethane) = √30/16 r(methane) / r(ethane) = 1.369 Result: Methane diffuses 1.369 tomes more faster than ethane. Example#14 Ratio of the rates of diffusion of gas A and B is 5. if molecular mass of gas B is 42 then find out the mass of A. Data: ratio of rate of diffusion of gas A and B = 5 molar mass of gas B = 42 molar mass of gas A = ? Solution: Since, by Graham’s law: rA / rB = √MB / MA 5 = √42/MA square on both sides: 52 = √42/MA2 25 = 42/MA MA = 42/25 MA = 1.68g Result: Mass of gas A is calculated as 1.68g. Example#15 At a specific temperature and pressure it takes 290sec for 1.5dm3 sample of He to effuse. Under similar conditions, if 1.5dm3 of an unknown gas “X” takes 1085sec to effuse, calculate molar mass of gas “X”. Page 20 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Data: Time for He = 290sec Volume of He = 1.5dm3 Volume of unknown gas (X) = 1.5dm3 Time for unknown gas (X) = 1085sec molar mass of unknown gas (X) = ? Solution: Since, by Graham’s law: rHe / rX = √MX / MHe VHe / tHe = √ MX / MHe VX / tX 1.5/290 = √MX / 4 1.5/1085 3.746 = √MX / 4 square on both sides: (3.746) 2 = MX / 4 MX = 56g/mol Result: Molar mass of gas X is calculated as 56g/mol. Example#16 If it takes 8.5sec for 5cm3 sample of CO2 to effuse. How long would it take for 5cm3 of SO2 to effuse from same container. Data: Time for CO2 = 8.5sec Volume of CO2 = 5cm3 / 1000 = 5x10-3dm3 Time for SO2 = ? Page 21 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Volume of SO2 = 5cm3 / 1000 = 5x10-3dm3 molar mass of CO2 = 44g/mol molar mass of SO2 = 64g/mol Solution: Since, by Graham’s law: rCO2 / rSO2 = √MSO2 / MCO2 VCO2 / tCO2 = √ 64 / 44 VSO2 / tSO2 5x10-3/8.5 = 1.2060 5x10-3/ tSO2 5x10-3 tSO2 = 1.2060 X 8.5 5x10-3 5x10-3 x tSO2 = 1.2060 0.0425 5x10-3 x tSO2 = 1.2060 x 0.0425 5x10-3 x tSO2 = 0.0512 tSO2 = 10.255sec Result: time required for SO2 is 10.255sec. Liquefaction Of Gases: Liquefaction: “At very high temperature and low pressure, attractive forces among gases become active and molecules come close to each other as a result of which gases convert in to liquid. This is known as liquefaction”. Critical temperature: “All gases has a specific temperature above which gas can’t convert in to liquid no matter how much pressure is applied, this temperature is known as critical temperature”. Page 22 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Critical pressure: “The pressure required at critical temperature to convert gas in to liquid is called critical pressure”. Joule-Thomson effect: “When a compressed gas move from high pressure to low pressure by a nozzle, it expands and causes a fall in temperature. This is known as Joule-Thomson effect”. Methods For Liquefaction Of Gases: Linde’s method of liquification of gases: Principle: This method works on the principle of Joule-Thomson effect. Working: In this, dry air is compressed to 200atm pressure by compressor. Produced heat is removed by passing through cooler. Now this gas is allowed to pass through spiral pipe having nozzle. When the gas comes out from nozzle to expansion chamber, its pressure reduces from 200atm to 1atm. By this fall of temperature occur. This cooled gas, cool the incoming compressed gas. This process repeat again and again, till the gas liquefied. Diagram: Liquification of gas by Linde’s method Uses of Liquified Gas: S.No Liquefied gas Uses 1. Liquefied natural gas (L.N.G) Used as fuel. 2. Liquefied petroleum gas (L.P.G) Used as a fuel for portable heaters. 3 Liquefied ammonia Used as refrigerant. 4. Liquefied oxygen Used for welding purpose. 5. Liquefied nitrogen Used for medical science. Page 23 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Fourth State Of Matter (Plasma): Plasma: “Plasma is a mixture of positive ions, electrons and neutral particles. It is neutral because of having equal number of positive and negative charges”. It was first identified by William Crooks (1879), during discharge tube experiment. More than 99% of visible universe is made up of plasma. It is present in sun, stars and ionosphere. On earth it’s present in flames, welding arcs and fluorescent lights etc. Properties of plasma: 1. Characteristic color: It show characteristic color in discharge tube, depending on the nature of gas. e.g: hydrogen gives green, oxygen red and nitrogen purple. 2. Effect of magnetic field: It get influenced by magnetic field as it possess positive and negative charge particles. 3. Shape and volume: It has no definite shape nor definite volume. Uses of plasma: 1. In electronic devices: It is used in computer chips, T.V, cell phones, lasers, lamps, rocket propulsion and pulsed power switches etc. 2. Coating material: Plasma spray is used to coat a material on to another surface e.g: diamond coated films. 3. In medical: It is used in destroying biological hazards, cleaning the environment, healing wounds and sterilization of medical tools. 4. In industries: Plasma torches are used to cut metals, also used in semiconductors. 5. In lamps: Lightning of fluorescent lamp and neon sign is due to glow of plasma (when the bulb is turned on, gas becomes excited and creates glowing plasma which lightens surrounding). Numerical Questions: Page 24 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS 1. The atmospheric pressure is 83Kpa, convert it in psi and atm units? Data: Kpa = 83 psi = ? atm = ? Solution: We know, 101.325Kpa = 14.7psi 83Kpa = 14.7 x 83 / 101.325 psi 83Kpa = 12.0414psi And, 101.325Kpa = 1atm 83Kpa = 1 x 83 / 101.325 83Kpa = 0.8191atm Result: 83Kpa is equals to 12.0414psi and 0.8191atm. 2. Calculate the volume occupied by 8g of methane gas at 40oC and 842torr. Data: Volume = ? Mass = 8g Temperature = 40oC + 273 = 313K Pressure = 842torr / 760 = 1.1078atm Solution: We know, PV = nRT 1.1078 x V = (m / mol.mass) 0.0821 x 313 1.1078 x V = (8/16) 0.0821 x 313 Page 25 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS 1.1078 x V = 12.8486 V = 12.8486 / 1.1078 V = 11.58dm3 Result: The calculated volume is 11.58dm3. 3. At 35oC, oxygen gas has a volume of 456cm3 and 0.85atm pressure. Calculate the pressure when this gas is transferred to 1dm3 cylinder where it cooled to 20oC. Data: T1 = 35oC + 273 = 308K V1 = 456cm3 /1000 = 0.456dm3 P1 = 0.85atm P2 = ? V2 = 1 dm3 T2 = 20oC + 273 = 293K Solution: We know, P1V1/T1 = P2V2/T2 0.85 x 0.456 / 308 = P2 x 1 / 293 1.25 x 10-3 = P2 x 1 / 293 1.25 x 10-3 x 293 = P2 0.368 = P2 P2 = 0.368atm Result: The calculated pressure is 0.368atm. 4. Compare the rates of diffusion of following pairs of gases. i) H2 and D2 ii) He and SO2 Data: Page 26 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS i) Rate of diffusion H2 and D2 = ? ii) Rate of diffusion He and SO2 = ? Solution: i)We know, rH2 / rD2 = √MD2/MH2 rH2 / rD2 = √ 4/2 rH2 / rD2 = 1.41/1 ii)We know, rHe/ rSO2 = √MSO2/MHe rHe/ rSO2 = √ 64 / 4 rHe/ rSO2 = 4/1 Result: The rate of diffusion of rH2 / rD2 is 1.41/1 and rHe/ rSO2 is 4/1. 5. A 500cm3 vessel has hydrogen gas at 400torr pressure and another 1dm3 vessel contains oxygen gas at 600torr pressure. If under the similar condition of temperature these gases are transferred to 2dm3 empty vessel, calculate the pressure of the mixture of gases in new vessel. Data: Volume of Hydrogen gas = V1 = 500cm3 /1000 = 0.5dm3 Pressure of Hydrogen gas = P1 = 400torr Volume of Oxygen gas = V1 = 1dm3 Pressure of Oxygen gas = P1 = 600torr V2 = 2dm3 P2 =? Solution: We know, (for hydrogen gas) P1V1 = P2V2 400 x 0.5 = P2 x 2 Page 27 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS P2 = 100torr And, (for oxygen gas) P1V1 = P2V2 600 x 1 = P2 x 2 P2 = 300torr Result: The calculated pressures are 100 and 300torr. 6. If 16cm3 of hydrogen effuses in 30sec. what volume of SO2 will effuse in same time under same conditions? Data: Volume of H2 = 16cm3/1000 = 0.016dm3 Time = 30sec Volume of SO2 = ? Time = 30sec Solution: i)We know, rH2 / rSO2 = √MSO2/MH2 V H2 / t H2 = √ 64 / 2 VSO2 / tSO2 0.016/30 = 5.6568 VSO2/ 30 0.016 / VSO2 = 5.6568 0.016 / 5.6568 = VSO2 VSO2 = 2.828 x 10-3dm3 OR 2.828cm3 Page 28 of 29  DHACSS DEGREE COLLEGE FOR BOYS AND GIRLS Result: The calculated volume is 2.828 x 10-3dm3 OR 2.828cm3. 7. 40dm3 of hydrogen gas was collected over water at 1257torr pressure at 23 oC. What would be the volume of dry hydrogen gas at standard conditions? The vapour pressure of water is 21torr of Hg (n=1mol). Data: Volume of Hydrogen gas = V = 40dm3 Pressure of moist gas = Pmoist = 1257torr / 760 = 1.6539atm Temperature = T = 23oC + 273 = 296K Vapour Pressure of water = Pwater = 21torr / 760 = 0.0276atm Volume of dry gas = Vdry = ? Solution: We know, Pdry = Pmoist - Pwater Pdry = 1.6539 - 0.0276 Pdry = 1.6262atm now, PV = nRT 1.6262 x V = 1 x 0.0821x 296 V = 14.943dm3 Result: The calculated volume is 14.943dm3. Page 29 of 29