State of Matter PDF

# UNIT 7 ## State of Matter - There are three physical states of matter: - Gaseous State - Liquid State - Solid State - For eg. Water has following three physical states: - Ice ($H_2O (s)$) - Liquid ($H_2O (l)$) - Steam ($H_2O (g)$) - **Gaseous State:** - The state of...

# UNIT 7 ## State of Matter - There are three physical states of matter: - Gaseous State - Liquid State - Solid State - For eg. Water has following three physical states: - Ice ($H_2O (s)$) - Liquid ($H_2O (l)$) - Steam ($H_2O (g)$) - **Gaseous State:** - The state of matter in which it has neither fixed shape nor fixed volume is called gaseous state. ## Gas Laws - Boyle's Law - Charle's Law - Combined gas Law - Dalton's law of partial pressure - Grahm's Law of diffusion ## # Boyle's Law - In 1662, Robert Boyle gave a mathematical relationship between volume of a gas & pressure of the gas at constant temperature. - Boyle's Law states that "The volume of the given mass of a gas is inversely proportional to pressure at constant temperature". - If P be the pressure of the gas & V be the volume of the gas at given temperature. - Then, According to Boyle's law: $V\propto\frac{1}{P}$ when temperature remains constant. - Again, - If P₁ & V₁ be the initial pressure and volume of a given mass of gas & P₂, V₂ are final pressure & volume of same mass of the gas at same temperature. - Applying Boyle's law at initial condition, $V_1\propto\frac{1}{P_1}$ at temperature T. - or, $V_1 = \frac{K}{P_1}$ - or, $K = P_1V_1 - (i)$ - Where K is proportionality constant which value depends upon amount of gas and temperature. - Similarly, - Applying Boyle's law at final condition $V_2 \propto \frac{1}{P2} $ at temperature T. - or, $V_2 = \frac{K}{P2} $ - or, $K = P_2V_2 - (ii) $ - From eqⁿ (i) and (ii) we get, $P_1V_1= P_2V_2=K$ - .. $P_1V_1=P_2V_2$ This is the formulae for Boyle's law. ## 8. At what condition the product of volume & pressure of a given mass of a gas remain constant? - At Constant temperature, the product of volume and pressure of a given mass of a gas remains constant. - According to Boyle's law, $PV=K$ at constant temperature. ## Graphical representation of Boyle's law: - Boyle's law can be verified by plotting a graph between pressure & volume of a given mass of a gas: - Curve like hyperbola $P \uparrow$ $PV = constant$ $V$ $0 \rightarrow$ - Straight line passing through origin. - Straight line parallel to x-axis ## # Chrale's law - In 1787, the relationship between volume & temperature of a given mass of a gas at constant pressure was given by Jacques Alexandre Cesar Charle. - This law states that, "The volume of given mass of a gas increases or decreases by $1/273$ of its initial volume at 0℃ for every each(1) degree rise or fall in temperature at constant pressure". - Let V_o ml is the volume of a gas at 0℃ - Then, according to challe's law, - On heating at 1°C, Volume of gas becomes = $V_o + \frac{1}{273}V_o$ = $V_о\left(1+\frac{1}{273}\right)$ - Similarly, Volume of gas at 2°C = $V_о\left(1+\frac{2}{273}\right)$ - Volume of gas at t°C = $V_o\left(1+\frac{t}{273}\right)$ - If V₁ ml & V₂ ml be the volume of a gas at t₁℃ & t₂℃ respectively. - Then , according to charle's law, - V₁ = $V_o\left(1+\frac{t1}{273}\right) - (i)$ - and V₂ = $V_o\left(1+\frac{t2}{273}\right) - (ii)$ - Dividing (i) by (ii) $\frac{V_1}{V_2} = \frac{273+t_1}{273+t_2}$. - $\frac{V_1}{V_2} = \frac{T_1}{T_2}$ where T₁= 273 + t₁ which is temperature in Kelvin scale. - This can be written as, $\frac{V_1}{V_2} = \frac{T_1}{T_2}$ (at constant pressure) - This is Charle's law formula. - Again, - V₁ = constant - or, V₁= constant x T - Charle's law can be stated as, "at Constant pressure the volume of given mass of a gas is directly proportional to absolute temperature". ## Concept of absolute temperature: - Let V_o ml be the volume of a gas at 0°C. According to charle's law, - On cooling the gas, - Volume of gas at -1°C = $\left[V_о - \frac{1}{273}x V_о\right]$ ml. - Volume of gas at -273°C = $\left[V_о - \frac{273}{273}x V_о\right]$ ml = 0 ml. - .. The hypothetical temperature at which volume of gas becomes zero is called absolute zero temperature. It is -273°C. It is equal to 0 Kelvin. At this temperature all gases have been converted into liquid or solid state. ## Verification of charle's law: - Volume - -273°C 0 t°C - - Fig: V vs T ## Charle's law can be verified by plotting graph between volume occupied by a gas & temperatures at constant pressure: - A straight line is obtained which on extra plototion the line meets. - At -273°C which shows that volume of a gas at -273°C is 0. ## # Combined gas equation $P_1V_1 / T_1 = P_2V_2 / T_2$ - A gas equation which is obtained by combining Boyle's law & charle's law is combined gas equation. It gives the simultaneous effect of temperature & pressure on volume of a gas. - It can be derived as: - Let V ml be the volume of a gas at pressure P & temperature T. - According to Boyle's law, at constant temperature volume of a given mass of a gas is inversely proportional to pressure. - ie $V\propto\frac{1}{P}$ (at constant temperature ) - (i) - According to charle's law, at constant pressure volume of a given mass of a gas is directly proportional to absolute temperature. - ie $V\propto T$ (at constant pressure) - (ii) - Combining (i) and (ii), we get. - $V \propto \frac{T}{P}$ (both T & P varies) - or, $V = KP$ where K is proportionality constant which value depends on amount of gas taken. - or $PV = KT$ - or $K = \frac{PV}{T} - (iii)$ - Equation (iii) is one form of Combined gas equation. If P_1V_1 & T_1 are initial pressure, volume and temperature of a certain mass of a gas respectively. & P_2V_2 & T_2 are final. Then, - According to equation of state. - $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}=K$ - $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ - Above egn is combined gas equation. ## Concept of Ideal gas & real gas - A gas which obey the Boyle's law, charle's law and combined gas egn at all temperature and pressure conditions is called ideal gas or perfect gas. - There are no ideal gas in nature but known gasses follows gas laws at low pressure high temperature. - A gas which doesnot obey Boyle's law, Charle's law & Combined gas equation in low temperature & high pressure is called real gas. Foreg. H₂, N₂, CO₂ ## Ideal gas equation [PV=nRT] - Where, - P = pressure of gas - V= volume of gass - n=no. of mole of gas - R = universal gas constant - T = temperature of gas ## Derivation of ideal gas equation: - According to Boyle's law - $V_1 \propto \frac{1}{P_1}$ [T is constant] - According to charle's law - $V\propto T$ [P is constant] - Similarly, - According to Avogadro's law, - $V \propto n$ [T & P are kept constant] - Combining these three equation (i), (ii) & (iii), we get - $V \propto \frac{nT}{P}$ - or, $V = \frac{RnT}{P}$ [R is proportionality constant(K) known as Universal gas constant] - .. PV = nRT - This equation is well known equation and is called ideal gas equation. ## Value of R: - The Universal gas constant R may be defined as: - R = $\frac{PV}{nT}$ - At NTP : 1 mole gas occupies 22.4 litre. - Given, n= I mole, V=22.4 litre, T=273K,P=1 atm. - We know. - R = $\frac{PV}{nT}$ = $\frac{1 atm x 22.4 ltr}{1 mole x 273 K}$ - .. R = 0.0821 atm. ltr. mole^-1 K^-1 ## 9.3(A) Two vessel of capacity 1.5 litre and 2 litre contain hydrogen gas and oxygen fas respectively under a pressure of 750 mm and 100 mm. the gases are mixed together in a 5 litre vessel. What will be the final pressure of mixture? - **Solution:** - For hydrogen gas; - Volume of H₂ (V₁)=1.5 litre - Pressure of H₂ (P₁)= 750 mm of Hg. - Total volume of gas(V₂)=5 litre - Pressure (P₂)=? - Now, - P₁V₁=P₂V₂ - P₂ = $\frac{PLV_1}{V2}=\frac{750x1.5}{5}$ - .. P₂ = 225 mm of Hg. - For Oxygen gas; - Volume of O₂ (V₁)=2 litre. - Pressure of O₂ (P₁)= 100 mm of Hg. - Total volume of gas (V₂)=5 litre - Pressure (P₂)=? - Now, - P₁V₁=P₂V₂ - P₂= $\frac{PIVI}{V2}=\frac{100x2}{5} $ - .. P₂ = 40 mm of Hg. - Again; - Total pressure of gaseous mixture: P_H2 + P_O2 = 225+40 = 265 mm of Hg. H ## 9.4(A) A fire extinguisher having capacity 3 litres. Contains 4.4 Kg. CO₂. How much CO₂ gas will it deliver at NTP? - **Solution:** - Given; mass of CO₂ (m) = 4.4 kg. = 4.4x1000 = 4400 gram. - Molecular weight of CO₂ (M)=44 - V₂=% (At NTP) - T=273 K, P= 1 atm, R= 0.0821. atm litre K^-1 mole^-1 - From Ideal gas equation: - PV=nRT - or $1 x V = \frac{M}{44} * 0.0821*273$ - or, $V = 4400 x 22.4173 = 2241.37 $ litres - 3 litres remains in vessel. - Hence, It will deliver = 2241.33-3 = 2238.33 litres H ## 9. 4(B) The mass of 525 ml of a gaseous compound at 28°C and 730 mm Hg. pressure was found to be 0.9 g. what will be the Volume of 2g of the gas at 30°c and 760 mm Hg. pressure? [Given; R = 0.0822 litre atm K^-1 mol^-1] - **Solution:** - **PART 1^st** - mass(m) = 0.9 g.m. - Volume(v) = 525 ml = 0.525 Ltr. - Temperature (T)=273+28=301 K - pressure of gas (P) = 730 mm Hg = 0.96 atm - R= 0.0821 l atm LK^-1 mol^-1 - molecular mass (M)=? - We know, - PV=nRT - or. PV=M/RT - or. M = $\frac MRT = \frac {0.9 x 0.0821 x 301} {0.96 x 0.525} = 44.12$ g.m. - **PART 2^nd** - mess(m) = 2 g.m., (T)=30°C=303K,P=760mmHg=1atm - Molecular mass(M) = 44.12 g.m., R= 0.0821 L atm K^-1 mol^-1 - V=8, - We know, PV=nRT - or, V- $\frac{MRT}{MP}= 2 x 0.0822 x 303 = 1.127 $ litre. - Thus, the final volume is 1.127 litre H ## 9. 5(A) A flask of 0.3 litre capacity was weighted after it had been evacuated. It was then filled with a gas of unknown molecular mass at 1.0 atm pressure and temperature of 300 k. The increase in mass of the flask was 0.977 g. Calculate the molecular mass of the gas? - **Solution:** - Given: - Votume (V) = 0.3 litre - Pressure (p) = 1 atm, Temperature (T) = 300k - weight of gas (w) = 0.977 g. - Molecular mass (M) = ? - Now - According to ideal gas equation; - PV=nRT - or, PV = w/M RT - or. M= $\frac WRT / PV = \frac{0.977 * 0.0821 * 300}{1 * 0.3}$ - .. M= 80.2 g.m. H ## 9.6(A) When 2g of a gas A is introduced into an evacuated flask kept at 25°c. the pressure is found to be one atmosphere. If 3g of another gas B is added to the same flask, the total pressure becomes 1.5 atmosphere at the same temperature. Assuming ideal behaviour of gases. Calculate the ratio of molecular masses. - **Solution:** - For Gas A - Given: weight of gas(w) = 2 g.m. - Temp (T) = 25°C = (25+273)k=298K. - pressure (P) = 1 atm - Volume of two gasses flasks is same = 1 Ltr.. - Moleculas mass(M) = ? - We know, $PV = \frac{MRT}{M}$ - or, $M = \frac{WRT}{PV}$ - $M = \frac{2 * 0.0821 * 298} {1 *1} = 48.93$ g.m. - Hence, The moleculas mass of gas A=48.93 = M_A - For gas B: - m=3g.M., T=25℃=298K., P=0.5 atm, R=0.08211 atm K^-1 mol-¹ V=1 Ltr. - Now PV=nRT - or. PV=RT/M - $M = \frac{MRT}{PV} = \frac{3 * 0.0821 * 298} {0.5 *1} = 146.79$ g.m. - Again, M_A:M_B= 48.93: 146.79 = 1:3 - Thus, the ratio of molecular masses is 1:3 H ## 9.7(A) Two moles of ammonia are enclosed in a five litres flask at 27°C. Calculate the pressure extected by gas assuming that the gas behaves like an ideal gas. - **Solution: ** - number of mole (n)= 2, - R = 0.0821 L atm K^-1 mol^-1 - Temperature (T)=27℃=(273+27) K=1300K - pressure (P) = ? - volume of ammonia (V)=5 Ltr. - Now - According to ideal gas equation; - PV=nRT - or, $P =\frac{nRT}{V} =2 * 0.0821 * 300 / 5 = 9.852$ atm H ## 9.8(A) 0.50 g.m of a volatile liquid was introduced into a globe of 500 ml capacity. The globe was heated to 118°C. so that all the liquid vapourized exerted a pressure of 190 mm Hg. - **Sol: ** - Given; - W = 0.50 g.m, V=1000ml = 1ltr, T=91°C=364K, P=190mm Hg. = 0.252 atm, R= 0.0821 atm L.K^-1 mol^-1 & M=? - Now PV=nRT or, PV=RT/M or, M = $\frac{ WRT}{PV}$ - or, M= $ \frac{0.50 x 0.0821 x 364}{0.25 x 1} = 59.76$ g.m - Thus, Molecular mass of given liquid is 59.76# ## 9.9(A) The ballon can hold 975 (cc) (ml) of air at 5°c will it brust when it is taken into a home set 25°c? Assume that the pressure of the gas in the ballon remains Constant. - **Som:** - For intial: - Volume (V) = 975 ml = 0.975 ltr. - Tempo (T) = 5°c = (273+5) K = 278K - For Final: - Volume (V₂)=? - Temp (T₂)=25°c=298 K - .. NOW, - According to charle's law, - $\frac{V_1}{V_2} = \frac{T_1}{T_2}$ - V₂ = $\frac{0.975 x 298}{278}$ = 1.045143855 ltr - ... Ballon will brust, V₂= 1045.14 ml. H ## Q.10(A) Find the number of moles of molecules present in 50 ml of an ideal gas exerting a pressure of 770 mm at 25°c. (R=0.0821.2tmL K^-1 mol-¹) - **Son** - Stiven; no. of moles (n)=?, Volume (V) = 50ml=0.05 ltr. - pressure (P) = 770 mm H.g. = 1.013 atm, Temp²(T)=25°=298K - R= 0.0821 1 atm K^-1 mol^-1 - Now - PV=nRT - or n= PV/RT = 1.013 x 0.05 / 0.0821 x 298 = 2.07 x10^-3 - Thus, The number of moles is 2.07 x10^-3 H ## # Dalton's law of partial pressure:- - Dalton's law of partial pressure states that, "In a mixture of two or more non-reacting gases, the total pressure of gases is equal to Sum of partial pressure of component gas al constant temperature". - Mathematically, - P_T = P₁+ P₂+ P₃+ ----- Where P₁, P₂, P₃ at partial pressure of gas 1, Gas 2, Gas 3 and so on and P_T is total pressure of mixture. ## Derivation: - Daltons law of partial pressure can be derived by using ideal gas equation let n₁, n₂ and n₃ are the no. of motes of component gases of a mixture contained in N volume of vessel and n is the total no. of motes of mixture. - Then, n=n₁+n₂ + n₃ + ----- - From ideal gas equation; - PV = nRT - or P_TV = (n₁ + n₂ + n₃)RT - .. P_T V = n₁RT + n₂RT + n₃RT - P_T = P₁ + P₂ + P₃ ## A aqueous tension is the pressure exerted by water vapour present in moist gas at only particular temperature. It is denoted by f. It's Value depends upon temperature. ## #GRAHM'S LAW OF DIFFUSION: - It States that "Under Similar Condition of temperature and pressure, the rate of diffusion of gases is inversely proportional to square root of their densities." - Mathematically, $r \propto \sqrt{\frac{1}{d}}$ - (i) - Where, r= rate of diffusion of gas - d = density of gas ## DERIVATION - According to the Graham's law of diffusion $r \propto \frac{1}{d}-(i)$ - If r₁ and r₂ be the rates of diffusion of two gases and d₁, d₂ be their corresponding density then, r₁ $\propto \sqrt{\frac{1}{d_1}}$ (at constant T & P) - or, r₁= $k\sqrt{\frac{1}{d_1}} - (ii)$ - Similarly, r₂ $\propto \sqrt{\frac{1}{d_2}}$ - or, r₂=$k \sqrt{\frac{1}{d_2}} - (iii)$ - Dividing (ii) by (iii), we get, - $\frac{r_1}{r_2} = \sqrt{\frac{d_2}{d_1}} - (iv)$ - Since, molecular weight of a gas is equal to twice of Vapour density. - M = 2 x v.d. - or, v.d=M/2 - $d_1 = \frac{M_1}{2}$ - $d_2 = \frac{M_2}{2}$ - From eqⁿ (iv) and (iv) - $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} - (v)$ - Where M₁ and M₂ are molecular weight of two gases. - Again, - Rate of diffusion means volume of gas diffused per unit time. - ie γ = $\frac{Volume of a gas (V)}{time taken (t)}$ - or r = $\frac{V}{t}$ - or V₁ = t₁ r₁ - V₁ = t₂ r₂ - Putting in eqⁿ(V) - $\frac{V_1}{t_1} = \sqrt{\frac{M_2}{M_1}}$ - or, $\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}$ - Time taken for diffusion of two gases of equal to volume is directly proportional to square root of their densities and molecular weight. ## #EFFUSION:- - It is difficult to determine the rate of diffusion but is is easy to determine rate of effusion is defined as the phenomenon in which a certain volume of gas is allowed to escape through a small pin hote. - The rate of diffusion is always equal to rate of effusion. - if; t₁=t₂=t $r = \frac{V_1}{t_1}=\frac{V_2}{t_2}= \frac{V_1+V_2}{t_1+t_2} (vi)$ - $r = \frac{V_1}{t_1} = \frac{V_2}{t_1} (vi)$ - $r = \frac{V_1}{t_1} = \sqrt{\frac{d}{d}}=\sqrt{\frac{M_1}{M_2}} (vii)$ - Main, - $r = \frac{V_1*t_2}{t_1*V_2} = \sqrt{\frac{d_2}{d_1}} = \sqrt{\frac{M_2}{M_1}}$ ## 9.1(A) How long will it take some of hydrogen gas to diffuse through a partition if 250 ml of oxygen diffuse in 50 minutes under similar condition? - **Solution:** - For bydrogen gas, - Volm (V₁) = 500 ml, Time (t₁)=? - molecular mass (m₁)=2 g. - Rate of diffusion of hydrofen(r₁) = $ \frac{V_1}{t1}$ - For Oxygen gas! - Voym (V₂) = 250 ml. - time (t₂) = 50 minutes - molecular mass (m₂) = 32 g - Prate of diffusion of oxygen (r₂) = $ \frac{V_2}{t_2}$ = 250/50 = 5 - We know that, - $\frac{r_1}{r_2} = \sqrt{\frac{m_2}{m_1}}$ - or, $\frac{500/t_1}{5} = \sqrt{\frac{32}{2}}$ - or, $\frac{500}{t_1}=\sqrt{16}$ - $\frac{500}{t_1}=4$ - $t_1 = \frac{500}{4}=25$ minutes - Hence, the time will take by hydrofen gas is 25 minutes. ## 9.2(A) A Saturated hydrocarbon CnH2n+2 diffuse through a porous membrane twice as fast as sulfur dioxide. Xetermine the molecullar formlila of the hydrocarbon. - **Solution:** - The rate of diffusion of SO₂ fas (r₁)=r - The rate of diffusion of hydrocarbon (r₂)=2r - molecular mass of SO₂(m₁)=32+32=64 - molecular mass of hydrocarbon (m₂)=? - We know that, - $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$ - or. $\frac{r}{2r} = \sqrt{\frac{M_2}{64}}$ - or (0.5)² = $\frac{M_2}{64}$ - or, 0.25= $\frac{M_2}{64}$ - or, M₂ = 16 - Hence, the molecular mass of hydrocarbon = 16 - Now, - Cn H_2n+2=16 - or, 12xn+1(2n+2)=16 - or, 12n+2n+2=16 - or 14 n=14 - n=1 - Molecular formula of hydrocarbon is C₁H_2x1+2 = CH₄ - ie. methane. ## 9.2 [B] Hydrogen gas diffuse five times faster than another gas A. Find the molecular wt. of gas A. - **Solution:** - Given: Rate of diffusion of gas A (r₁)=r - " " " hydrogen (r₂) = 5r - molecular wt - of hydrofen (M₂)=2 - " " " gas A (M₁) = ? - We know, - $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$ - or, $\frac{r}{5r} = \sqrt{\frac{2}{M_1}}$ - or, $\frac{1}{5} = \sqrt{\frac{2}{M_1}}$ - or 25 M₁ = 2 - or, M₁ = $\frac{2}{25}$ - :. M₁: M₂= 1:25 # ## 9.7(A) Calculate the volume occupied by 5.0 g of acetylene gas at 50°C and 740mm Hg. pressure. - **Solution:** - molar mass of acetylene (C₂H₂)(M)=(12×2)+(1x2)=26 g/mol - mass of acetylene (m) = 5 g.m. - Temp (T) = 50°C = 273 + 50 = 323 K - pressure (P) = 740 mm Hg. = 740/760 = 0.9737 atm. - Volume (V)=? - According to ideal gas eqn, - PV=nRT=MRT/M - .. V = $\frac{MRT}{PM}=\frac{5 * 0.0821 * 323}{0.9737 * 26}$ - V = 8.51 ltr. # ## 9.7(B) One mole of a gas Occupies a volm. of 1 litre at 27°C. What will be the pressure of the gas? - **Solution** - No. of mole of gas (n) = 1 mol. - vorm of gas (V) = 1 Ltr. - Temp of gas (T)=27℃=300.K - pressure of gas (P) = ? - R= 0.0821 L. atm K^-1 mol^-1 - We know, $PV=nRT$ - $P = \frac{nRT}{V} = 1 * 0.0821 * 300 / 1 = 24.63$ atm - Hence, pressure of gas is 24.63 atm. ## Q.7(C) Calculate the mass of oxygen whose gas whose volume is 320 ml at 17°C and 2 atm pressure? - **Solution:** - mass of O₂ = ?=(g). - molecular mass of O₂ (M) = 32 - Volume of O₂ (V) = 320 ml = 0.320 Ltr. - pressure of O₂ (P) = 2 atm - Temp (T) = 17°C = (17+273)K= 290.K - R= 0.0821 L. atm K^-1 mol^-1 - We know; - PV=nRT = $ \frac {MRT}{M}= MRT$ - or, $ m = \frac{PVM}{RT} = \frac{2 * 0.320 * 32}{0.0821 * 290} = 0.860$ g.m. - Hence, mass of Oxygen is 0.860 g.m ## 9. 8(A) A gas occupies 350 ml at 40°C. Calculate the temp at which the volm of the gas becomes 400 ml assume that the pressure remains constant? - **Solution**: - V₁ = 350 ml - V₂ = 400 ml - T₁ = 40°C = (40+273)K=313 K - T₂ = ? - According to the Charle's law, - $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ - $ \frac{350}{313}= $ $\frac{400}{T_2}$ - T₂ = $\frac{ 400 * 313}{350} $ - T₂ = 357.71 K