Gaseous State Solved Examples

GASEOUS STATE SOLVED EXAMPLES Ex. 1 A gas column is trapped between closed end of a tube and a mercury column of length (h) when this tube is placed with its open end upwards the length of gas column is (1), the length of gas column becomes (2) when open end of tube is held downwards. Find atmospheric pressure in terms of height of Hg column. Sol. for gas P1 = (PO + h) P2 = (PO – h) 2 V1 = r 1 V2 = r22 at const T. and moles. P1V1 = P2V2 ; (PO + h) r21 = (PO – h) r22 PO2 + h1 = PO2 – h2 PO2 – Po1 = h, + h2 P0 = Ex. 2 h (l 1 + l 2 ) cm of Hg column Ans. (l 2 – l 1) The diameter of a bubble at the surface of a lake is 4 mm and at the bottom of the lake is 1 mm. If atmospheric pressure is 1 atm and the temperature of the lake water and the atmosphere are equal. what is the depth of the lake ? (The density of the lake water and mercury are 1 g/ml and 13.6 g/ml respectively. Also neglect the contribution of the pressure due to surface tension) Sol. P1V1 = P2V2 (760 mm × 13.6 × g) 4 4  (4 mm/2)3 = (760 mm × 13.6 × g + h × 1 × g)  (1 mm/2)3 3 3 760 × 13.6 × 64 = (760 × 13.6 + h) h = 64 × 760 × 13.6 – 760 × 13.6 h = 63 × 760 × 13.6 mm h= Ex. 3 Sol. 63×760×13.6 km = 0.6511 km = 651.1 m Ans. 1000×1000 A gas is initially at 1 atm pressure. To compress it to 1/4 th of initial volume, what will be the pressure required ? P1 = 1 atm V1 = V P2 = ? P1V1 = P2V2 P2 = V2 = V 4 at const. T & n P1V1 1 atm × V = = 4 atm Ans. V V2 4 Ex. 4 Find the lifting power of a 100 litre balloon filled with He at 730 mm and 25°C. (Density of air = 1.25 g /L). Sol. Since, PV = nRT PV = W RT T M  W= PVM 730 100× 4 = × g RT 760 0.082× 298 i.e., Wt. of He = 15.72 g Wt. of air displaced = 100 × 1.25 g/L = 125 g Lifting power of the balloon = 125 g – 15.72 g = 109.28 g Ans.  Ex. 5 A weather balloon filled with hydrogen at 1 atm and 300 K has volume equal to 12000 litres. On ascending it reaches a place where temperature is 250 K and pressure is 0.5 atm. The volume of the balloon is : (A) 24000 litres (B) 20000 litres (C) 10000 litres (D) 12000 litres Sol. Using P1V1 P2V2 = ; T1 T2 1 atm × 12000 L 0.5 atm × V2 = 300 K 250 K  V2 = 20,000 L Hence Ans. (B) Ex. 6 If water is used in place of mercury then what should be minimum length of Barometer tube to measure normal atmospheric pressure. Sol. PHg = PH O = P . 2 atm 0.76 m × 13.6 × g = h H2O × 1 × g h H2O = 0.76 × 13.6 = 10.336 m Ans. Ex. 7 Sol. A car tyre has a volume of 10 litre when inflated. The tyre is inflated to a pressure of 3 atm at 17°C with air. Due to driving the temperature of tyre increases to 47°C. (a) What would be pressure at this temperature ? (b) How many litres of air measured at 47°C and pressure of 1 atm should be left out to restore the tyre to 3 atm at 47°C V = 10L P, = 3 atm T, = 17oC = 290K T2 = 47oC = 320K (a)  Here v & n are constant  Using P1 P1T2 3×320 P2 =  P2 = = atm T1 T1 T2 290 i.e. P2 = 3.3103 atm i.e. Pressuse at 47oC = 3.3103 atm Ans. (b) Now, P1 = 3 atm ; V1 = 10L ; T1 = 17oC = 290K Final P2 = 3atm ; V2 = ? ; T2 = 47oC = 320 K 10L ×320K V1 V2 V ×T =  V2 = 2 2 = = 11.035L T2 290K T1 T1  Volume of gas should be removed at 47oC & 3atm = (11.035 – 10) L = 1.035L Now this vol at 1atm be v then 1.0351 × 3atm = v × 1atm  v = 3.105L  Reqd vol = 3.105 L Ans. GASEOUS STATE Ex. 8 Sol. A tube of length 50 cm is initially in open atmosphere at a pressure 75 cm of Hg. This tube is dipped in a Hg container upto half of its length. Find the level of mercury column in side the tube. If after dipping the tube, the length of air column be x cm (situation shown in the adjoining figure). Then by using, Pi Vi = Pf Vf We have. 75 cm Hg × A = Pf × x × A ....... (1) & also, Pf = 75 cm Hg + (x – l ) 2 (= 50 cm) ....... (2) (2) & (1)  [75 + (x – 25)] × x = 75 × 50  x2 + 50 x – 3750 = 0  x = 41.14 or – 91.14 But, x can't be – ve  x = 41.14  mercury column inside the tube = (50 – 41.14) cm = 8.86 cm Ans. Ex. 9 Sol. An open container of volume V contains air at temperature 27ºC or 300 K.The container is heated to such a temperature so that amount of gas coming out is 2/3 of (a) amount of gas initially present in the container. (b) amount of gas finally remaining in the container. Find the temperature to which the container should be heated. (a) Here , P & V are constant, n & T are changing. Let, initially the amount of gas present be n & temp is 27ºC or 300K. Finally amount of gas present in container = n – 1  2 n =  × n  & final temp. be T. Then using 3 3  n1T1 = n2T2 , we have, n × 300 = n × T2  T2 = 900K 3 i.e., final temp = 900K Ans. Sol. Ex. 10 Sol. (b) Let there be x moles of gas remaining in the container, 2 5x x) = n  =n 3 3  (x + x=  Using n1 T1 = n2 T2  T2 = 500K final temperature = 500 K Ans. 2 of x come out 3 3n 5 n × 300 K = 3n × T2 5 Four one litre flasks are separately filled with the gases, O2, F2, CH4 and CO2 under the same conditions. The ratio of number of molecules in these gases : (A) 2 : 2 : 4 : 3 (B) 1 : 1 : 1 : 1 (C) 1 : 2 : 3 : 4 (D) 2 : 2 : 3 : 4 According to avogadro’s hypothesis. All the flasks contains same no. of molecules  Ratio of no. of molecules of O2, F2, CH4 & CO2 = 1 : 1 : 1 : 1 Ans (B) Ex. 11 Sol. A sample of water gas has a composition by volume of 50% H2, 45% CO and 5% CO2. Calculate the volume in litres at STP at which water gas which on treatment with excess of steam will produce 5 litre of H2. The equation for the reaction is : CO + H2O  CO2 + H2 If x L CO in needed then  x  ×50%  L volume of H2 in water gas =   0.45  1 x  x × L = L = 0.9  0.45 2  But, from equation : CO + H2O  CO2 + H2 & Gay-Lussac’s law, we get, that the volume of H2 produced = volume of CO taken.  Volume of H2 due to reaction = x L   x  + x L= 5 L Total volume of H2 =   0.9   1.9x =5L 0.9  x=  Volume of water gas = 0.9× 5 1.9 x 0.9×5 L= L = 5.263 L Ans. 0.45 1.9× 0.45 Ex. 12 5 ml of H2 gas diffuses out in 1 sec from a hole. Find the volume of O2 that will diffuse out from the same hole under identical conditions in 2 sec. Sol. Rate of diffusion of H2 = 5ml = 5ml/s = rH 2 (say) 1 sec 1 1 = 5ml/s × 4 4  Volume of O2 diffused in 2.0 seconds  ro 2 = rH 2 × = Ex. 13 5 × 2 ml = 2.5 ml Ans. 4 Ans. The partial pressure of hydrogen in a flask containing two grams of hydrogen and 32 gm of sulphur dioxide is : (A) 1/16th of the total pressure (B) 1/9th of the total pressure (C) 2/3 of the total pressure (D) 1/8th of the total pressure (C) Sol. n H2 = 2g = 1mol. 2g / mol  PH2 =  (C) n n H2 H2 + nSO2  nSO 2 = × PT = 32g = 0.5 mol 64g / mol 1 2 × PT = P . (1+ 0.5) 3 T GASEOUS STATE Ex. 14 Ans. Sol. Equal volume of two gases which do not react together are enclosed in separate vessels. Their pressures are 10 mm and 400 mm respectively. If the two vessels are joined together, then what will be the pressure of the resulting mixture (temperature remaining constant) : (A) 120 mm (B) 500 mm (C) 1000 mm (D) 205 mm (D) Let, vol of containers be V & temperature be T P1 = 10mm P2 = 400mm  n1 = P1V RT  n1 + n2 = & n2 = P2V RT (P1 + P2 )× V RT After joining two containers final vol = (V+V) = 2V(for gases)  Pfinal = =  n1 + n 2  RT Vfinal 10 + 400  mm 2  P1 + P2  RT (P1 + P2 )× V × = 2 2V RT = = 205 mm. Ex. 15 A vessel contains H2 & O2 in the molar ratio of 8 : 1 respectively. This mixture of gases is allowed to diffuse through a hole, find composition of the mixture coming out of the hole. Sol. Here, nH2 : n O 2 = 8 : 1   rH 2 rO2 = 8 × 1 & rH 2 rO2 = n H2 M O2 n O2 M H2 32 32 = 1 2  no. of molesof H 2 coming out  / Δt  no. of molesof O2 coming out  / Δt = 32 1  Required composition of H2 : O2 coming out = 32 : 1 Ans. Ex. 16 Sol. (i) Calculate the pressure exerted by 1023 gas molecules, each mass 10–22 g in a container of volume one litre. The rms velocity is 105 cm/sec. (ii) What is the total kinetic energy (in cal) of these particles ? (iii) What must be the temperature ? (i) Here N = 1023 molecules m = 10–22g , V = 1L = 1000 cm3 rms velocity = 105 cm/second 1  PV = Nm U 2 3 1  P = Nm U 2 3 = 1 × 1023 × 10–22g × (105)2cm2/s2 3×1000cm 3 = 3.33 × 107 dyne/cm2 Ans. 446 (ii) Total K . E of the molecules 1 Nm U 2 2 1 = × 1023 × 10–25 kg × (103m/s)2 2 10000J = = 5000 J  1195.0 Cal Ans. 2 1 3 (iii) Avg K . E of one molecule = m 2 = kT 2 U 2 = T= 1 mU 2 × 3 k   2 –25 3 1 10 Kg × 10 m / s × 6.023×10 = × 3 8.314 K R   Q k = NA    = 2414.8 K Ans. Ex. 17 The rate of diffusion of a sample of a ozonized oxygen is 0.98 times than that of oxygen. Find the % (by volume of ozone in the ozonized sample). Sol. Let, rate of diffusion of ozonized oxygen be rg & Let, reat of diffusion of oxygen be rO 2  but rg rO 2 rg rO 2 = 0.98 (given)  M O2 =   Mg  (0.98)2 = 1/2     Form (1)  0.98 = ......... (1) (mean molar mass of ozonized oxygen = Mg) 32 Mg 32 32  Mg = = 33.32 Mg  0.98 2 Let % of O3 be x  % O2 = (100 – x) in ozonized oxygen.  48x + 100 – x  × 32 100 = 33.32  3200 + 16x = 33.32  x = i.e. % of O3 = 8.25 % Ans. 132 = 8.25 % 16 GASEOUS STATE Ex. 18 Sol. A gaseous mixture containing CO, methane CH4 & N2 gas has total volume of 40 ml. This mixture is exploded with excess of oxygen on cooling this mixture a contraction of 30 ml is observed & when this mixture is exposed to aqueous KOH a further contraction of 30 ml is observed .find the composition of the mixture. Let vol of CO be x mL vol of CH4 be y mL vol of N2 be z mL On explosion with excess of oxygen the following reactions takes place 1 O (g)  CO2 (g) (By Gaylussac's law of combing volume) 2 2 xmL x mL CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) y mL y mL 2 ymL N2 remains unreacted On cooling H2O (g) liquifies hence volume reduction of 30 mL is observed  2y = 30  y = 15 But, vol of CO2 obtained = (x + y) mL This is absorbed in KOH & vol reduction of 30 mL is observed.  x + y = 30  x = 30 – y = (30 – 15) = 15 and, x + y + z = 40  z = 40 – x – y = 40 – 15 – 15 = 10  Composition of mixture is vol of CO = 15 mL Ans. vol of CH4 = 15 mL Ans. vol of N2 = 10 mL Ans. CO(g) + Ex. 19 Under critical states of a gas for one mole of a gas, compressibility factor is : (A) Sol. 3 8 (B) 8 3 (C) 1 (D) 1 4 PC VC (Under critical condition) RTC 8a , VC = 3b , TC = 27Rb For 1 mole of gas Z = But, PC = a 27b 2 3  a  3b 27Rb  Z=  × = 2  × R 8a 8  27b  Ans. (A) Ex. 20 One litre of oxygen gas is passed through a ozonizer & the final volume of the mixture becomes 820 ml. If this mixture is passed through oil of turpentine. Find the final volume of gas remaining. Sol. The reaction that takes place in ozonizer as 3O2  2O3. If Vml of O2 out of 1000 ml is ozonized then vol of O3 obtained =  Final vol of ozonized oxygen 2 = (1000 – V + V) mL = 820 mL 3 1  1000 – V = 820 3 V  = 180  V = 540 3  O2 remaining = (1000 – 540) mL = 460 mL Ans. 2 V 3 ( O3 is absorbed in oil of turpentine) Ex. 21 If for two gases of molecular weights MA and MB at temperature TA and TB ; TAMB = TBMA, then which property has the same magnitude for both the gases. (A) Density Sol. (B) Pressure Given that TA MB = TB MA  But, r.m.s. = 3RT M  r.m.sA = 3RTA MA (C) KE per mol (D) RMS speed TB TA = MB MA & r.m.sB = 3RTB MB  r.m.sA = r.m.sB  Ans. (D) Ex. 22 It has been considered that during the formation of earth H2 gas was available at the earth. But due to the excessive heat on the earth this had been escaped. What was the temperature of earth during its formation? (The escape velocity is 1.1 x 106 cm/s) Sol. Escape velocity of H2 should be equal to average velocity of H2.  Avg velocity of H2 = 1.1 × 106 cm/s = 1.1 × 104 m/s But, avg. velocity = 8RT πM 8×8.314× T π × 2×10 –3  1.1 × 104 = 1.1×10  T= 4 2 × π × 2×10 –3 8×8.314 M H2   2g  2  10 – 3 kg K = 11430.5 K = 11157.5oC Ans. 450