Gaseous States Solved Problems

Page # 7 GASEOUS STATES SOLVED PROBLEMS Ex.1 Sol. A gas occupies 300 ml at 27ºC and 730 mm pressure. What would be its volume at STP(A) 162.2 ml (B) 262.2 ml (C) 362.2 ml (D) 462.2 ml (B) Given at T1 = 300 K, T2 = 273 K (STP)  300   litre, V1 = 300 ml =  1000  atm. P2 = 1 atm.,  730   P1 =  760  V2 = ?  P1V1 P2 V2 1x V2 730 x 300 = , = T1 T2 273 760 x1000 x 300  V2 = 0.2622 litre = 262.2 ml. Ex.2 A truck carrying oxygen cylinders is filled with oxygen at -23ºC and at a pressure of 3 atmosphere in Srinagar, Kashmir. Determine the internal pressure when the truck d rives thro ug h M adras. Tamilnadu. Where the temperature is 30ºC (A) 2.64 atm. (B) 1.64 atm. (C) 1 atm. (D) 3.64 atm. Sol. (D) P1 = 3 atm., P2 = ? T1 = – 23 + 273 = 250 K T2 = 273 + 30 = 303 K. P1 P2 P2 3 = T1 T2 , 250 = 303 P2 = 3 x 303 = 3.64 atm. 250 Ex.3 The density of a gas at -23ºC and 780 torr is 1.40 gram per litre. Which one of the following gases is it (A) CO2 (B) SO2 (C) Cl2 (D) N2 Sol. (D) PV = nRT 780 x 1 = n x .082 x (273 - 23) 760 n = 780 n = 0.0501 moles. 760  0.082  250  .0501 moles of the gas weigh = 1.40 gm.  1 mole of gas weigh = So gas is N2. 140 . = 28 gram. .0501 Ex. 4 Calculate the weight of CH4 in a 9 litre cylinder at 16 atm and 27ºC (R = 0.08 lit. atm/k) (A) 96 gm (B) 86 gm (C) 80 gm (D) 90 gm Sol. (A) Given P = 16 atm, V = 9 litre. T = 300 K, mCH4 = 16, R = 0.08 litre atm/k. PV = w/m × R × T w 16 x 9 = × 0.08 x 300 w = 96 gm. 16 Ex. 5 What is the density of sulphur dioxide (SO2) at STP (A) 2.86 gm/lit. (B) 1.76 gm/lit (C) 1.86 gm/lit (D) None of these. Sol. (A) The gram molecular weight of SO2 = 64 gm/ mole. Since 1 mole of SO2 occupies a volume of 22.4 litres at S.T.P. 64 = 2.86 gm/lit. 22.4 5gm of XeF4 gas was introduced into a vessel of 6 litre capacity at 80ºC. What is the pressure of the gas in atmosphere (A) .21 atm (B) .31 atm (C) .11 atm (D) .41 atm. (C) Given V = 6 litre, T = 353 K, R= 0.082, W = 5gm. m= 207.3 W PV = × R × T m 5 P × 6 = × 0.082 × (273 + 80) 207.3 5 x.082 x 353 P = = .11 atm. 6 x 207.3 Density of SO2 = Ex.6 Sol. Ex.7 Calculate the temperature at which 28 gm N2 occupies a volume of 10 litre at 2.46 atm(A) 300 K (B) 320 K (C) 340 K (D) 280 K Sol. (A) Given wN2 = 28gm , P = 2.46 atm, V = 10 litre. mN2 = 28,   w PV =   RT m  28  2.46 x 10 =   x 0.0821 x T 28 T = 300 K. Page # 8 GASEOUS STATES Ex.8 A mixture of gases at 760 mm pressure contains 65% nitrogen, 15% oxygen and 20% Carbondioxide by volume. What is the partial pressure of each in mm (A) 494, 114, 252 (B) 494, 224, 152 (C) 494, 114, 152 (D) None of these. Sol. (C) 65 P'N2 = 760 × = 494 mm 100 P'O2 15 = 760 × = 114 mm 100 20 = 152 mm. 100 Ex.9 0.45 gm of a gas 1 of molecular weight 60 and 0.22 gm of a gas 2 of molecular weight 44 exert a total pressure of 75cm of mercury. Calculate the partial pressure of the gas 2 (A) 30 cm of Hg (B) 20 cm of Hg (C) 10 cm of Hg (D) 40 cm of Hg. Sol. (A) No. of moles of gas 1 = n1 P'CO2 = 760 × w1 0.45 = m = = 0.0075 60 1 No. of moles of gas 2 = n2 w2 0.22 = m = = 0.0050 44 2 Total no. of moles = n 1 + n 2 = 0.0075 + 0.0050 = 0.0125 P2 partial pressure of gas 2 = 0.0050 x 75 = 30 cm of Hg. 0.0125 Ex.10 The total pressure of a sample of methane collected over water is 735 torr at 29ºC. The aqueous tension at 29ºC is 30 torr. What is the pressure exerted by dry methane (A) 605 torr (B) 205 torr (C) 405 torr (D) 705 torr Sol. (D) Ptotal = Pdry methane + Pwater 735 = Pdry methane + 30  Pdry methane = 735 - 30 = 705 torr. Ex.11 The odour from a gas A takes six seconds to reach a wall from a given point. If the molecular weight of gas A is 46 grams per mole and the molecular weight of gas B is 64 grams per mole. How long will it take for the odour from gas B to reach the same wall from the same point. Approximately - Sol. (A) 6 Sec (C) 8 Sec (B) rA rB MB MA = (B) 7 Sec (D) 9 Sec rA , r B = 64 46 = 1.18 Time taken for the odour of B to reach the wall = 1.18 x 6 = 7.08 sec  7 sec. Ex.12 1 litre of oxygen effuses through a small hole in 60 min. and a litre of helium at the same temperature and pressure effuses through the same hole in 21.2 min. What is the atomic weight of Helium (A) 2.99 (B) 3.99 (C) 2.08 (D) 1.99 Sol. (B) rO2 rHe = 1000 / 60 212 . = = 1000 / 21. 2 60 Squaring both of sides MHe = MHe MO 2 (212 . )2 (60) 2 = = MHe 32 MHe 32 (212 . ) 2 x 32 = 3.99 (60) 2 Since Helium is monoatomic so Atomic weight = Molecular weight = 3.99 Ex.13 What is the temperature at which oxygen molecules have the same r.m.s. velocity as the hydrogen molecules at 27ºC (A) 3527ºC (B) 4227ºC (C) 4527ºC (D) 4000ºC Sol. (C) r.m.s. velocity C = For oxygen Co2 = For CH2 at 27ºC = 3RT M 3RT 32 3 xR x 300 2 When Co2 = CH2 3RT = 32 3RT 3 xR x 300 3R x 300 , = 32 2 2 300 x 32 = 4800 K. 2 Then in ºC T = 4800 - 273 = 4527ºC T = Ex.14 Calculate the total kinetic energy in joules, of the molecules in 8 gm of methane at 27ºC(A) 1770.5 Joule (B) 1870.5 joule (C) 1970.5 joule (D) 1670.5 joule Page # 9 GASEOUS STATES Sol. (B) 3 EK (For 1 mole) = RT = 3741 Joule. 2 Total energy of 8 gm of methane 1 = mole of methane 2 3741 = = 1870.5 Joule. 2 Ex.15 Calculate the root mean square velocity of SO2 at S.T.P. (A) 3.26 × 104 cm/sec (B) 1.26 × 102 cm/sec (C) 1.26 × 10 4 cm/sec (D) 3.26 × 102 cm/sec Sol. (A)  = 3RT M vrms of SO2 = 3 x 8.314 x10 7 x 273 = 3.26 x 104 cm/sec. 64 Ex.16 Calculate the number of atoms in 1 g of helium (A) 1.506 × 10 23 atoms (B) 1.605 × 1032 atoms (C) 1.056 × 1025 atoms (D) None of these Sol. (A) Atomic mass of He = 4 4 g of He contain = 6.023 x 1023 atoms  1 g of He contains = 6.023 x 10 23 = 1.506 x 1023 atoms. 4 Ex.17 7.00 g. of a gas occupies a volume of 4.1 lit res at 300 K and 1 at mosphere pressure. Calculate the molecular mass of the gas (A) 40 g mol–1 (B) 42 g mol–1 –1 (C) 48 g mol (D) 45 g mol–1 Sol. (B) PV = nRT  n = PV/RT (1 atm.) (4.1 L) n = e0.082 L atm K 1 1 mol 1 300K = 6 mol jb g Mass of gas 7 1 now n = = = Mol. mass of gas 6 Mol. mass of gas Thus, molecular mass of gas = 7 x 6=42g mol –1 Ex.18 Calculate density of ammonia at 30ºC and 5 atm. pressure (A) 3.03 g/litres. (B) 3.82 g/litres (C) 3.42 g/litres. (D) 4.42 g/litres. Sol. (C) m m RT PV = nRT, or PV = RT or P = x M V M 1 or P = d x RT M d = MP/RT ; d = 17 x 5 = 3.42 g/litres. 0.082 x 303 Ex.19 3 moles of a gas are present in a vessel at a temperature of 27ºC . Calculate the value of gas constant (R) in terms of kinetic energy of the molecules of gas (A) 7.4 x 10–4 KE per degree kelvin. (B) 4.5 x 10–4 KE per degree kelvin. (C) 7.4 x 10–5 KE per degree kelvin. (D) None of these Sol. (A) K.E. for 1 mole = 3/2 RT 2 K.E. for 3 moles = 9/2 RT. or R = KE 9T = 2 KE = 7.4x10–4KE per degree kelvin. 9 ( 300) Ex.20 In the following diagram, container of NH3 gas and container of HCl gas, connected through a long tube, are opened simultaneously at both ends; the white NH4Cl ring first formed will be at Q point. If OP = 40cm, then find OQ - White fumes NH3 Gas Sol. Q P O HCl Gas (A) 35 cm (B) 23.74cm (C) 30 cm (D) 31.25 cm (B) Let OQ = x cm so QP = (40 – x) cm Diffused volume of NH3 gas = Area of T.S. of tube × Distance travelled by NH3 gas VNH3 = A × OQ = Ax {Where A is area of T.S. of tube} Similarly in the same time, Diffused volume of HCl gas = Area of T.S. of tube × Distance travelled by HCl gas VHCl = A × QP = A (40 – x) From Graham’s Law of diffusion   rNH3 rHCl  VNH3 / t VHCl / t MHCl MNH3 = 36.5 = 1.46 17  x = 1.46 ( 40  x )   x = 23.74 cm OQ = 23.74 cm