IIT JAM Follow
September 7, 2020 5:54 pm 30 pts
thook of Physical Chemistry CELLANEOUS NUMERICALs Ongmole of steam is condensed reversibly to liquid water at 373 K and 101.325 kPa ressure Ahe heat of vaporization of water is 2.256 8 kJ g. Assuming that the steam behaves as an ideal gas, calculate w, q, A,U, A,S, A,A and A,G for the condensation process. The process is H,O(g. 101.325 kPa, 373 K) H,O(1, 101.325 kPa, 373 K) and AvapH (2.256 8 kJ g) (18 g mol) = 40.624 kJ mol Since the process takes place at constant pressure, therefore 9p-AvapH=- 40.624 kJ mol w =- p AV=- p(m,1 m, g p(Vmg where m.e is the molar volume of the gas at 373 K. Hence w = (101.325 kPa) {22.414 dm mol) (373 K/273K)} = 3 103.0 kPa dm' mol = 3 103 J mol A H= q, =- 40.622 4 kJ mol 40 622.4 J mol AS=- =-108.907 J K'mol 373 K A,G =0 A,G = AA+(Av )RT, we have Since AA =- (Av)RT= ---1) (8.314 J K mol) (373 K) = 3 101 J mol A,U= q +w = - 40.622 4 kJ mol+3.101 kJ mol=- 37.52 kJ molr
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Reshav It is PV formula only...nothing new
Priyanshu kumar
The molar volume of a gas is the volume of one mole of a gas at STP. At STP, one mole (6.02 × 1023 representative particles) of any gas occupies a volume of 22.4 L . A mole of any ... 