Question

A closed system consists of 0.5kg of superheated steam initially at 1MPa, 300 ⁰C. The system undergoes a reversible isothermal process during which the steam condenses to a saturated liquid whilst exchanging heat wih a reservoir at 30⁰C. Show the process on a T-s diagram and calculate the change in enthropy (kJ/K) of the system, the reservoir (surroundings) and the universe.

Answer 1

mass of steam = 0.50 Kg = 0.50 kg x (1000 g / 1 Kg) = 500 g

Hence moles of steam, n = 500 g / 18.0 g/mol = 27.78 mol

T = 300 K, P1 = 1 MPa = 9.87 atm

Initial volume of steam, V1 = nRT/P1 = 27.78 mol x 0.0821 L.atm.mol-1K-1 x 300K / 9.87 atm

= 69.32 L

After the reversible isothermal condensation the steam is completely condensed to water.

Hence final volume, V2 = 500 g / 1.00 g/mL = 500 mL = 0.500 L

Entropy change during reversible isothermal process can be calculated as

Hence entropy change during the process, DeltaS = 2.303nRlogVf/Vi

= 2.303 x 27.78 mol x 8.314 JK-1mol-1 x log (0.500L / 69.32L) = - 1139.3 J/K = - 1.139 KJ/K (answer)

T = 300 C = 300 + 273 = 573 K

work done during isothermal process = 2.303nRTlogVf/Vi

=  2.303 x 27.78 mol x 8.314 JK-1mol-1 x 573K x log (0.500L / 69.32L) = - 652.82 KJ

Surrounding temperature, T = 30 C = 30+273 = 303 K

Heat exchanged to the surrounding = - (- 652.82 KJ) = + 652.82 KJ

Hence entropy change of the surrounding = S(surrounding) = + 652.82 KJ / 303K = 2.1545 KJ/K (answer)

S(universe) = S(system) +  S(surrounding)

= - 1.139 KJ/K + 2.1545 KJ/K = 1.015 KJ/K (answer)