Problem on Adiabatic expansion

Question

Calculate the change in entropy for the system for each of the following cases. Explain the sign that you obtain by a physical argument

a) A gas undergoes a reversible, adiabatic expansion from an initial state at 500 K, 1 MPa, and 8.314 L to a final volume of 16.628 L.

b) One mole of methane vapor is condensed at its boiling point, 111 K; Δh_v = 8.2 [kJ/mol].

c) One mole of liquid water is cooled from 100°C to 0°C. Take the average heat capacity of water to be 4.2 JK^-1g^-1.

d) Two blocks of the same metal with equal mass are at different temperatures, 200°C and 100°C. These blocks are brought together and allowed to come to the same temperature. Assume that these blocks are isolated from their surroundings. The average heat capacity of the metal is 24 JK^-1mol^-1.

Answer 1

(a) Since the heat transfer, ΔQ = 0, in reversible adiabatic process, the entropy change,

ΔS = ΔQ/T = 0

(b) ΔS = Δhv/T = (-8.2 kJ/mol)/111 K = -0.074 kJ/(mol.K) = -74 J/(mol.K)

Since one mol is condensed, -74J/K is the entropy change, and this heat taken up by surrounding whose entropy change is positive 74J/K, and hence the entropy change of system plus surrounding is zero, in confirmation with the second law of thermodynamics.

(c) ΔS = ΔQ/T = ∫_cp,avgdT/T = _cp,avg ∫dT/T = cp,avg ln (T₂/T₁) = 4.2 ln(273/373) = = -1.31 J/(gK).

But we have 1 mol of water, i.e. 18 gm of water. Hence ΔS = -1.31 x 18 = -23.58 J/K

The negative sign implies that heat is lost or transferred from system to surrounding.

In other words water is cooled, by transferring the heat, hence the change in entropy is negative, while the surrounding gain the same amount of heat and for it the change in entropy is positive, hence the total change in entropy is zero, i.e. System + Surroundings.

(d) Let the equilibrium temperature be T,

mC_p(200 – T) = mC_p(T – 100)
(200 – T) = (T – 100)
T = 150^oC

Total change in entropy of the system,

        = change in entropy of 1^st block + change in entropy of 2^nd block

        = c_p ln (T₂/T₁) + cp ln (T₂/T₁)

        = 24ln (423/473) + 24ln (423/373)

        = 0.338 J/mol.K

Thus the entropy change is positive in this case, implying there are more configurations when the two blocks are allowed to interact.