A sample containing 2.50 mol of an ideal gas at 298 K is expanded from an initial volume of 10.0 L to a final volume of 50.0 L. Calculate ΔG and ΔA for this process for:

(a) an isothermal reversible path and (b) an isothermal expansion against a constant external pressure of 0.750 bar. Explain why ΔG and ΔA do or do not differ from one another.

a) ΔG=ΔA = - 33.452 J

b) ΔG=ΔA=0

Explanation:

a) according to the first law of thermodynamics the change in Gibbs free energy and Helmholtz free energy is

dG=VdP - SdT

dA = - PdV - SdT

for an isothermal path dT=0, therefore

ΔG=∫VdP = ∫ (nRT/P) dP = nRT ln P2/P1

ΔA = ∫ (-P) dV = ∫ (-nRT/V) dV = nRT ln V1/V2

also from the ideal gas equation from an isothermal process

P1*V1 = nRT = P2*V2 → P2/P1 = V1/V2

ΔG = nRT ln V1/V2 = 2.5 mol * 8.314 J/mol K P ln (10L/50L) = - 33.452 J

for Helmholtz free energy is also

ΔA = nRT ln V1/V2 = - 33.452 J

since the final energy is lower than the initial one, the process is spontaneous

for an isothermal process and isobaric process, dT=0 and dP=0, therefore

ΔG = VdP - SdT = 0

since → P2/P1 = V1/V2 = constant → dV=0

ΔA=0

ΔG=ΔA for an ideal gas in an isothermal process since G and A differ only in the H and U terms, but since these ones depends only on temperature for an ideal gas (H (T) = U (T) + PV = U (T) + nRT), they stay constant under an isothermal process and therefore ΔG=ΔA.