Asample 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.

2b2t

explanation:

2b2t