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7 March, 02:59

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.

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  1. 7 March, 03:12
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    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.
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