G=H−TS G=U+PV−TSΔG=ΔU+PΔV+VΔP−TΔS−SΔTFrom the first and second laws, TΔS=ΔU+PΔV∴ ΔG=VΔP−SΔTG=H+TΔGΔT ORG=H+T(∂G∂T)P−HT2=−GT2+1T(∂G∂T)P =[∂(G/T)∂T]PH=−T2[∂(G/T)∂T]P]

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$T^{2} {[\frac{\partial (G / T)}{(\partial T)}]}_{P}$

$- T^{2} {[\frac{\partial (G / T)}{(\partial T)}]}_{P}$

$T^{2} {[\frac{\partial (G / T)}{(\partial T)}]}_{V}$

$- T^{2} {[\frac{\partial (G / T)}{(\partial T)}]}_{V}$

answer is B.

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Detailed Solution

$G = H - T S \begin{array}{l} G = U + P V - T S \\ Δ G = Δ U + P Δ V + V Δ P - T Δ S - S Δ T \end{array}$

From the first and second laws,

$\begin{array}{l} T Δ S = Δ U + P Δ V \\ ∴ Δ G = V Δ P - S Δ T \end{array}$

$\begin{array}{l} G = H + T \frac{Δ G}{Δ T} \\ O R \\ G = H + T {(\frac{\partial G}{\partial T})}_{P} \\ - \frac{H}{T^{2}} = - \frac{G}{T^{2}} + \frac{1}{T} {(\frac{\partial G}{\partial T})}_{P} \\ = {[\frac{\partial (G / T)}{\partial T}]}_{P} \\ H = - T^{2} {[\frac{\partial (G / T)}{\partial T}]}_{P}] \end{array}$