Question Number 23984 by Tinkutara last updated on 10/Nov/17
![Which of the following relation is/are correct? (1) ΔG = ΔH − TΔS (2) ΔG = ΔH + T[((δ(ΔG))/(δT))]_P (3) ΔG = ΔH + TΔS (4) ΔG = ΔH + ΔnRT](Q23984.png)
$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{relation}\:\mathrm{is}/\mathrm{are} \\ $$$$\mathrm{correct}? \\ $$$$\left(\mathrm{1}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:−\:\mathrm{T}\Delta\mathrm{S} \\ $$$$\left(\mathrm{2}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:+\:\mathrm{T}\left[\frac{\delta\left(\Delta\mathrm{G}\right)}{\delta\mathrm{T}}\right]_{\mathrm{P}} \\ $$$$\left(\mathrm{3}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:+\:\mathrm{T}\Delta\mathrm{S} \\ $$$$\left(\mathrm{4}\right)\:\Delta\mathrm{G}\:=\:\Delta\mathrm{H}\:+\:\Delta\mathrm{nRT} \\ $$
Answered by Zezo9970010@gmail.com last updated on 10/Nov/17
$${number}\:\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{correct} \\ $$
Commented by Tinkutara last updated on 11/Nov/17
$${There}\:{are}\:\mathrm{2}\:{answers}. \\ $$
Answered by Tinkutara last updated on 12/Nov/17
![To prove: [((δ(ΔG))/(δT))]_P =−ΔS 1^(st) law: ΔU=q+w dU=dQ+dw dU=Tds−Pdv ...(1) ΔG=ΔH−TΔS dG=dH−d(TS) dG=dH−Tds−sdT ...(2) H=U+PV dH=dU+d(PV) dH=dU+PdV+VdP ...(3) dH=Tds−PdV+PdV+VdP dH=Tds+VdP ...(4) dG=Tds+VdP−Tds−sdT dG=VdP−sdT ...(5) [(dG/dT)]_P =0−s(dT/dT) [(dG/dT)]_P =−S](Q24081.png)
$${To}\:{prove}:\:\left[\frac{\delta\left(\Delta{G}\right)}{\delta{T}}\right]_{{P}} =−\Delta{S} \\ $$$$\mathrm{1}^{\boldsymbol{{st}}} \:\boldsymbol{{law}}:\:\Delta{U}={q}+{w} \\ $$$${dU}={dQ}+{dw} \\ $$$$\boldsymbol{{dU}}=\boldsymbol{{Tds}}−\boldsymbol{{Pdv}}\:...\left(\mathrm{1}\right) \\ $$$$\Delta{G}=\Delta{H}−{T}\Delta{S} \\ $$$${dG}={dH}−{d}\left({TS}\right) \\ $$$$\boldsymbol{{dG}}=\boldsymbol{{dH}}−\boldsymbol{{Tds}}−\boldsymbol{{sdT}}\:...\left(\mathrm{2}\right) \\ $$$${H}={U}+{PV} \\ $$$${dH}={dU}+{d}\left({PV}\right) \\ $$$$\boldsymbol{{dH}}=\boldsymbol{{dU}}+\boldsymbol{{PdV}}+\boldsymbol{{VdP}}\:...\left(\mathrm{3}\right) \\ $$$${dH}={Tds}−{PdV}+{PdV}+{VdP} \\ $$$$\boldsymbol{{dH}}=\boldsymbol{{Tds}}+\boldsymbol{{VdP}}\:...\left(\mathrm{4}\right) \\ $$$${dG}={Tds}+{VdP}−{Tds}−{sdT} \\ $$$$\boldsymbol{{dG}}=\boldsymbol{{VdP}}−\boldsymbol{{sdT}}\:...\left(\mathrm{5}\right) \\ $$$$\left[\frac{{dG}}{{dT}}\right]_{{P}} =\mathrm{0}−{s}\frac{{dT}}{{dT}} \\ $$$$\left[\frac{\boldsymbol{{dG}}}{\boldsymbol{{dT}}}\right]_{\boldsymbol{\mathrm{P}}} =−\boldsymbol{{S}} \\ $$