1.

Which of the following thermodynamic relation is correct

Answer»

dG=VdP-SdT
DE=PdV+TdS
dH=VdP+TdS
dG=VdP+SdT

Solution :`G=H-TS`
`because H = E + PV`
`therefore G = E + PV - TS`
Upon differentiation
`dG=dE+PdV+VdP-TdS-SdT"....(i)"`
dq=dE-dw
If the work done is only due to expansion, then,
-dw=PdV
for a reversible PROCESS,
`dS=(dq)/(T)or TdS = dq"....(II)"`
and ALSO dq `= dE + PdV ".....(iii)"`
Combining eqs. (i) , (ii) and (iii), we have
`dG=VdP-SdT`.


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