See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(a) V dE nC dT,  (b) P dH nC dT  (c) G H TS   , H E PV   Differentiating we get     dG d E PV d TS     dE PdV VdP TdS SdT.......... i      From 1st law dq = dE + PdV  work of expansion dq and dS= entropy change T  dq TdS  Form eq (i) dG dq VdP dq SdT      dG VdP SdT ........... ii   Equation (ii) at const. pressure p dG S dT       And at const. Temperature T dG V dP      