Chemistry - Thermochemistry Question with Solution | TestHub

(A) ${\mathrm{H}}_2\mathrm{O}(\mathrm{l})\rightleftharpoons {\mathrm{H}}_2\mathrm{O}(\mathrm{s})1\mathrm{atm}$ and 273 K .

∵ Exohtermic reaction $\therefore \mathrm{q}<0$

also volume increase $\therefore \mathrm{w}<0$

${\mathrm{H}}_2\mathrm{O}(\mathrm{s})$ has lesser entropy as comparted to ${\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

$\therefore \Delta {\mathrm{S}}_{\mathrm{sys}}<0\Delta \mathrm{U}<0$

Also since reaction is at equilibirum $\therefore \Delta \mathrm{G}=0$

Hence in (R) & (T)

(B) ∵ isolated condition ∴ $\mathrm{q}=0$, also expansion against vacuum $\therefore \mathrm{w}=0\therefore \Delta \mathrm{U}=0$, also I.G.

∴ Temperature remains constant.

Also, entropy increases

Also for physical process

$\mathrm{dG}=\mathrm{VdP}-\mathrm{SdT}$

∴ it will be $>0$

∴ Ans. is (P), (Q), (S)

(C) ∵ isolated contain $\therefore \mathrm{q}=0$ and $\mathrm{w}=0$

∵ Gases are getting mixed ∴ entropy of system is increasing

Also $\Delta \mathrm{U}=0(\because \Delta \mathrm{U}=\mathrm{q}+\mathrm{w})$ Also ∵ mixing is spontaneous

$\therefore \Delta \mathrm{G}<0$

∴ Ans. is (P), (Q), (S)

(D) ∵ Reversible heating and cooling

∴ initial a find state will be same.

hence atl state functions will remain same. Also q and w for heating and cooling will be opposite of each other

∴ overall $\mathrm{q}=0$

∴ overall $\mathrm{w}=0$

Also Q mixing is spontaneous

$\therefore \Delta \mathrm{G}<0$ ∴ Ans. is (P), (Q), (S), (T)