Chemistry - Thermodynamics - I Question with Solution | TestHub

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

∵ Endothermic reaction $\therefore \mathrm{q}>0$

also volume decrease $\therefore \mathrm{w}>0$

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

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

(Q) ∵ 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

(R) $\because$ isolated container $\therefore \mathrm{q}=0$ and $\mathrm{w}=0$

∵ Gases are gettting mixed ∴ entropy of system is increasing

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

Also ∵ mixing is spontaneous $\therefore \Delta \mathrm{G}<0$

(S) ∵ Reversible heating and cooling

∴ initial a find state will be same.

hence all state functions will remain same.

Also q and w for heating and cooling will be opposite of each other

∴ overall $\mathrm{q}=0$

∴ overall $\mathrm{w}=0$