Chemistry - Thermodynamics - II Question with Solution | TestHub

(A) Mixing of ideal gases is spontaneous.

${\mathrm{w}}_{\text{net }}=0$ andq $=0.\therefore \Delta \mathrm{U}=0$

Temperature remains constant.

$\Delta {\mathrm{S}}_{\text{sys }}>0$ and $\Delta {\mathrm{S}}_{\text{surr }}>0$

(B) ${\mathrm{PV}}^2=$ Constant

${\mathrm{C}}_{\mathrm{m}}={\mathrm{C}}_{\mathrm{V}}+\frac{\mathrm{R}}{1-\mathrm{x}}=\frac{3\mathrm{R}}{2}-\mathrm{R}=\frac{\mathrm{R}}{2}$

TV=constant. So, 'T' decreases on increasing volume.

Gas lost heat and process is reversible. $\Delta {\mathrm{S}}_{\text{univ }}=0$.

(C) ${\mathrm{PV}}^{3/2}=$ constant

${\mathrm{C}}_{\mathrm{m}}={\mathrm{C}}_{\mathrm{V}}+\frac{\mathrm{R}}{1-\mathrm{x}}=\frac{3\mathrm{R}}{2}+\frac{\mathrm{R}}{1-(3/2)}=-\frac{\mathrm{R}}{2}$.

On increasing volume, T will decrease. ( $\because {\mathrm{TV}}^{1/2}=$ constant).

The gas gains heat and process is reversible. $\Delta {\mathrm{S}}_{\text{univ }}=0$.

(D) Temperature increases between 1 and 2, work is done by the system.

∴ Gas gains the heat and process is reversible. $\Delta {\mathrm{S}}_{\text{univ }}=0$.