Gas A (mol . Mass =16) and gas B (mol . Mass =4) are confined in to...

At $27^{{}^{{}^{\circ}}} \mathrm{C}$, total pressure $\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}=2+2=4 \mathrm{~atm}\frac{r_A}{r_B}=\frac{n_A}{n_B}=\frac{P_A}{P_B} \sqrt{\frac{M_B}{M_A}} \Rightarrow \frac{n_A}{n_B}=\frac{1}{2}\therefore$ moles diffused out at $27^{{}^{{}^{\circ}}} \mathrm{C}$ are in the ratio $1: 2$ Initial moles a a moles left $a-x \quad a-2 x$ Total moles after diffusion $=a-x+a-2 x=a$ ( $\because$ P drops from 4 atm to $2 \mathrm{~atm} \Rightarrow$ moles should decrease from 2 a to a) $\therefore \mathrm{a}=3 \mathrm{x}\therefore$ After diffusion moles of $A$ and $B$ are $3 \mathrm{x}-\mathrm{x}=2 \mathrm{x}$ and $3 \mathrm{x}-2$\mathrm{x}=\mathrm{x}$\therefore$ moles of $A$ and $B$ left are in the ratio $2: 1\therefore \quad$ Partial pressures after diffusion at $27^{{}^{{}^{\circ}}} \mathrm{C}$ For A it is $\frac{2}{3} \times \mathrm{P}=\frac{2}{3} \times 2=\frac{4}{3}$ For $B$ it is $\frac{1}{3} \times \mathrm{P}=\frac{1}{3} \times 2=\frac{2}{3}$ When the vessel is heated to $127^{{}^{{}^{\circ}}} \mathrm{C}$, Partial pressure of $\mathrm{A}=\frac{4}{3} \times \frac{4}{3}=\frac{16}{9}$ atm $\left(\because \frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2}\right)$ Partial pressure of $B=\frac{2}{3} \times \frac{4}{3}=\frac{8}{9} \mathrm{~atm}$ Now $$\begin{aligned} & \mathrm{A} \\ & \begin{array}{l}\mathrm{16} \\ \frac{16}{9}\end{array} \\ & \frac{\mathrm{~B} \rightleftharpoons \mathrm{C}}{9}-\mathrm{P} \\ & \frac{16}{9} \quad \frac{8}{9}-\mathrm{P} \quad \mathrm{P} \\ & \mathrm{K}_{\mathrm{P}}=0.1=\frac{\mathrm{P}}{\left(\frac{16}{9}-\mathrm{P}\right)\left(\frac{8}{9}-\mathrm{P}\right)} \Rightarrow \text { partial pressure of } \mathrm{C}, \mathrm{P}=0.126 \mathrm{~atm}\end{aligned}$$