JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Equilibrium

Let the partial pressures be $P_{\mathrm{B}}$, $P_{\mathrm{C}}$ and $P_{\mathrm{E}}$, remembering that $\mathrm{C}$ is produced by both solids. For the first solid $K_{p_1}=P_{\mathrm{B}}{P}_{\mathrm{C}}=x$ and for the second $K_{p_2}=P_{\mathrm{E}}{P}_{\mathrm{C}}=y$. If the amounts formed give $P_{\mathrm{B}}=a$, $P_{\mathrm{E}}=b$ and $P_{\mathrm{C}}=a+b$, since every $\mathrm{B}$ and every $\mathrm{E}$ is matched by a $\mathrm{C}$, then $x=a(a+b)$ and $y=b(a+b)$, and adding gives $x+y=(a+b{)}^2$, so $a+b=\sqrt{x+y}$. The total pressure is $P_{\mathrm{B}}+P_{\mathrm{C}}+P_{\mathrm{E}}=a+(a+b)+b=2(a+b)=2\sqrt{x+y}$ atm.