Question

Two identical containers joined by a small pipe initially contain the same gas at pressure $P_0$ and absolute temperature $T_0$ One container is now maintained at the same temperature while the other is heated to $2T_0$. The common pressure of the gasses will be-

• A. $\left(1\right)\frac{3}{2}{P}_0$
• B. $\left(2\right)\frac{4}{3}{P}_0$
• C. $\left(3\right)\frac{5}{3}{P}_0$
• D. $\left(4\right)2P_0$

Answer 1

The correct option is B $\left(2\right)\frac{4}{3}{P}_0$

Apply conserulation of mass (media)

{B} $\begin{array}{c}PV=nRT\\ \frac{P_0{V}_0}{T_0}+\frac{P_0}{T_0}=\frac{P_2{V}_0}{T_0}+\frac{P_2{V}_0}{2T_0}\\ \frac{2P_0{V}_0}{T_0}=\frac{3}{2}\frac{P_2{V}_0}{T_0}\Rightarrow P_2=\frac{4}{3}{P}_0\end{array}$