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 $2 T_{0}$. The common pressure of the gases will be:

• A. $\frac{3}{2} p_{0}$
• B. $\frac{4}{3} p_{0}$
• C. $\frac{5}{3} p_{0}$
• D. $2 p_{0}$

Answer 1

Answer: (B)

By gas law $p_{f} v_{1} =n_{1} R T_{1}$
$p_{f} v_{2} =n_{2} R T_{2}$
$n_{1}+n_{2} =n$
$\Rightarrow \frac{p_{f} v}{R T_{0}}+\frac{p_{f} v}{R\left(2 T_{0}\right)} =\frac{p_{0}(2 v)}{R T_{0}}$
$\Rightarrow p_{f}+\frac{p_{f}}{2} =2 p_{0}$
$\Rightarrow \frac{3 p_{f}}{2} =2 p_{0}$
$p_{f}=\frac{4 p_{0}}{3}$