Question

Two identical containers each of volume $V_0$ are joined by a small pipe. The containers contain identical gases at temperature $T_0$ and pressure $P_0$. One container is heated to temperature $2T_0$ while maintaining the other at the same temperature $T_0$. The common pressure of the gas is $P$ and $n_1$ is the number of moles of gas in container at temperature $2T_0.$ Then

• A. $P=2P_0$
• B. $P=\frac{4}{3}{P}_0$
• C. $n_1=\frac{2}{3}\frac{P_0{V}_0}{R{T}_0}$
• D. $n_1=\frac{3}{2}\frac{P_0{V}_0}{R{T}_0}$

Answer 1

Answer: (B), (C)

The correct options are
B $P=\frac{4}{3}{P}_0$
C $n_1=\frac{2}{3}\frac{P_0{V}_0}{R{T}_0}$

Initially for container $A,$
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For container $B$
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Total number of moles $=n_0+n_0=2n_0$
Since, even on heating, the total number of moles is conserved,
We have $n_1+n_2=2n_0$

Let $P$ be the common pressure. Then for container $A,$
$\begin{array}{c}P{V}_0=n_{1}R2T_0\\ \therefore n_1=\frac{P{V}_0}{2R{T}_0}\end{array}$
And for container $B,P{V}_0=n_{2}R{T}_0$
$\therefore n_2=\frac{P{V}_0}{R{T}_0}$
Substituting the values of $n_0,n_1$ and $n_2$ in equation (i), we get
$\frac{P{V}_0}{2R{T}_0}+\frac{P{V}_0}{R{T}_0}=\frac{2\cdot P_0{V}_0}{R{T}_0}\Rightarrow P=\frac{4}{3}{P}_0$
Number of moles in container (at temperature $2T_0)$
$=n_1=\frac{P{V}_0}{2R{T}_0}=\left(\frac{4}{3}{P}_0\right)\frac{V_0}{2R{T}_0}=\frac{2}{3}\frac{P_0{V}_0}{R{T}_0}[\text{As}P=\frac{4}{3}{P}_0]$