Question

Two identical containers $A$ and $B$ each of volume $V_0$ are joined by a small pipe. The containers contain identical gases at temperature $T_0$ and pressure $P_0$. Container $A$ is heated to temperature $2T_0$ while maintaining $B$ at the same temperature. If the common pressure of the gas is $P$ and $n_1$ and $n_2$ are the number of moles of the gas in $A$ and $B$ respectively, then choose the correct option(s):

• A. $P=\frac{4}{3}{P}_0$
• B. $P=\frac{2}{3}{P}_0$
• C. $n_1=\frac{2}{3}\frac{P_0{V}_0}{R{T}_0}$
• D. $n_2=\frac{4}{3}\frac{P_0{V}_0}{R{T}_0}$

Answer 1

Answer: (A), (C), (D)

$n_2=\frac{4}{3}\frac{P_0{V}_0}{R{T}_0}$

From the data given in the question,
Initial condition for container A
$P_0{V}_0=n_{0}R{T}_0$
$\Rightarrow$ Number of moles of the gas in container A is
$n_0=\frac{P_0{V}_0}{R{T}_0}$
Initial condition for container B
$P_0{V}_0=n_{0}R{T}_0$
$\Rightarrow$ Number of moles of the gas in container B is
$n_0=\frac{P_0{V}_0}{R{T}_0}$

Final condition:
For A, $P{V}_0=n_{1}R\left(2T_0\right)$
$\Rightarrow$ Number of moles of the gas in container A is $n_1=\frac{P{V}_0}{2R{T}_0}$
For B. $P{V}_0=n_{2}R{T}_0$
$\Rightarrow$ Number of moles of the gas in container B is $n_2=\frac{P{V}_0}{R{T}_0}$
As the quantity of gas remains constant, we can say that
Total number of moles of the gas in both containers after joining = Total number of moles of the gas in both containers before joining
i.e $n_1+n_2=2n_0$
$\Rightarrow \frac{P{V}_0}{2R{T}_0}+\frac{P{V}_0}{R{T}_0}=\frac{2P_0{V}_0}{R{T}_0}$
$\Rightarrow P=\frac{4}{3}{P}_0$
Also, $n_1=\frac{2}{3}\frac{P_0{V}_0}{R{T}_0}$
& $n_2=\frac{4}{3}\frac{P_0{V}_0}{R{T}_0}$
Thus, options (a), (c) and (d) are correct answers.