Question

A partition divides a container having insulated walls into two compartments I and II as shown in the figure. The same gas fills the two compartments whose initial parameters are given. The partition is a conducting wall which can move freely without friction. Which of the following statement(s) is /are correct with reference to the final equilibrium position?

• A. The pressure in the two compartments is equal.
• B. Volume of compartment I is $\frac{3V}{5}$.
• C. Volume of compartment II is $\frac{12V}{5}$.
• D. Pressure in compartment I is $\frac{5P}{3}$.

Answer 1

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

Pressure in compartment I is $\frac{5P}{3}$.

At equilibrium position, the net force on the partition will be zero.
Hence, the gases in both the compartments are in mechanical equilibrium i.e pressure on both sides is the same.
Hence, option (a) is correct.
Given, $P,2P$ are the initial pressures and $V,2V$ are the initial volumes of compartment I and II.
Also, both the compartments are at the same temperature $T$.
From the ideal gas equation $PV=nRT$, we can write that no of moles in compartment I and II are:
$n_1=\frac{P_1{V}_1}{R{T}_1}=\frac{PV}{RT}$
$n_2=\frac{P_2{V}_2}{R{T}_2}=\frac{\left(2P\right)\left(2V\right)}{RT}=4\frac{PV}{RT}$
$\Rightarrow n_2=4n_1$

Since the walls are closed, the number of moles remains conserved.
At mechanical equlibrium, pressure becomes equal in both parts
i.e $P_1^{\prime }=P_2^{\prime }$
Using $P_1^{\prime }{V}_1^{\prime }=n_{1}R{T}_1$
$P_2^{\prime }{V}_2^{\prime }=n_{2}R{T}_2$
$\because P_1^{\prime }=P_2^{\prime }\&T_1=T_2=T$
[because the partition is conducting, both sides will remain in thermal equilibrium]
We can write that, $\frac{V_1^{\prime }}{V_2^{\prime }}=\frac{n_1}{n_2}=\frac{1}{4}$
$\Rightarrow V_2^{\prime }=4V_1^{\prime }.....\left(1\right)$

Also, $V_1^{\prime }+V_2^{\prime }=3V$
[total volume of container remains constant]
Using equation $\left(1\right)$, we get
$\Rightarrow V_1^{\prime }+4V_1^{\prime }=3V$
$\Rightarrow V_1^{\prime }=\frac{3}{5}V$ and $V_2^{\prime }=\frac{12}{5}V$
Hence, (b) and (c) are correct.
In compartment (I):
$P_1^{\prime }{V}_1^{\prime }=n_{1}R{T}_1$
$\Rightarrow P_1^{\prime }\left(\frac{3V}{5}\right)=\left(\frac{PV}{RT}\right)R\left(T\right)$
$\Rightarrow P_1^{\prime }=\frac{5PV}{3V}=\frac{5}{3}P$

Thus, options (a), (b), (c) and (d) are correct answers.