A partition divides a container having insulated wall, into two compartments- I and II. 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 statements is/are correct, with reference to the final equilibrium position?

The pressure in the two compartments are equal

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Volume of compartment I is

\frac{3 V}{5}

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Volume of compartment II is

\frac{12 V}{5}

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Final pressure in compartment I is

\frac{5 P}{3}

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Solution

The correct options are
A The pressure in the two compartments are equal
B Volume of compartment I is $\frac{3 V}{5}$
C Volume of compartment II is $\frac{12 V}{5}$
D Final pressure in compartment I is $\frac{5 P}{3}$
In the equilibrium position the net force on the partition will be zero.
Hence pressure on both sides are same.
Hence, (A) is correct.
Initially, $P V = n R T$
$n_{1} = \frac{P_{1} V_{1}}{R T_{1}} = \frac{P V}{R T}$
$n_{2} = \frac{(2 P) (2 V)}{R T} = 4 \frac{P V}{R T} \Rightarrow n_{2} = 4 n_{1}$
Number of moles remains conserved.
Finally, pressure becomes equal in both parts. Since, conductive partition is present, temperature in both the compartment will be equal.

Using,
$P_{1} V_{1} = n_{1} R T_{1}$
$P_{2} V_{2} = n_{2} R T_{2}$

$∵ P_{1} = P_{2} & T_{1} = T_{2}$

$∴ \frac{V_{1}}{V_{2}} = \frac{n_{1}}{n_{2}} = \frac{1}{4} \Rightarrow V_{2} = 4 V_{1}$

Also
$V_{1} + V_{2} = 3 V \Rightarrow V_{1} + 4 V_{1} = 3 V$
$\Rightarrow V_{1} = \frac{3}{5} V$ and $V_{2} = \frac{12}{5} V$
Hence (B) and (C) are correct.
In compartment (I):
$P_{1} V_{1} = n_{1} R T_{1}$
$P_{1} (\frac{3 V}{5}) = (\frac{P V}{R T}) R (T)$
$P_{1} = \frac{5 P V}{3 V} = \frac{5}{3} P$
Hence (D) is also correct.

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