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

3 V / 5

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Volume of compartment II is

12 V / 5

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Temperature in both the compartment will be equal.

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Solution

The correct options are
A The pressure in the two compartments are equal.
B Volume of compartment I is $3 V / 5$ .
C Volume of compartment II is $12 V / 5$ .
D Temperature in both the compartment will be equal.
In the equilibrium position, the net force on the partition will be zero.
Hence pressure on both sides is same,
Therefore, $(a)$ is correct.

$n_{1} = \frac{P_{1} V_{1}}{R T_{1}} = \frac{P V}{R T}$

$n_{2} = \frac{(2 P) (2 V)}{R T} = 4 \frac{P V}{R T} \Rightarrow n_{2} = 4 n_{1}$

Moles remain conserved.
Finally pressure becomes equal in both the parts. Since, conducting partition is present, therefore temperature in both the compartment will become equal. Therefore, option (d) is also correct.

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

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

$P_{1} = P_{2}$ and $T_{1} = T_{2}$

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

$V_{2} = 4 V_{1}$

Also, $V_{1} + V_{2} = 3 V \Rightarrow V_{1} + 4 V_{1} = 3 V$

$V_{1} = \frac{3}{5} V$ and $V_{2} = \frac{12}{5} V$

Hence, options (b) and (c) are correct.

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