An ideal gas (relative permittivity $\in_{r} = 1)$ having f degrees of freedom is kept inside a thermally insulated vessel, the vessel has two large pistons each with charge $+ Q a n d - Q$ . Assume charges are uniformly distributed. Pistons can move without friction inside the vessel. At an instant, the charge of pistons is increased k times with the help of an external agent. Then which of the following statement(s) is/are correct.

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The ratio of final to initial temperature of gas in equilibrium is $\frac{K^{2} + f}{2 + f}$

Work done by the agent in the process is $(K^{2} - 1) n R T_{1}$ where $T_{1}$ is initial temperature.

Work done by the agent in the process is $(K^{2} - 1) \frac{Q^{2} L_{1}}{2 A \in_{0}}$ where $L_{1}$ is initial separation between pistons

The ratio of final to initial temperature of gas in equilibrium is $\frac{2 K^{2} + f}{2 + f}$

answer is B, C, D.

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Detailed Solution

Stage 1

$P_{1} A L_{1} = n R T_{1}$

$\frac{Q^{2}}{2 A \in_{0}} = P_{1} A = \frac{n R T_{1}}{L_{1}}$ (Force Balancing)

Stage 2
Charge is suddenly changed
Question Image

System is not in equilibrium
Stage 3

Finally system in equilibrium

$P_{2} A L_{2} = n R T_{2} \frac{K^{2} Q^{2}}{2 A \in_{0}} = \frac{n R T_{2}}{L_{2}}$
Energy of the system is conserved between stages 2 and 3
$\Rightarrow \frac{1}{2} \in_{0} E_{2}^{2} A L_{1} + \frac{f}{2} n R T_{1} = \frac{1}{2} \in_{0} E_{2}^{2} A L_{2} + \frac{f}{2} n R T_{2}$
On solving all the above equation
$\frac{T_{2}}{T_{1}} = \frac{2 K^{2} + f}{2 + f} W_{e x t} = (K^{2} - 1) \frac{Q^{2} L_{1}}{2 A \in_{0}} = (K^{2} - 1) n R T_{1}$