Two pistons having low thermal conductivity divide an adiabatic container in three equal parts as shown. An ideal gas is present in the three parts A, B & C having initial pressures as shown and same temperatures. Now the pistons are released. Then the final equilibrium length of part A after long time will be

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The container is adiabatic and the pistons have low thermal conductivity, so no heat exchange occurs between the parts.

All three parts initially have the same temperature.

Let the final equilibrium pressure be $P_{f}$ .

Since the pistons are free to move, mechanical equilibrium requires:

$P_{A} = P_{B} = P_{C} = P_{f}$

Initial pressures:

$P_{A} = P, P_{B} = 2 P, P_{C} = P$

Initial lengths:

$L_{A} = L_{B} = L_{C} = \frac{L}{3}$

Using the ideal gas law for each part:

$P V = n RT$

At equilibrium, temperature remains the same for all parts, hence:

$P_{i} V_{i} = P_{f} V_{f}$

For part A:

$P \cdot \frac{L}{3} = P_{f} L_{A}$

For part B:

$2 P \cdot \frac{L}{3} = P_{f} L_{B}$

For part C:

$P \cdot \frac{L}{3} = P_{f} L_{C}$

Adding all three:

$P_{f} (L_{A} + L_{B} + L_{C}) = \frac{4 P L}{3}$

But:

$L_{A} + L_{B} + L_{C} = L$

So,

$P_{f} = \frac{4 P}{3}$

Now for part A:

$L_{A} = \frac{P}{P _{f}} \cdot \frac{L}{3} = \frac{P}{4 P /3} \cdot \frac{L}{3} = \frac{L}{4}$

$L_{A} = \frac{L}{4}$

Correct option: (B)

Physics - Heat & Thermo Question with Solution | TestHub

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