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GATE 2024 CH – Question 56

Thermodynamics · Laws of thermodynamics, open and closed systems, entropy and chemical potential · 2 marks · Numerical answer

An isolated system consists of two perfectly sealed cuboidal compartments $A$ and $B$ separated by a movable rigid wall of cross-sectional area 0.1 m$^2$ as shown in the figure. Initially, the movable wall is held in place by latches $L_1$ and $L_2$ such that the volume of compartment $A$ is 0.1 m$^3$. Compartment $A$ contains a monoatomic ideal gas at 5 bar and 400 K. Compartment $B$ is perfectly evacuated and contains a massless Hookean spring of force constant 0.3 N m$^{-1}$ at its equilibrium length (stored elastic energy is zero). The latches $L_1$ and $L_2$ are released, the wall moves to the right by 0.2 m, where it is held at the new position by latches $L_3$ and $L_4$. Assume all the walls and latches are massless. The final equilibrium temperature, in K, of the gas in compartment $A$, rounded off to 1 decimal place, is _________

a closed box divided by a movable wall. Compartment A (gas) is on the left, and compartment B on the right holds a spring between the wall and the right end. The latches L1 and L2 hold the wall, and L3 and L4 are 0.2 m to the right.

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Show answer and explanation

Correct answer: 398.00 to 402.00

Explanation

The system is isolated, so the energy is conserved. The only work done by the gas goes into compressing the spring: $\frac{1}{2}kx^2 = \frac{1}{2} \times 0.3 \times 0.2^2 = 0.006$ J. The gas holds $n = \frac{PV}{RT} = \frac{5 \times 10^5 \times 0.1}{8.314 \times 400} = 15.0$ mol, with $C_v = 12.47$ J mol⁻¹ K⁻¹. So $\Delta T = -\frac{0.006}{15.0 \times 12.47} = -3.2 \times 10^{-5}$ K, which is negligible: the final temperature is 400.0 K (the spring constant is so small that the gas expands almost freely).