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QPages 62-64: Q21: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q21 Solution: Figure shows a cylindrical tube of volume V with adiabatic…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often confuse a diathermic wall (which allows heat transfer) with a movable piston (which allows work), incorrectly calculating non-zero work for the gas. Since the wall is explicitly fixed, the volume of each compartment remains constant, ensuring that the work done by or on the gas is strictly zero ($W = 0$).

Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5nRT1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p1,T1p_1, T_1 on the left and p2,T2p_2, T_2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides. a) How much work has been done by the gas on the left part? b) Find the final pressures on the two sides. c) Find the final equilibrium temperature. d) How much heat has flown from the gas on the right to the gas on the left?

Step-by-Step Verified Solution

**Answer:** (a) W=0W = 0 (b) P1′=P2′=(P1+P2)/2P_1' = P_2' = (P_1 + P_2)/2 (c) T=P1V+P2V2nRT = \frac{P_1V + P_2V}{2nR} (d) Heat flow depends on specific heats. **Step-by-Step Explanation:** (a) The wall is fixed, so no work is done: W=0W = 0. (b) After equilibrium, pressures equalise: P1′=P2′=(P1+P2)/2P_1' = P_2' = (P_1 + P_2)/2. (c) Final temperature T=(P1+P2)V2nRT = \frac{(P_1 + P_2)V}{2nR}. (d) Heat flows from the hotter side to the cooler side through the diathermic wall.

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