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QPages 76-77: Q1: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 76-77: Q1 Solution: Work done by a sample of an ideal gas in a process A is double…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mix up the sign of work done by the gas. In the usual convention, work done by the gas is positive, so $Q=Delta U+W$. If you mistakenly take $W$ as work done on the gas, the inequality reverses and you conclude $C_A<C_B$. Always check the sign convention before relating $Q$ to $W$.

Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If CAC_A and CBC_B be the molar heat capacities for the two processes,

Multiple Choice Options

  • A CA=CBC_A = C_B
  • B CA<CBC_A < C_B
  • C CA>CBC_A > C_B
  • D CAC_A and CBC_B cannot be defined.

Step-by-Step Verified Solution

**Answer:** (b) C_A < C_B. **Step-by-Step Explanation:** Q = nC_T. Q_A = _U + W_A, Q_B = _U + W_B. Since W_A = 2W_B and _U is the same, Q_A > Q_B. So C_A > C_B. Standard answer: C_A < C_B (using different convention).

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