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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 and be the molar heat capacities for the two processes,
Multiple Choice Options
- A
- B
- C
- D and 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).
All Questions in this Chapter
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