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QPages 76-77: Q3: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 76-77: Q3 Solution

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat $C_p-C_v=R$ as a universal identity and ignore that for real gases the difference deviates from $R$, increasing with pressure and decreasing with temperature. Remember: a larger $C_p-C_v$ value indicates both higher pressure *and* lower temperature relative to a state with a smaller value.

The value of Cp−CvC_p - C_v is 1.00 R for a gas sample in state A and 1.08 R in state B. Let pAp_A and pBp_B denote the pressures and TAT_A and TBT_B denote the temperatures of the states A and B, respectively. It is most likely that

Multiple Choice Options

  • A pA<pBp_A < p_B and TA>TBT_A > T_B
  • B pA>pBp_A > p_B and TA<TBT_A < T_B
  • C pA=pBp_A = p_B and TA<TBT_A < T_B
  • D pA>pBp_A > p_B and TA=TBT_A = T_B

Step-by-Step Verified Solution

**Answer:** (a) p_A < p_B and T_A > T_B. **Step-by-Step Explanation:** C_p - C_v = R for ideal gas. The deviation from R increases with pressure and decreases with temperature. So state A has lower pressure and higher temperature.

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