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QPage 77: Q5: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Page 77: Q5 Solution: Three identical adiabatic containers A, B and C contain helium, …
⚠️ Common Exam Traps & Conceptual Pitfalls
Students frequently treat the adiabatic exponent $\gamma$ as a constant for all gases, so they apply $P_f=P_i(V_i/V_f)^\gamma$ and $T_f=T_i(V_i/V_f)^{\gamma-1}$ with the same $\gamma$ for He, Ne and O$_2$. This leads to the mistaken belief that final pressures or temperatures must be identical. Remember that $\gamma=C_p/C_v$ differs for monatomic (He, Ne) and diatomic (O$_2$) gases, and must be used separately for each gas in adiabatic calculations.
Three identical adiabatic containers A, B and C contain helium, neon and oxygen, respectively, at equal pressure. The gases are pushed to half their original volumes.
Multiple Choice Options
- A The final temperatures in the three containers will be the same.
- B The final pressures in the three containers will be the same.
- C The pressures of helium and neon will be the same but that of oxygen will be different.
- D The temperatures of helium and neon will be the same but that of oxygen will be different.
Step-by-Step Verified Solution
**Answer:** (b), (d).
**Step-by-Step Explanation:**
For adiabatic compression, P_f = P_i(V_i/V_f)^_. Since _ is different for different gases, final pressures differ. Also, final temperatures differ because _ differs.
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