Verified Solution • 2025–26 Session moderate

QPages 77-80: Q21: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q21 Solution: Consider a given sample of an ideal gas ) having initial…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the original volume $V_0$ into the adiabatic formula $pV^gamma= ext{const}$ even though the adiabatic step starts from a different state. For an adiabatic change from $p_i$ to $p_f$, use $V_f=V_i,(p_i/p_f)^{1/gamma}$ where $V_i$ is the volume *at the start of that adiabatic step*. Mixing up $V_i$ with $V_0$ or using the wrong exponent ($gamma$ instead of $1/gamma$) leads to an incorrect final volume. Always track the state after each process and apply the appropriate relation to that state.

Consider a given sample of an ideal gas Cp/Cv=γC_p/C_v = \gamma) having initial pressure p0p_0 and volume V0V_0. a) The gas is isothermally taken to a pressure p0/2p_0/2 and from there, adiabatically to a pressure p0/4p_0/4. Find the final volume. b) The gas is brought back to its initial state. It is adiabatically taken to a pressure p0/2p_0/2 and from there, isothermally to a pressure p0/4p_0/4. Find the final volume.

Step-by-Step Verified Solution

**Answer:** Final volume = 2(1+γ)/γV02^{(1+\gamma)/\gamma}V_0 in both cases. **Step-by-Step Explanation:** Given: initial P0P_0, V0V_0. (a) Isothermal to P0/2P_0/2: V=2V0V = 2V_0. Adiabatic to P0/4P_0/4: V=2V0×21/γ=2(1+γ)/γV0V = 2V_0 \times 2^{1/\gamma} = 2^{(1+\gamma)/\gamma}V_0. (b) Adiabatic to P0/2P_0/2: V=21/γV0V = 2^{1/\gamma}V_0. Isothermal to P0/4P_0/4: V=2×21/γV0=2(1+γ)/γV0V = 2 \times 2^{1/\gamma}V_0 = 2^{(1+\gamma)/\gamma}V_0 [citation:16].

All Questions in this Chapter

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