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QPages 77-80: Q13: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q13 Solution: Half mole of an ideal gas ) is taken through the cycle abcda,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently apply $C_p$ to all processes or fail to identify that $bc$ and $da$ are isochoric (constant volume) processes, requiring $C_v$ instead. Always verify the constraint (constant $V$ or $P$) before selecting the molar heat capacity, as $Q = nC\Delta T$ is only valid for these specific paths.

Half mole of an ideal gas γ=5/3\gamma = 5/3) is taken through the cycle abcda, as shown in the figure. Take R=253 J K−1 mol−1R = \frac{25}{3} \text{ J K}^{-1} \text{ mol}^{-1}. a) Find the temperature of the gas in the states a, b, c and d. b) Find the amount of heat supplied in the processes ab and bc. c) Find the amount of heat liberated in the processes cd and da.

Step-by-Step Verified Solution

**Answer:** (a) Ta=120T_a = 120 K, Tb=240T_b = 240 K, Tc=480T_c = 480 K, Td=240T_d = 240 K (b) Qab=1250Q_{ab} = 1250 J, Qbc=1500Q_{bc} = 1500 J (c) Qcd=−2500Q_{cd} = -2500 J, Qda=−750Q_{da} = -750 J. **Step-BY-Step Explanation:** Given: n=0.5n = 0.5 mol, R=25/3R = 25/3 J/mol-K, γ=5/3\gamma = 5/3. (a) Using PV=nRTPV = nRT: - Ta=PaVanR=100×103×5000×10−60.5×25/3=120T_a = \frac{P_aV_a}{nR} = \frac{100 \times 10^3 \times 5000 \times 10^{-6}}{0.5 \times 25/3} = 120 K - Tb=240T_b = 240 K, Tc=480T_c = 480 K, Td=240T_d = 240 K (b) Qab=nCpΔT=0.5×γRγ−1×120=1250Q_{ab} = nC_p\Delta T = 0.5 \times \frac{\gamma R}{\gamma-1} \times 120 = 1250 J. Qbc=nCvΔT=0.5×Rγ−1×240=1500Q_{bc} = nC_v\Delta T = 0.5 \times \frac{R}{\gamma-1} \times 240 = 1500 J. (c) Qcd=−nCpΔT=−2500Q_{cd} = -nC_p\Delta T = -2500 J. Qda=−nCvΔT=−750Q_{da} = -nC_v\Delta T = -750 J [citation:12].

All Questions in this Chapter

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