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QPages 77-80: Q25: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q25 Solution: 1 litre of an ideal gas ) at 300 K is suddenly compressed to…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the adiabatic work formula as $W=\frac{P_2V_2-P_1V_1}{\gamma-1}$ without checking the sign convention, treating $W$ as work done by the gas. In a compression the gas does negative work, so the correct expression for work *by* the gas is $W=\frac{P_1V_1-P_2V_2}{\gamma-1}$; using the opposite sign flips the sign of $\Delta U$ and all subsequent results.

1 litre of an ideal gas γ=1.5\gamma = 1.5) at 300 K is suddenly compressed to half its original volume. a) Find the ratio of the final pressure to the initial pressure. b) If the original pressure is 100 kPa, find the work done by the gas in the process. c) What is the change in internal energy? d) What is the final temperature? e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. f) The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas. g) Calculate the total work done in the cycle.

Step-by-Step Verified Solution

**Answer:** (a) P2/P1=21.5=2.83P_2/P_1 = 2^{1.5} = 2.83 (b) W=−82W = -82 J (c) ΔU=82\Delta U = 82 J (d) T2=424T_2 = 424 K (e) W=−41.4W = -41.4 J (f) W=103W = 103 J (g) Wnet=−20.4W_{net} = -20.4 J. **Step-by-Step Explanation:** Given: γ=1.5\gamma = 1.5, T1=300T_1 = 300 K, V1=1V_1 = 1 L, V2=0.5V_2 = 0.5 L, P1=100P_1 = 100 kPa. (a) P2=P1(2)1.5=283P_2 = P_1(2)^{1.5} = 283 kPa. Ratio = 2.83. (b) W=P1V1−P2V2γ−1=−82W = \frac{P_1V_1 - P_2V_2}{\gamma-1} = -82 J. (c) ΔU=−W=82\Delta U = -W = 82 J. (d) T2=T1(2)0.5=424T_2 = T_1(2)^{0.5} = 424 K. (e) Isobaric cooling: W=PΔV=−41.4W = P\Delta V = -41.4 J. (f) Isothermal expansion: W=nRTln⁡2=103W = nRT\ln 2 = 103 J. (g) Wnet=−82−41.4+103=−20.4W_{net} = -82 - 41.4 + 103 = -20.4 J [citation:5].

All Questions in this Chapter

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