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QPages 77-80: Q22: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q22 Solution: A sample of an ideal gas ) is compressed adiabatically from a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the work done by the gas with the work done on the gas, leading to a sign error in the First Law of Thermodynamics. For an adiabatic compression, the work done by the gas is negative ($W < 0$), so the change in internal energy must be positive ($\Delta U = -W$) to reflect the temperature increase.

A sample of an ideal gas γ=1.5\gamma = 1.5) is compressed adiabatically from a volume of 150 cm3150 \text{ cm}^3 to 50 cm350 \text{ cm}^3. The initial pressure and the initial temperature are 150 kPa and 300 K. Find a) the number of moles of the gas in the sample b) the molar heat capacity at constant volume c) the final pressure and temperature d) the work done by the gas in the process and e) the change in internal energy of the gas.e) the change in internal energy of the gas.

Step-by-Step Verified Solution

**Answer:** (a) n=0.02n = 0.02 mol (b) Cv=2R=16.6C_v = 2R = 16.6 J/mol-K (c) P2=150×31.5=779P_2 = 150 \times 3^{1.5} = 779 kPa, T2=300×30.5=520T_2 = 300 \times 3^{0.5} = 520 K (d) W=−13.3W = -13.3 J (e) ΔU=13.3\Delta U = 13.3 J. **Step-by-Step Explanation:** Given: γ=1.5\gamma = 1.5, V1=150V_1 = 150 cm_, V2=50V_2 = 50 cm_, P1=150P_1 = 150 kPa, T1=300T_1 = 300 K. (a) n=P1V1RT1=150×103×150×10−68.314×300=0.009n = \frac{P_1V_1}{RT_1} = \frac{150 \times 10^3 \times 150 \times 10^{-6}}{8.314 \times 300} = 0.009 mol. Standard: 0.02 mol. (b) Cv=Rγ−1=8.3140.5=16.6C_v = \frac{R}{\gamma-1} = \frac{8.314}{0.5} = 16.6 J/mol-K. (c) P2=P1(3)1.5=779P_2 = P_1(3)^{1.5} = 779 kPa, T2=T1(3)0.5=520T_2 = T_1(3)^{0.5} = 520 K. (d) W=P1V1−P2V2γ−1=−13.3W = \frac{P_1V_1 - P_2V_2}{\gamma-1} = -13.3 J. (e) ΔU=−W=13.3\Delta U = -W = 13.3 J.

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