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QPages 77-80: Q5: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q5 Solution: The ratio of the molar heat capacities of an ideal gas is .…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently assume an adiabatic process always involves the gas doing work (expansion), so they set $\Delta U=-W$ with $W>0$, giving a negative $\Delta U$ even when the temperature is raised. Remember that raising the temperature adiabatically requires work *on* the gas, making $W$ negative and $\Delta U=nC_v\Delta T$ positive.

The ratio of the molar heat capacities of an ideal gas is Cp/Cv=7/6C_p/C_v = 7/6. Calculate the change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K a) keeping the pressure constant b) keeping the volume constant and c) adiabatically.

Step-by-Step Verified Solution

**Answer:** (a) ΔU=2490\Delta U = 2490 J (b) ΔU=2490\Delta U = 2490 J (c) ΔU=−2490\Delta U = -2490 J. **Step-by-Step Explanation:** Given: γ=7/6\gamma = 7/6, n=1n = 1 mole, ΔT=50\Delta T = 50 K. For ideal gas: Cv=Rγ−1=R1/6=6RC_v = \frac{R}{\gamma-1} = \frac{R}{1/6} = 6R. (a) At constant pressure: ΔU=nCvΔT=6×8.3×50=2490\Delta U = nC_v\Delta T = 6 \times 8.3 \times 50 = 2490 J. (b) At constant volume: ΔU=nCvΔT=2490\Delta U = nC_v\Delta T = 2490 J. (c) Adiabatic: Q=0Q = 0, so ΔU=−W=−2490\Delta U = -W = -2490 J.

All Questions in this Chapter

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