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QPages 77-80: Q7: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q7 Solution: An ideal gas expands from to at a constant pressure of when 50…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently insert the given ΔU into $U=\frac{f}{2}nRT$ as if it were the change in internal energy, forgetting that the formula gives the total internal energy at a temperature. The correct relation is $\Delta U=\frac{f}{2}nR\Delta T$, and one must first determine $\Delta T$ (or know the gas’s degrees of freedom) before using it; also, never assume the gas is monatomic unless the problem states it.

An ideal gas expands from 100 cm3100 \text{ cm}^3 to 200 cm3200 \text{ cm}^3 at a constant pressure of 2.0×105 Pa2.0 \times 10^5 \text{ Pa} when 50 J of heat is supplied to it. Calculate a) the change in internal energy of the gas b) the number of moles in the gas if the initial temperature is 300 K c) the molar heat capacity CpC_p at constant pressure and d) the molar heat capacity CvC_v at constant volume.

Step-by-Step Verified Solution

**Answer:** (a) ΔU=30\Delta U = 30 J (b) n=0.008n = 0.008 mol (c) Cp=20.8C_p = 20.8 J/mol-K (d) Cv=12.5C_v = 12.5 J/mol-K. **Step-by-Step Explanation:** Given: V1=100V_1 = 100 cm_, V2=200V_2 = 200 cm_, P=2×105P = 2 \times 10^5 Pa, Q=50Q = 50 J. (a) W=PΔV=2×105×100×10−6=20W = P\Delta V = 2 \times 10^5 \times 100 \times 10^{-6} = 20 J. ΔU=Q−W=50−20=30\Delta U = Q - W = 50 - 20 = 30 J. (b) For monatomic gas, U=32nRTU = \frac{3}{2}nRT: 30=32n×8.3×300⇒n=0.00830 = \frac{3}{2}n \times 8.3 \times 300 \Rightarrow n = 0.008 mol. (c) Cv=ΔUnΔT=300.008×300=12.5C_v = \frac{\Delta U}{n\Delta T} = \frac{30}{0.008 \times 300} = 12.5 J/mol-K. Cp=Cv+R=12.5+8.3=20.8C_p = C_v + R = 12.5 + 8.3 = 20.8 J/mol-K [citation:12].

All Questions in this Chapter

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