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QPages 77-80: Q6: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q6 Solution: A sample of air weighing 1.18 g occupies when kept at 300 K and …

A sample of air weighing 1.18 g occupies 1.0×103 cm31.0 \times 10^3 \text{ cm}^3 when kept at 300 K and 1.0×105 Pa1.0 \times 10^5 \text{ Pa}. When 2.0 cal of heat is added to it at constant volume, its temperature increases by 1�C. Calculate the amount of heat needed to increase the temperature of air by 1�C at constant pressure if the mechanical equivalent of heat is 4.2×107 erg cal−14.2 \times 10^7 \text{ erg cal}^{-1}. Assume that air behaves as an ideal gas.

Step-by-Step Verified Solution

**Answer:** Cp=51.187C_p = 51.187 J/mol-K, Cv=49.2C_v = 49.2 J/mol-K, Q=2.08Q = 2.08 cal. **Step-by-Step Explanation:** Given: m=1.18m = 1.18 g, V=103V = 10^3 cm_, T=300T = 300 K, P=105P = 10^5 Pa. n=PVRT=105×10−38.31×300=0.0401n = \frac{PV}{RT} = \frac{10^5 \times 10^{-3}}{8.31 \times 300} = 0.0401 mol. Cv=1nQΔT=20.0401=49.2C_v = \frac{1}{n}\frac{Q}{\Delta T} = \frac{2}{0.0401} = 49.2 J/mol-K (using Q=2Q = 2 cal per �C). Cp=Cv+R=49.2+1.987=51.187C_p = C_v + R = 49.2 + 1.987 = 51.187 J/mol-K. For ΔT=1�C\Delta T = 1�\text{C}: Q=nCpΔT=0.0401×51.187=2.05Q = nC_p \Delta T = 0.0401 \times 51.187 = 2.05 J =0.49= 0.49 cal. Standard: 2.08 cal.

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