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QPages 77-80: Q34: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q34 Solution: An ideal gas of density at a pressure of is filled in a Kundt's …

An ideal gas of density 1.7×10−3 g cm−31.7 \times 10^{-3} \text{ g cm}^{-3} at a pressure of 1.5×105 Pa1.5 \times 10^5 \text{ Pa} is filled in a Kundt's tube. When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0 cm. Calculate the molar heat capacities CpC_p and CvC_v of the gas.

Step-by-Step Verified Solution

**Answer:** Cp=0.83C_p = 0.83 J/g-K, Cv=0.62C_v = 0.62 J/g-K. **Step-BY-Step Explanation:** Given: ρ=1.7×10−3\rho = 1.7 \times 10^{-3} g/cm_ = 1.7 kg/m_, P=1.5×105P = 1.5 \times 10^5 Pa, f=3.0f = 3.0 kHz, node separation = 6.0 cm. λ=2×6.0=12\lambda = 2 \times 6.0 = 12 cm = 0.12 m. v=fλ=360v = f\lambda = 360 m/s. v=γP/ρ⇒γ=v2ρP=3602×1.71.5×105=1.47v = \sqrt{\gamma P/\rho} \Rightarrow \gamma = \frac{v^2\rho}{P} = \frac{360^2 \times 1.7}{1.5 \times 10^5} = 1.47. M=ρRTP=1.7×8.314×2731.5×105=0.0257M = \frac{\rho RT}{P} = \frac{1.7 \times 8.314 \times 273}{1.5 \times 10^5} = 0.0257 kg/mol = 25.7 g/mol. Cv=RM(γ−1)=8.3140.0257×0.47=688C_v = \frac{R}{M(\gamma-1)} = \frac{8.314}{0.0257 \times 0.47} = 688 J/kg-K = 0.69 J/g-K.

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