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Question

4.0 gm4.0 \text{ gm} of gas occupies 22.4 litres22.4 \text{ litres} at NTP. The specific heat capacity of the gas at a constant volume is 5.0 J K1mol15.0 \text{ J K}^{-1}\text{mol}^{-1}. If the speed of sound in the gas at NTP is 952 ms1952 \text{ ms}^{-1}, then the molar heat capacity at constant pressure will be: (R=8.31 J K1mol1R=8.31 \text{ J K}^{-1}\text{mol}^{-1})

A

8.0 J K1mol18.0 \text{ J K}^{-1}\text{mol}^{-1}

B

7.5 J K1mol17.5 \text{ J K}^{-1}\text{mol}^{-1}

C

7.0 J K1mol17.0 \text{ J K}^{-1}\text{mol}^{-1}

D

8.5 J K1mol18.5 \text{ J K}^{-1}\text{mol}^{-1}

Step-by-Step Solution

At NTP (Normal Temperature and Pressure), 22.4 L22.4 \text{ L} of an ideal gas corresponds to 1 mole1 \text{ mole}. Therefore, the molar mass of the gas is M=4.0 g/mol=4.0×103 kg/molM = 4.0 \text{ g/mol} = 4.0 \times 10^{-3} \text{ kg/mol}. The speed of sound in a gas is given by the formula: v=γRTMv = \sqrt{\frac{\gamma RT}{M}} Squaring both sides and solving for γ\gamma: γ=v2MRT\gamma = \frac{v^2 M}{RT} Given v=952 ms1v = 952 \text{ ms}^{-1}, R=8.31 J K1mol1R = 8.31 \text{ J K}^{-1}\text{mol}^{-1}, and T=273 KT = 273 \text{ K}: γ=(952)2×4.0×1038.31×273=906304×0.0042268.631.6\gamma = \frac{(952)^2 \times 4.0 \times 10^{-3}}{8.31 \times 273} = \frac{906304 \times 0.004}{2268.63} \approx 1.6 Alternatively, using density ρ=MassVolume=4.0×103 kg22.4×103 m3\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{4.0 \times 10^{-3} \text{ kg}}{22.4 \times 10^{-3} \text{ m}^3} and pressure P=1.013×105 PaP = 1.013 \times 10^5 \text{ Pa}: γ=v2ρP=(952)2×(4.022.4)1.013×1051.6\gamma = \frac{v^2 \rho}{P} = \frac{(952)^2 \times \left(\frac{4.0}{22.4}\right)}{1.013 \times 10^5} \approx 1.6 The ratio of molar heat capacities is γ=CpCv\gamma = \frac{C_p}{C_v}. Given the molar heat capacity at constant volume Cv=5.0 J K1mol1C_v = 5.0 \text{ J K}^{-1}\text{mol}^{-1}: Cp=γCv=1.6×5.0=8.0 J K1mol1C_p = \gamma C_v = 1.6 \times 5.0 = 8.0 \text{ J K}^{-1}\text{mol}^{-1}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVE OPTICSoccupieslitresspecificcapacityconstant

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