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QPages 77-80: Q1: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q1 Solution: A vessel containing one mole of a monatomic ideal gas molecular …

A vessel containing one mole of a monatomic ideal gas molecular weight = 20 g mol−120 \text{ g mol}^{-1}) is moving on a floor at a speed of 50 m s−150 \text{ m s}^{-1}. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.

Step-by-Step Verified Solution

**Answer:** Rise in temperature = 2 K. **Step-by-Step Explanation:** Given: n=1n = 1 mole, M=20M = 20 g/mol = 0.02 kg/mol, v=50v = 50 m/s. Kinetic energy of vessel = internal energy of gas:
12mv2=12×0.02×502=25 J\frac{1}{2}mv^2 = \frac{1}{2} \times 0.02 \times 50^2 = 25 \text{ J}
For monatomic gas, U=32nRTU = \frac{3}{2}nRT:
25=32×1×8.31×T⇒T=5024.93≈2 K25 = \frac{3}{2} \times 1 \times 8.31 \times T \Rightarrow T = \frac{50}{24.93} \approx 2 \text{ K}

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