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Q.

A light container having a monoatomic ideal gas enclosed within is moving with speed v1. Its speed is suddenly reduced to v2. Find the resulting change in temperature. (Mass of each molecule of gas = m, k = Boltzmann constant).

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answer is 3.

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Detailed Solution

Loss in kinetic energy of container is converted to random motion of molecules. 12mv12−12mv22=32kΔT (Since monoatomic gas has three degrees of freedom).⇒ΔT=mv12−v223k

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