First slide

Kinetic theory of ideal gases

Question

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).

Moderate

Solution

Loss in kinetic energy of container is converted to random motion of molecules.
$\frac{1}{2} m v_{1}^{2} - \frac{1}{2} m v_{2}^{2} = \frac{3}{2} k (Δ T)$
(Since monoatomic gas has three degrees of freedom).
$\Rightarrow Δ T = \frac{m (v_{1}^{2} - v_{2}^{2})}{3 k}$

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