Question:
A monoatomic gas of mass $4.0 u$ is kept in an insulated container. Container is moving with velocity $30 \mathrm{~m} / \mathrm{s}$. If container is suddenly stopped then change in temperature of the gas ( $\mathrm{R}=$ gas constant) is $\frac{x}{3 R}$. Value of $x$ is
Solution:
$(3600)$
$\Delta \mathrm{K}_{\mathrm{E}}=\Delta \mathrm{U}$
$\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}$
$\frac{1}{2} \mathrm{mv}^{2}=\frac{3}{2} \mathrm{n} \mathrm{R} \Delta \mathrm{T}$
$\frac{\mathrm{mv}^{2}}{3 \mathrm{nR}}=\Delta \mathrm{T}$
$\frac{4 \times(30)^{2}}{3 \times 1 \times \mathrm{R}}=\Delta \mathrm{T}$
$\Delta \mathrm{T}=\frac{1200}{\mathrm{R}}$
$\frac{\mathrm{x}}{3 \mathrm{R}}=\frac{1200}{\mathrm{R}}$
$\mathrm{x}=3600$