First slide

Law of equipartition of energy

Question

An insulated container containing monoatomic gas of molar mass m is moving with a velocity $v_{o}$ . If the container is suddenly stopped, find the change in temperature.

Moderate

A

$\frac{{mv}_{0}^{2}}{2 R}$
B

$\frac{{mv}_{0}^{2}}{3 R}$
C

$\frac{3 {mv}_{0}^{2}}{2 R}$
D

$\frac{5 {mv}_{0}^{2}}{3 R}$

Solution

When, the container stops, its total kinetic energy is transferred to gas molecules in the form of translational kinetic energy, thereby increasing the
absolute temperature.
Assuming n = number of moles
Given, m = molar mass of the gas.
$If ∆ T = change in absolute temperature$
Then, kinetic energy of molecules due to velocity $v_{o}$ ,
$∆ K_{motion} = \frac{1}{2} (mn) {v_{0}}^{2}$ -----------------(i)
Increase in translational kinetic energy
$∆ K_{translation} = n \frac{3}{2} R (∆ T)$ ------------(ii)
According to kinetic theory Eqs. (i) and (ii) are equal
$⇒ \frac{1}{2} (mn) {v_{0}}^{2} = n \frac{3}{2} R (∆ T)$
$(mn) {v_{0}}^{2} = n 3 R (∆ T)$
$\Rightarrow ∆ T = \frac{(mn) {v_{0}}^{2}}{3 nR} = \frac{{mv}_{0}^{2}}{3 R}$

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