First slide

Law of equipartition of energy

Question

Internal energy of $n_{1}$ moles of hydrogen of temperature T is equal to the internal energy of $n_{2}$ mole of helium at temperature 2T.Then the ratio $\frac{n_{1}}{n_{2}}$ is

Moderate

A

$\frac{3}{5}$
B

$\frac{2}{3}$
C

$\frac{6}{5}$
D

$\frac{3}{7}$

Solution

Internal energy of a n moles of an ideal gas at temperature T is given by
$U = \frac{f}{2} nRT [f = degress of freedom]$
$U_{1} = U_{2}$
$f_{1} n_{1} T_{1} = f_{2} n_{2} T_{2}$
$∴ \frac{n_{1}}{n_{2}} = \frac{f_{2} T_{2}}{f_{1} T_{1}} = \frac{3 \times 2}{5 \times 1} = \frac{6}{5}$
Here, $f_{2} = degrees of freedom of He = 3$ and $f_{1} = degrees of freedome of H_{2} = 5$

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