Question

The internal energy of $n_{1}$ mol of hydrogen of temperature $T$ is equal to the internal energy of $n_{2}$ mol of helium at temperature $2T$ . What will be the value of the ratio $\frac{n_{1}}{n_{2}}$ ?

Answer 1

The internal energy of $n$ moles of an ideal gas at temperature T is given by
$\textit{U}=\frac{f }{2}\textit{nRT}$ ( $f =$ degrees of freedom)
$U_{1}=U_{2}$
$f _{1}n_{1}T_{1}=f⁡_{2}n_{2}T_{2}$
$\therefore \frac{n_{1}}{n_{2}}=\frac{f _{2} T_{2}}{f ⁡_{1} T_{1}}=\frac{3 \times 2}{5 \times 1}=\frac{6}{5}$ $=1.2$
Here, $f _{2}$ = degrees of freedom of He = 3
and, $f _{1}$ = degrees of freedom of H₂ = 5