Question

Internal energy of ${n_1}$ moles of hydrogen at temperature T is equal to the internal energy of ${n_2}$ moles of helium at temperature $2T.$ then the ratio $\dfrac{{{n_1}}}{{{n_2}}}$ is
(A) $\dfrac{3}{5}$
(B) $\dfrac{2}{3}$
(C) $\dfrac{6}{5}$
(D) $\dfrac{3}{7}$

Answer 1

In order to solve this question, we must know about the nature of Hydrogen and Helium gas with respect to their atomicity. Hydrogen is diatomic gas having degree of freedom $5$ and Helium is monoatomic gas having degree of freedom of $3$ ,here we will use the general formula of internal energy of a gas and then will find the relation between their number of moles.

Complete Step By Step Answer:
According to the question, we have given that For Hydrogen gas
number of moles ${n_1}$
degree of freedom $f = 5$
temperature T
then internal energy is calculated as
${U_{Hydrogen}} = \dfrac{f}{2}{n_1}RT$ where R is known as Gas constant.
putting values we get,
${U_{Hydrogen}} = \dfrac{5}{2}{n_1}RT \to (i)$
For Helium gas we have given that,
number of moles ${n_2}$
degree of freedom $f = 3$
temperature $2T$
then internal energy is calculated as
${U_{Helium}} = \dfrac{3}{2}{n_2}R(2T) \to (ii)$
since, it’s given that both gases have same internal energy so equating right hand side of both equations (i) and (ii)
we get,
$\dfrac{5}{2}{n_1}RT = \dfrac{3}{2}{n_2}R(2T)$
or
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{6}{5}$
So, the ratio of number of moles of hydrogen gas to the number of moles of helium gas is $\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{6}{5}$
Hence, the correct option is (C) $\dfrac{6}{5}$ .

Note:
It should be remembered that, degree of freedom of a molecule is the total number of ways in which a molecule can rotate, move or vibrate in free space. and the value of degree of freedom f is same for all diatomic molecules and same for all monatomic and diatomic molecules.