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

Hint: 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.