Question

Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? (R = 8.3 J/mol K)
(A) 17.4 J/mol K (B) 15.7 J/mol K
(C) 19.7 J/mol K (D) 21.6 J/mol K

Answer 1

fmix=n1+n2n1f1+n2f2=52×3+3×5=521

Cv=2fR=521×2R=17.4 J/mol K.

Answer 2

The molar specific heat of mixture at constant volume $= 17.4 \frac{J}{mol K}$.

Option (A) is correct.

Explanation:

General formula for specific heat capacity at constant volume $= \frac{fR}{2}$

Where $f =$ degree of freedom, for mono atomic $f = 3$, for diatomic $f = 5$

Since helium gas is mono atomic and hydrogen molecule is diatomic.

Given :

Specific heat at constant volume for helium gas $C_{v1} = \frac{3R}{2}$

Specific heat at constant volume for hydrogen molecule $C_{v2} = \frac{5R}{2}$

No. of moles for helium gas $n_{1} = 2$

No. of moles for hydrogen gas $n_{2} = 3$

Gas constant $R = 8.3$

For mixture of specific heat  $C_{Vmix} = \frac{n_{1}C_{v1}+n_{2}C_{v2} }{n_{1}+n_{2} }$

   $C_{V mix} =\frac{3R + 7.5 R}{5}$

    $C_{Vmix} = 17.4 \frac{J}{mol K}$

Thus, the molar specific heat of mixture at constant volume $= 17.4 \frac{J}{mol K}$