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 ? $(R=8.3\frac{\text{J}}{\text{mol K}})$

• A. $21.6\frac{\text{J}}{\text{mol K}}$
  
• B. $15.7\frac{\text{J}}{\text{mol K}}$
  
• C. $17.4\frac{\text{J}}{\text{mol K}}$
• D. $19.7\frac{\text{J}}{\text{mol K}}$

Answer 1

The correct option is C $17.4\frac{\text{J}}{\text{mol K}}$

The molar specific heat of the mixture at constant volume is given by,

$C_{V_{mix}}=\frac{n_1\left[C_{V_1}\right]+n_2\left[C_{V_2}\right]}{n_1+n_2}$

For helium gas (monatomic)

$C_{V_1}=\frac{3}{2}R$

For hydrogen molecule (diatomic)

$C_{V_2}=\frac{5}{2}R$

$\Rightarrow C_{V_{mix}}=\left[\frac{(2\times \frac{3R}{2})+(3\times \frac{5R}{2})}{2+3}\right]$

$\Rightarrow C_{V_{mix}}=2.1R$

$\Rightarrow C_{V_{mix}}=2.1\times 8.3$

$\Rightarrow C_{V_{mix}}=17.4\frac{\text{J}}{\text{mol K}}$

Hence, option $\left(C\right)$ is correct.