Question

Two moles of a monoatomic gas is mixed with three moles of a diatomic gas. The molar specific heat of the mixture at a constant volume is

• A. $1.6R$
• B. $2.1R$
• C. $6.25R$
• D. $3.6R$

Answer 1

The correct option is B $2.1R$

Given, no. of moles of monoatomic gas $n_1=2$
& For a monoatomic gas, $C_{V_1}=\frac{3}{2}R$
No. of moles of diatomic gas, $n_2=3$
& For a diatomic gas, $C_{V_2}=\frac{5}{2}R$
We know that,
Molar specific heat of the mixture
$C_V^{\prime }=\frac{n_1\times C_{V_1}+n_2\times C_{V_2}}{n_1+n_2}$
From the data given in the question,
$C_V^{\prime }=\frac{2\times \frac{3R}{2}+3\times \frac{5R}{2}}{2+3}=\frac{21R}{10}=2.1R$
Thus, option (b) is the correct answer.