Question

Two moles of a monoatomic gas are mixed with three moles of a diatomic gas. What will be the molar specific heat of the mixture at a constant volume?

• A. $1.6\mathrm{R}$
• B. $2.1\mathrm{R}$
• C. $3.6\mathrm{R}$
• D. $6.25\mathrm{R}$

Answer 1

The correct option is B

$2.1\mathrm{R}$

Explanation for correct option

(B) $\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{R}$

Molar specific heat

• The amount of heat required to increase the temperature of $1\mathrm{mol}$ of a gas through $1\mathrm{K}$ at constant volume is referred to as molar specific heat at constant volume.
• Formula is as follows:

$C_{V,m}=\frac{n_1{C}_{V_1}+n_2{C}_{V_2}}{n_1+n_2}$

• $C_{V,m}$ is the molar specific heat of mixture.
• $n_1\mathrm{and}{n}_2$ are the number of moles of mono and diatomic gases.
• $C_{v_1}\mathrm{and}{C}_{v_2}$ are the specific heat of mono and diatomic gases.

Calculation for the molar specific heat of the mixture

• For monoatomic gas, $C_{V_1}=\frac{3}{2}\mathrm{R}$
• For diatomic gas, $C_{V_2}=\frac{5}{2}\mathrm{R}$
• The molar specific heat for the given mixture of gases is as follows:

$\begin{array}{rcl}{C}_{V,m}& =& \frac{n_1{C}_{V_1}+n_2{C}_{V_2}}{n_1+n_2}\\ & =& \frac{2\left({\displaystyle \frac{3}{2}}\mathrm{R}\right)+3\left({\displaystyle \frac{5}{2}}\mathrm{R}\right)}{2+3}\\ & =& \frac{21\mathrm{R}}{10}\\ & =& 2.1\mathrm{R}\end{array}$

Explanation for incorrect options

• The molar specific heat of the mixture at pressure is $2.1\mathrm{R}$.
• Hence all other options (A), (C) and (D) are incorrect.

Therefore, the correct option is (B) $\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{R}$.