A container holds a mixture of three non reacting gases:$\mathbf{2}\mathbf{.}\mathbf{40}\mathit{m}\mathit{o}\mathbf{\text{l}}$of gas 1 with${\mathit{C}}_{\mathbf{V}\mathbf{1}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathit{J}\mathbf{/}\mathbf{(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{.}\mathbf{K}\mathbf{)}\mathbf{}$,$\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathit{m}\mathit{o}\mathbf{\text{l}}$of gas 2 with${\mathit{C}}_{\mathbf{V}\mathbf{2}}\mathbf{=}\mathit{J}\mathbf{/}\mathbf{(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{.}\mathbf{K}\mathbf{)}$andof gas 3 with${\mathit{C}}_{\mathbf{V}\mathbf{3}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathit{J}\mathbf{/}\mathbf{(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{.}\mathbf{K}\mathbf{)}$. What is${\mathit{C}}_{\mathbf{V}}$of the mixture?

Mole,

$$\begin{gathered} n_1=2.40\text{\hspace{0.17em}mol} \\ {n}_2=1.50\text{\hspace{0.17em}mol} \\ {n}_3=3.20\text{mol} \end{gathered}$$

Specific heat capacity at constant volume;

$$\begin{gathered} C_{V1}=12.0\text{\hspace{0.17em}J}/\text{mol.K} \\ {C}_{V2}=12.8\text{\hspace{0.17em}J}/\text{mol.K} \\ {C}_{V3}=20.0\text{\hspace{0.17em}J}/\text{mol.K} \end{gathered}$$

The expression for the amount of energy transferred to the gas is given by,

$Q=n{C}_V\Delta T$

Here Q is the amount of energy transferred to gas, n is the number of moles, $C_v$is the specific heat capacity at constant volume and$\Delta T$ is the temperature difference.

The expression for the work done by the gas is given by,

$W=p\Delta V$

Here W is the work done by the gas, P is the pressure and$\Delta V$ is the change in the volume.

According to the first law of thermodynamics

$\Delta E_{int}=Q-W$…… (i)

But, work done by the gas is

$W=P\Delta V=0$……. (ii)

From equation (i) and equation (ii)

$\therefore \Delta E_{int}=Q$

We know that,

$Q=n{C}_V\Delta T$

$\therefore \Delta E_{int}=n{C}_V\Delta T$……. (iii)

Change in the internal energy of the mixture is

$\begin{array}{c}\Delta E_{int}=(n_1{C}_{V1}+n_2{C}_{V2}+n_3{C}_{V3})\Delta T\\ =((2.40\times 12.0)+(1.50\times 12.8)+(3.20\times 20.0))\Delta T\\ =112\Delta T\end{array}$

Substitute$112\Delta T$ for$\Delta E_{int}$ into the equation (iii)

$\therefore C_V=\frac{112}{n}$

Substitute the value of total number of moles

$\begin{array}{c}{C}_V=\frac{112}{2.40+1.50+3.20}\\ =15.77\\ \approx 15.8\frac{\text{J}}{\text{mol.K}}\end{array}$

Therefore, the value of of $C_V$the mixture is $15.8\frac{\text{J}}{\text{mol.K}}$.