91Ó°ÊÓ

Mole,

$$\begin{gathered} n_1=2.40\text{″¾´Ç±ô} \\ {n}_2=1.50\text{″¾´Ç±ô} \\ {n}_3=3.20\text{mol} \end{gathered}$$

Specific heat capacity at constant volume;

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

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

$Q=n{C}_V\mathrm{Δ}T$

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$\mathrm{Δ}T$ is the temperature difference.

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

$W=p\mathrm{Δ}V$

Here W is the work done by the gas, P is the pressure and$\mathrm{Δ}V$ is the change in the volume.

According to the first law of thermodynamics

$\mathrm{Δ}{E}_{int}=Q-W$…… (i)

But, work done by the gas is

$W=P\mathrm{Δ}V=0$……. (ii)

From equation (i) and equation (ii)

$∴\mathrm{Δ}{E}_{int}=Q$

We know that,

$Q=n{C}_V\mathrm{Δ}T$

$∴\mathrm{Δ}{E}_{int}=n{C}_V\mathrm{Δ}T$……. (iii)

Change in the internal energy of the mixture is

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

Substitute$112\mathrm{Δ}T$ for$\mathrm{Δ}{E}_{int}$ into the equation (iii)

$∴C_V=\frac{112}{n}$

Substitute the value of total number of moles

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

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