JEE Physics 2024 Question with Solution

Given:

• Moles of monoatomic gas: $n_1 = 3$, $\gamma_1 = \frac{5}{3}$
• Moles of diatomic gas: $n_2 = 2$, $\gamma_2 = \frac{7}{5}$

Find: The adiabatic exponent $\gamma_{\text{mix}}$ for the mixture.

For a mixture of ideal gases,

$$\gamma_{\text{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n_1 C_{v,1} + n_2 C_{v,2}}$$

and

$$C_v = \frac{R}{\gamma - 1}, \qquad C_p = \frac{\gamma R}{\gamma - 1}$$

For the monoatomic gas,

$$C_{v,1} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3}{2}R$$ $$C_{p,1} = \frac{\frac{5}{3}R}{\frac{2}{3}} = \frac{5}{2}R$$

For the diatomic gas,

$$C_{v,2} = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5}{2}R$$ $$C_{p,2} = \frac{\frac{7}{5}R}{\frac{2}{5}} = \frac{7}{2}R$$

Now compute the total heat capacities:

$$3 \left(\frac{5}{2}R\right) + 2 \left(\frac{7}{2}R\right) = \frac{15}{2}R + \frac{14}{2}R = \frac{29}{2}R$$ $$3 \left(\frac{3}{2}R\right) + 2 \left(\frac{5}{2}R\right) = \frac{9}{2}R + \frac{10}{2}R = \frac{19}{2}R$$

Therefore,

$$\gamma_{\text{mix}} = \frac{\frac{29}{2}R}{\frac{19}{2}R} = \frac{29}{19} \approx 1.52$$

So, the adiabatic exponent of the mixture is $1.52$. The correct option is C.

Given: The two gases are mixed in amounts $3$ moles and $2$ moles with adiabatic exponents $\frac{5}{3}$ and $\frac{7}{5}$ respectively.

Find: The value of $\gamma$ for the mixture.

The first approach shown in the source tries a direct weighted average of $\gamma$, but the correct method is to average the heat capacities first and then take their ratio. That is why the final result becomes $\frac{29}{19}$ and not the intermediate incorrect arithmetic shown earlier.

Use

$$\gamma_{\text{mix}} = \frac{C_{p,\text{mix}}}{C_{v,\text{mix}}}$$

with

$$C_{v,\text{mix}} = \frac{n_1 C_{v,1} + n_2 C_{v,2}}{n_1+n_2}, \qquad C_{p,\text{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n_1+n_2}$$

So,

$$\gamma_{\text{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n_1 C_{v,1} + n_2 C_{v,2}}$$

Now calculate each gas heat capacity from its individual $\gamma$ value. For gas $1$:

$$C_{v,1} = \frac{R}{\gamma_1-1} = \frac{R}{\frac{5}{3}-1} = \frac{3}{2}R, \qquad C_{p,1} = \frac{5}{2}R$$

For gas $2$:

$$C_{v,2} = \frac{R}{\gamma_2-1} = \frac{R}{\frac{7}{5}-1} = \frac{5}{2}R, \qquad C_{p,2} = \frac{7}{2}R$$

Substitute into the mixture formula:

$$\gamma_{\text{mix}} = \frac{3 \cdot \frac{5}{2}R + 2 \cdot \frac{7}{2}R}{3 \cdot \frac{3}{2}R + 2 \cdot \frac{5}{2}R}$$ $$= \frac{\frac{15}{2}R + \frac{14}{2}R}{\frac{9}{2}R + \frac{10}{2}R} = \frac{\frac{29}{2}R}{\frac{19}{2}R} = \frac{29}{19} = 1.52$$

Hence, the correct option is C.