A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas (𝛾 = 5/3) and one mole of an ideal diatomic gas (𝛾 = 7/5). Here, 𝛾 is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of 66 Joule when heated at constant pressure. The change in its internal energy is ________ Joule.

 To determine the change in internal energy of the gas mixture, we'll follow these steps:

1. Determine the total number of moles and the effective $\gamma$ for the mixture.

2. Use the relationship between work done at constant pressure and the change in internal energy.

3. Calculate the change in internal energy.

Step 1: Determine the Total Number of Moles and Effective $\gamma$

The mixture consists of:

• 2 moles of a monatomic gas with ${\gamma }_1=\frac{5}{3}$

• 1 mole of a diatomic gas with ${\gamma }_2=\frac{7}{5}$

Total number of moles ($n_{\text{total}}$):

$$n_{\text{total}}=2+1=3\text{ moles}$$

Effective $\gamma$ for the mixture:

For a mixture of ideal gases, the effective $\gamma$ can be approximated using the mole fractions and the individual $\gamma$ values. However, a more precise approach involves calculating the effective heat capacities.

The specific heat at constant volume ($C_V$) and constant pressure ($C_P$) for each gas are:

• Monatomic gas: $C_{V1}=\frac{3}{2}R$, $C_{P1}=\frac{5}{2}R$

• Diatomic gas: $C_{V2}=\frac{5}{2}R$, $C_{P2}=\frac{7}{2}R$

Total heat capacities for the mixture:

$$C_V=n_1{C}_{V1}+n_2{C}_{V2}=2\left(\frac{3}{2}R\right)+1\left(\frac{5}{2}R\right)=3R+\frac{5}{2}R=\frac{11}{2}R$$$$C_P=n_1{C}_{P1}+n_2{C}_{P2}=2\left(\frac{5}{2}R\right)+1\left(\frac{7}{2}R\right)=5R+\frac{7}{2}R=\frac{17}{2}R$$

Effective $\gamma$:

$${\gamma }_{\text{eff}}=\frac{C_P}{C_V}=\frac{\frac{17}{2}R}{\frac{11}{2}R}=\frac{17}{11}\approx 1.545$$

Step 2: Relationship Between Work and Internal Energy Change

For an ideal gas undergoing a process at constant pressure, the work done ($W$) is related to the change in internal energy ($\mathrm{\Delta }U$) by:

$$W=P\mathrm{\Delta }V=nR\mathrm{\Delta }T$$$$\mathrm{\Delta }U=n{C}_V\mathrm{\Delta }T$$

Given that $W=66\text{ J}$, we can express $\mathrm{\Delta }T$ in terms of $W$:

$$\mathrm{\Delta }T=\frac{W}{nR}$$

Substitute $\mathrm{\Delta }T$ into the expression for $\mathrm{\Delta }U$:

$$\mathrm{\Delta }U=n{C}_V\left(\frac{W}{nR}\right)=\frac{C_V}{R}W$$

Step 3: Calculate the Change in Internal Energy

Using the total $C_V$ for the mixture:

$$C_V=\frac{11}{2}R$$$$\mathrm{\Delta }U=\frac{\frac{11}{2}R}{R}\times 66=\frac{11}{2}\times 66=11\times 33=363\text{ J}$$

However, let's verify the calculation:

Given:

$$\mathrm{\Delta }U=\frac{C_V}{R}W=\frac{\frac{11}{2}R}{R}\times 66=\frac{11}{2}\times 66=363\text{ J}$$

But let's cross-check with the effective $\gamma$:

$$\mathrm{\Delta }U=\frac{W}{\gamma -1}=\frac{66}{1.545-1}=\frac{66}{0.545}\approx 121.1\text{ J}$$

There seems to be a discrepancy. Let's re-evaluate the effective $\gamma$ calculation.

Re-evaluating Effective $\gamma$:

Using the mole fractions:

$${\gamma }_{\text{eff}}=\frac{n_1{\gamma }_1+n_2{\gamma }_2}{n_1+n_2}=\frac{2\times \frac{5}{3}+1\times \frac{7}{5}}{3}=\frac{\frac{10}{3}+\frac{7}{5}}{3}=\frac{\frac{50+21}{15}}{3}=\frac{\frac{71}{15}}{3}=\frac{71}{45}\approx 1.578$$

Now, using this $\gamma$:

$$\mathrm{\Delta }U=\frac{W}{\gamma -1}=\frac{66}{1.578-1}=\frac{66}{0.578}\approx 114.2\text{ J}$$

This still doesn't match the initial calculation. Let's consider another approach.

Alternative Approach:

For a mixture, the change in internal energy can be directly calculated using the total $C_V$:

$$\mathrm{\Delta }U=n_{\text{total}}{C}_V\mathrm{\Delta }T$$

Given:

$$W=n_{\text{total}}R\mathrm{\Delta }T\Rightarrow \mathrm{\Delta }T=\frac{W}{n_{\text{total}}R}$$

Thus:

$$\mathrm{\Delta }U=n_{\text{total}}{C}_V\left(\frac{W}{n_{\text{total}}R}\right)=\frac{C_V}{R}W$$

Using $C_V=\frac{11}{2}R$:

$$\mathrm{\Delta }U=\frac{\frac{11}{2}R}{R}\times 66=\frac{11}{2}\times 66=363\text{ J}$$

This calculation is consistent. Therefore, the change in internal energy is:

$$\boxed{{\displaystyle 363}}$$