Three moles of an ideal gas having `gamma = 1.67` are mixed with 2 moles of another ideal gas having `gamma = 1.4`. Find the equivalent value of `gamma` for the mixture.

Explanation

To find the equivalent value of \( \gamma \) for the mixture of two ideal gases, we can follow these steps:

### Step 1: Understand the given data
We have:
- Gas 1: \( n_1 = 3 \) moles, \( \gamma_1 = 1.67 \)
- Gas 2: \( n_2 = 2 \) moles, \( \gamma_2 = 1.4 \)

### Step 2: Use the formula for \( C_v \) of the mixture
The specific heat at constant volume \( C_v \) for the mixture can be calculated using the formula:
\[ C_{v, \text{mixture}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]

### Step 3: Relate \( C_p \) and \( C_v \) to \( \gamma \)
We know that:
\[ \gamma = \frac{C_p}{C_v} \]
From this, we can express \( C_p \) in terms of \( C_v \):
\[ C_p = \gamma C_v \]

### Step 4: Calculate \( C_{v1} \) and \( C_{v2} \)
Using the relationship:
\[ C_{p1} = \gamma_1 C_{v1} \quad \text{and} \quad C_{p2} = \gamma_2 C_{v2} \]
We can express \( C_{v1} \) and \( C_{v2} \) as:
\[ C_{v1} = \frac{R}{\gamma_1 - 1} = \frac{R}{1.67 - 1} = \frac{R}{0.67} \]
\[ C_{v2} = \frac{R}{\gamma_2 - 1} = \frac{R}{1.4 - 1} = \frac{R}{0.4} \]

### Step 5: Substitute \( C_{v1} \) and \( C_{v2} \) into the mixture formula
Now substituting these values into the formula for \( C_{v, \text{mixture}} \):
\[ C_{v, \text{mixture}} = \frac{3 \cdot \frac{R}{0.67} + 2 \cdot \frac{R}{0.4}}{3 + 2} \]
\[ = \frac{3R \cdot \frac{1}{0.67} + 2R \cdot \frac{1}{0.4}}{5} \]

### Step 6: Simplify the expression
Calculating the terms:
\[ = \frac{3R \cdot 1.4925 + 2R \cdot 2.5}{5} \quad \text{(using } \frac{1}{0.67} \approx 1.4925 \text{ and } \frac{1}{0.4} = 2.5\text{)} \]
\[ = \frac{4.4775R + 5R}{5} = \frac{9.4775R}{5} = 1.8955R \]

### Step 7: Calculate \( C_{p, \text{mixture}} \)
Now, using \( C_{p, \text{mixture}} = C_{v, \text{mixture}} + R \):
\[ C_{p, \text{mixture}} = 1.8955R + R = 2.8955R \]

### Step 8: Calculate \( \gamma \) for the mixture
Finally, we can find \( \gamma \) for the mixture:
\[ \gamma_{\text{mixture}} = \frac{C_{p, \text{mixture}}}{C_{v, \text{mixture}}} = \frac{2.8955R}{1.8955R} = \frac{2.8955}{1.8955} \approx 1.528 \]

### Conclusion
The equivalent value of \( \gamma \) for the mixture is approximately \( 1.528 \).

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