If one mole of a monatomic gas `(gamma=5/3)` is mixed with one mole of a diatomic gas `(gamma=7/5),` the value of gamma for mixture is

Options

A. 1.4
B. 1.5
C. 1.53
D. 3.07

Explanation

To find the value of gamma (γ) for the mixture of one mole of a monatomic gas and one mole of a diatomic gas, we can follow these steps:

### Step 1: Identify the specific heat capacities
For a monatomic gas:
- \( C_{p1} = \frac{5}{2}R \)
- \( C_{v1} = \frac{3}{2}R \)

For a diatomic gas:
- \( C_{p2} = \frac{7}{2}R \)
- \( C_{v2} = \frac{5}{2}R \)

### Step 2: Calculate the total heat capacities for the mixture
The total heat capacity at constant pressure (C_p) for the mixture can be calculated using the formula:
\[ C_{p \text{ mixture}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} \]
Where \( n_1 \) and \( n_2 \) are the number of moles of the monatomic and diatomic gases, respectively. Since we have 1 mole of each gas:
\[ C_{p \text{ mixture}} = \frac{1 \cdot \frac{5}{2}R + 1 \cdot \frac{7}{2}R}{1 + 1} \]
\[ C_{p \text{ mixture}} = \frac{\frac{5}{2}R + \frac{7}{2}R}{2} = \frac{\frac{12}{2}R}{2} = \frac{6}{2}R = 3R \]

### Step 3: Calculate the total heat capacities at constant volume for the mixture
The total heat capacity at constant volume (C_v) for the mixture can be calculated similarly:
\[ C_{v \text{ mixture}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]
\[ C_{v \text{ mixture}} = \frac{1 \cdot \frac{3}{2}R + 1 \cdot \frac{5}{2}R}{1 + 1} \]
\[ C_{v \text{ mixture}} = \frac{\frac{3}{2}R + \frac{5}{2}R}{2} = \frac{\frac{8}{2}R}{2} = \frac{4}{2}R = 2R \]

### Step 4: Calculate gamma for the mixture
Gamma (γ) is defined as the ratio of specific heat capacities:
\[ \gamma = \frac{C_p}{C_v} \]
Substituting the values we found:
\[ \gamma = \frac{C_{p \text{ mixture}}}{C_{v \text{ mixture}}} = \frac{3R}{2R} = \frac{3}{2} = 1.5 \]

### Final Answer
The value of gamma (γ) for the mixture is **1.5**.

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