If one mole each of a monoatomic and diatomic gases are mixed at low temperature then `C_(p)//C_(v)` ratio for the mixture is :

Options

A. 1.4
B. 1.428
C. 1.5
D. 1.33

Explanation

To solve the problem of finding the ratio \( \frac{C_p}{C_v} \) for a mixture of one mole each of a monoatomic and diatomic gas at low temperature, we can follow these steps:

### Step 1: Identify the specific heat capacities for monoatomic and diatomic gases.
- For a monoatomic gas:
- \( C_v = \frac{3}{2}R \)
- \( C_p = \frac{5}{2}R \)

- For a diatomic gas:
- \( C_v = \frac{5}{2}R \)
- \( C_p = \frac{7}{2}R \)

### Step 2: Calculate the total \( C_v \) for the mixture.
The total \( C_v \) for the mixture can be calculated as follows:
\[ C_{v, \text{mixture}} = \frac{C_{v, \text{mono}} + C_{v, \text{di}}}{2} \]
Substituting the values:
\[ C_{v, \text{mixture}} = \frac{\left(\frac{3}{2}R\right) + \left(\frac{5}{2}R\right)}{2} = \frac{\frac{8}{2}R}{2} = \frac{4R}{2} = 2R \]

### Step 3: Calculate the total \( C_p \) for the mixture.
The total \( C_p \) for the mixture can be calculated similarly:
\[ C_{p, \text{mixture}} = \frac{C_{p, \text{mono}} + C_{p, \text{di}}}{2} \]
Substituting the values:
\[ C_{p, \text{mixture}} = \frac{\left(\frac{5}{2}R\right) + \left(\frac{7}{2}R\right)}{2} = \frac{\frac{12}{2}R}{2} = \frac{6R}{2} = 3R \]

### Step 4: Calculate the ratio \( \frac{C_p}{C_v} \) for the mixture.
Now, we can calculate the desired ratio:
\[ \frac{C_p}{C_v} = \frac{C_{p, \text{mixture}}}{C_{v, \text{mixture}}} = \frac{3R}{2R} = \frac{3}{2} = 1.5 \]

### Conclusion
The ratio \( \frac{C_p}{C_v} \) for the mixture of one mole each of monoatomic and diatomic gases at low temperature is \( 1.5 \).