One mole of an ideal monoatomic gas is mixed with one mole of an equimolar mixture of monoatomic and diatomic ideal gases. Find the value of `lambda= (C_P /C_v)` for the final mixture

Options

A. `8/7`
B. `19/12`
C. `11/7`
D. `8/5`

Explanation

To find the value of \(\lambda = \frac{C_P}{C_V}\) for the final mixture of gases, we need to calculate the specific heat capacities \(C_P\) and \(C_V\) for the mixture.

### Step 1: Identify the specific heat capacities for each gas type
1. **Monoatomic gas**:
- \(C_P = \frac{5R}{2}\)
- \(C_V = \frac{3R}{2}\)

2. **Diatomic gas**:
- \(C_P = \frac{7R}{2}\)
- \(C_V = \frac{5R}{2}\)

### Step 2: Calculate the total \(C_P\) for the mixture
We have:
- 1 mole of monoatomic gas
- 1 mole of an equimolar mixture of monoatomic and diatomic gases (which means 0.5 moles of each).

The total \(C_P\) for the mixture can be calculated as follows:
\[ C_P = C_{P, \text{monoatomic}} + C_{P, \text{mixture}} \]
Where,
\[ C_{P, \text{mixture}} = \left(0.5 \times C_{P, \text{monoatomic}}\right) + \left(0.5 \times C_{P, \text{diatomic}}\right) \]
Substituting the values:
\[ C_{P, \text{mixture}} = \left(0.5 \times \frac{5R}{2}\right) + \left(0.5 \times \frac{7R}{2}\right) \]
Calculating:
\[ C_{P, \text{mixture}} = \frac{5R}{4} + \frac{7R}{4} = \frac{12R}{4} = 3R \]
Thus, the total \(C_P\) for the mixture is:
\[ C_P = \frac{5R}{2} + 3R = \frac{5R}{2} + \frac{6R}{2} = \frac{11R}{2} \]

### Step 3: Calculate the total \(C_V\) for the mixture
Similarly, we calculate \(C_V\):
\[ C_V = C_{V, \text{monoatomic}} + C_{V, \text{mixture}} \]
Where,
\[ C_{V, \text{mixture}} = \left(0.5 \times C_{V, \text{monoatomic}}\right) + \left(0.5 \times C_{V, \text{diatomic}}\right) \]
Substituting the values:
\[ C_{V, \text{mixture}} = \left(0.5 \times \frac{3R}{2}\right) + \left(0.5 \times \frac{5R}{2}\right) \]
Calculating:
\[ C_{V, \text{mixture}} = \frac{3R}{4} + \frac{5R}{4} = \frac{8R}{4} = 2R \]
Thus, the total \(C_V\) for the mixture is:
\[ C_V = \frac{3R}{2} + 2R = \frac{3R}{2} + \frac{4R}{2} = \frac{7R}{2} \]

### Step 4: Calculate \(\lambda\)
Now we can find \(\lambda\):
\[ \lambda = \frac{C_P}{C_V} = \frac{\frac{11R}{2}}{\frac{7R}{2}} = \frac{11}{7} \]

### Final Answer
Thus, the value of \(\lambda\) for the final mixture is:
\[ \lambda = \frac{11}{7} \]