Three moles of ideal gas A with `(C_(p))/(C_(v))=(4)/(3)` is mixed with two moles of another ideal gas B with `(C _(P))/(C_(v))=(5)/(3)` The `(C_(P))/(C_(v))` of mixture is (Assuming temperature is constant)

Options

A. ` 1.42 `
B. ` 1.50 `
C. ` 1.47 `
D. `1.45 `

Answer

To solve the problem of finding the \( \frac{C_P}{C_V} \) ratio for the mixture of two ideal gases A and B, we can follow these steps:

### Step 1: Identify the given data
- For gas A:
- Moles (\( n_1 \)) = 3
- \( \frac{C_P}{C_V} = \frac{4}{3} \)

- For gas B:
- Moles (\( n_2 \)) = 2
- \( \frac{C_P}{C_V} = \frac{5}{3} \)

### Step 2: Relate \( C_P \) and \( C_V \)
Using the relationship \( C_P = C_V + R \), we can express \( \frac{C_P}{C_V} \) as:
\[ \frac{C_P}{C_V} = 1 + \frac{R}{C_V} \]

### Step 3: Calculate \( C_V \) for both gases
For gas A:
\[ \frac{C_P}{C_V} = \frac{4}{3} \implies 1 + \frac{R}{C_{V1}} = \frac{4}{3} \]
\[ \frac{R}{C_{V1}} = \frac{4}{3} - 1 = \frac{1}{3} \implies C_{V1} = 3R \]

For gas B:
\[ \frac{C_P}{C_V} = \frac{5}{3} \implies 1 + \frac{R}{C_{V2}} = \frac{5}{3} \]
\[ \frac{R}{C_{V2}} = \frac{5}{3} - 1 = \frac{2}{3} \implies C_{V2} = \frac{3R}{2} \]

### Step 4: Use the formula for the mixture
The \( \frac{C_P}{C_V} \) ratio for the mixture can be calculated using:
\[ \frac{C_P}{C_V} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 C_{V1} + n_2 C_{V2}} \]
Where \( C_{P1} = C_{V1} + R \) and \( C_{P2} = C_{V2} + R \).

Now, substituting the values:
- \( C_{P1} = C_{V1} + R = 3R + R = 4R \)
- \( C_{P2} = C_{V2} + R = \frac{3R}{2} + R = \frac{5R}{2} \)

### Step 5: Substitute into the mixture formula
Now substituting into the mixture formula:
\[ \frac{C_P}{C_V} = \frac{3(4R) + 2\left(\frac{5R}{2}\right)}{3(3R) + 2\left(\frac{3R}{2}\right)} \]
Calculating the numerator:
\[ = \frac{12R + 5R}{9R + 3R} = \frac{17R}{12R} \]

### Step 6: Simplify the expression
\[ \frac{C_P}{C_V} = \frac{17}{12} \]

### Step 7: Final answer
Thus, the \( \frac{C_P}{C_V} \) ratio for the mixture is:
\[ \frac{C_P}{C_V} = \frac{17}{12} \approx 1.42 \]