Two moles the an ideal gas with `C_(v) = (3)/(2)R` are mixed with 3 of anthoer ideal gas with `C_(v) = (5)/(2)`R . The value of the `C_(p)` for the mixture is :

Options

A. 1.45 R
B. 2R
C. 3.1 R
D. 3.2 R

Explanation

To solve the problem of finding the molar specific heat at constant pressure (C_p) for a mixture of two ideal gases, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- For gas 1:
- Number of moles (n₁) = 2 moles
- Molar specific heat at constant volume (C_v₁) = (3/2)R
- For gas 2:
- Number of moles (n₂) = 3 moles
- Molar specific heat at constant volume (C_v₂) = (5/2)R

2. **Calculate C_p for Each Gas Using Mayer's Relation:**
- Mayer's relation states that C_p - C_v = R.
- For gas 1:
\[ C_{p1} = C_{v1} + R = \left(\frac{3}{2}R\right) + R = \frac{3}{2}R + \frac{2}{2}R = \frac{5}{2}R \]
- For gas 2:
\[ C_{p2} = C_{v2} + R = \left(\frac{5}{2}R\right) + R = \frac{5}{2}R + \frac{2}{2}R = \frac{7}{2}R \]

3. **Use the Formula for C_p of the Mixture:**
- The formula for the molar specific heat at constant pressure for the mixture is given by:
\[ C_{p \text{ mixture}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} \]
- Substitute the values:
\[ C_{p \text{ mixture}} = \frac{(2 \times \frac{5}{2}R) + (3 \times \frac{7}{2}R)}{2 + 3} \]

4. **Calculate the Numerator:**
- Calculate each term:
\[ 2 \times \frac{5}{2}R = 5R \]
\[ 3 \times \frac{7}{2}R = \frac{21}{2}R \]
- Combine these:
\[ 5R + \frac{21}{2}R = \frac{10}{2}R + \frac{21}{2}R = \frac{31}{2}R \]

5. **Calculate the Denominator:**
- The total number of moles is:
\[ n_1 + n_2 = 2 + 3 = 5 \]

6. **Final Calculation:**
- Substitute back into the formula:
\[ C_{p \text{ mixture}} = \frac{\frac{31}{2}R}{5} = \frac{31}{10}R = 3.1R \]

### Final Answer:
The molar specific heat at constant pressure for the mixture is \( C_{p \text{ mixture}} = 3.1R \).