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

Explanation

To find the value of \( C_p \) for the mixture of two ideal gases, we can follow these steps:

### Step 1: Identify the given values
We have:
- For gas 1: \( n_1 = 2 \) moles, \( C_{v1} = \frac{3}{2}R \)
- For gas 2: \( n_2 = 3 \) moles, \( C_{v2} = \frac{5}{2}R \)

### Step 2: Calculate the total number of moles
The total number of moles \( n \) in the mixture is:
\[ n = n_1 + n_2 = 2 + 3 = 5 \text{ moles} \]

### Step 3: Calculate the mixture's \( C_v \)
The formula for the specific heat at constant volume \( C_v \) for the mixture is given by:
\[ C_v = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]
Substituting the values:
\[ C_v = \frac{2 \cdot \frac{3}{2}R + 3 \cdot \frac{5}{2}R}{2 + 3} \]

### Step 4: Simplify the numerator
Calculating the numerator:
\[ 2 \cdot \frac{3}{2}R = 3R \]
\[ 3 \cdot \frac{5}{2}R = \frac{15}{2}R \]
Thus, the total in the numerator is:
\[ 3R + \frac{15}{2}R = \frac{6}{2}R + \frac{15}{2}R = \frac{21}{2}R \]

### Step 5: Calculate \( C_v \)
Now substituting back into the equation for \( C_v \):
\[ C_v = \frac{\frac{21}{2}R}{5} = \frac{21}{10}R \]

### Step 6: Calculate \( C_p \)
Using the relation \( C_p = C_v + R \):
\[ C_p = \frac{21}{10}R + R = \frac{21}{10}R + \frac{10}{10}R = \frac{31}{10}R \]

### Conclusion
Thus, the value of \( C_p \) for the mixture is:
\[ C_p = \frac{31}{10}R \]