Two moles of helium are mixed with n moles of hydrogen. If `(Cp)/(Cv)=(3)/(2)` for the mixture, then the value of n is :

Options

A. 1
B. 2
C. 3
D. `3//2`

Answer

To solve the problem, we need to find the value of \( n \) when two moles of helium are mixed with \( n \) moles of hydrogen, given that the ratio \( \frac{C_p}{C_v} \) for the mixture is \( \frac{3}{2} \).

### Step-by-Step Solution:

1. **Identify the properties of the gases:**
- For helium (He), which is a monatomic gas:
- \( C_v = \frac{3}{2} R \)
- \( C_p = C_v + R = \frac{3}{2} R + R = \frac{5}{2} R \)
- For hydrogen (H₂), which is a diatomic gas:
- \( C_v = \frac{5}{2} R \)
- \( C_p = C_v + R = \frac{5}{2} R + R = \frac{7}{2} R \)

2. **Set up the equation for the mixture:**
- The formula for the ratio \( \frac{C_p}{C_v} \) for a mixture of gases is given by:
\[ \frac{C_p}{C_v} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}} \]
- Here, \( n_1 = 2 \) (moles of helium), \( n_2 = n \) (moles of hydrogen), \( C_{p1} = \frac{5}{2} R \), \( C_{v1} = \frac{3}{2} R \), \( C_{p2} = \frac{7}{2} R \), and \( C_{v2} = \frac{5}{2} R \).

3. **Substituting the values into the equation:**
\[ \frac{3}{2} = \frac{2 \cdot \frac{5}{2} R + n \cdot \frac{7}{2} R}{2 \cdot \frac{3}{2} R + n \cdot \frac{5}{2} R} \]

4. **Simplifying the equation:**
- The \( R \) terms can be canceled out:
\[ \frac{3}{2} = \frac{5 + \frac{7}{2} n}{3 + \frac{5}{2} n} \]

5. **Cross-multiplying to eliminate the fraction:**
\[ 3(3 + \frac{5}{2} n) = 2(5 + \frac{7}{2} n) \]
- This simplifies to:
\[ 9 + \frac{15}{2} n = 10 + 7n \]

6. **Rearranging the equation:**
\[ 9 - 10 = 7n - \frac{15}{2} n \]
\[ -1 = 7n - \frac{15}{2} n \]
- Converting \( 7n \) to a fraction:
\[ -1 = \left( \frac{14}{2} n - \frac{15}{2} n \right) \]
\[ -1 = -\frac{1}{2} n \]

7. **Solving for \( n \):**
\[ n = 2 \]

Thus, the value of \( n \) is \( 2 \).