Two moles of helium are mixed with n moles of hydrogen. The root mean spure (rms) speed of the gas molecules in the mexture is `sqrt2` times the speed of sound in the mixture. Then value of n is

Options

A. 1
B. 3
C. 2
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, and the root mean square (rms) speed of the gas molecules in the mixture is \( \sqrt{2} \) times the speed of sound in the mixture.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have 2 moles of helium (He) and \( n \) moles of hydrogen (H₂).
- The rms speed of the gas mixture is \( \sqrt{2} \) times the speed of sound in the mixture.

2. **Formulas**:
- The rms speed \( v_{rms} \) of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas mixture.
- The speed of sound \( v_s \) in a gas is given by:
\[ v_s = \sqrt{\frac{\gamma RT}{M}} \]
where \( \gamma \) is the adiabatic index (ratio of specific heats).

3. **Setting Up the Relationship**:
- According to the problem, we have:
\[ v_{rms} = \sqrt{2} v_s \]
- Substituting the formulas for \( v_{rms} \) and \( v_s \):
\[ \sqrt{\frac{3RT}{M}} = \sqrt{2} \sqrt{\frac{\gamma RT}{M}} \]

4. **Simplifying the Equation**:
- Squaring both sides gives:
\[ \frac{3RT}{M} = 2 \cdot \frac{\gamma RT}{M} \]
- Canceling \( RT/M \) (assuming \( RT/M \neq 0 \)):
\[ 3 = 2\gamma \]
- Therefore, we find:
\[ \gamma = \frac{3}{2} \]

5. **Finding the Molar Mass of the Mixture**:
- The molar mass \( M \) of the mixture can be calculated using the formula:
\[ M = \frac{n_1M_1 + n_2M_2}{n_1 + n_2} \]
- Here, \( n_1 = 2 \) (moles of He), \( M_1 = 4 \, \text{g/mol} \) (molar mass of He), \( n_2 = n \) (moles of H₂), \( M_2 = 2 \, \text{g/mol} \) (molar mass of H₂):
\[ M = \frac{2 \cdot 4 + n \cdot 2}{2 + n} = \frac{8 + 2n}{2 + n} \]

6. **Calculating \( \gamma \) for the Mixture**:
- The expression for \( \gamma \) is given by:
\[ \gamma = \frac{C_p}{C_v} = 1 + \frac{R}{C_v} \]
- We can express \( C_v \) for the mixture:
\[ C_v = \frac{C_{v1}n_1 + C_{v2}n_2}{n_1 + n_2} \]
- Where \( C_{v1} = \frac{3}{2}R \) for He and \( C_{v2} = \frac{5}{2}R \) for H₂:
\[ C_v = \frac{\frac{3}{2}R \cdot 2 + \frac{5}{2}R \cdot n}{2 + n} = \frac{3R + \frac{5}{2}Rn}{2 + n} \]

7. **Substituting \( C_v \) into \( \gamma \)**:
- Now substituting \( C_v \) into the equation for \( \gamma \):
\[ \frac{3}{2} = 1 + \frac{R}{\frac{3R + \frac{5}{2}Rn}{2 + n}} \]
- Rearranging gives:
\[ \frac{1}{2} = \frac{R(2 + n)}{3R + \frac{5}{2}Rn} \]

8. **Cross Multiplying and Simplifying**:
- Cross-multiplying gives:
\[ (3R + \frac{5}{2}Rn) = 2R(2 + n) \]
- Expanding and simplifying:
\[ 3R + \frac{5}{2}Rn = 4R + 2Rn \]
- Rearranging terms:
\[ \frac{5}{2}Rn - 2Rn = 4R - 3R \]
- This simplifies to:
\[ \frac{1}{2}Rn = R \]
- Therefore:
\[ n = 2 \]

### Final Answer:
The value of \( n \) is \( 2 \).