The root mean square velocity of helium gas becomes the same as that of methane molecule at `327^@C` , when the temperature is

Options

A. 300 K
B. 450 K
C. 150 K
D. 1200 K

Answer

To solve the problem of finding the temperature at which the root mean square (rms) velocity of helium gas equals that of methane gas at 327°C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for root mean square velocity (vrms)**:
The formula for the root mean square velocity of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas.

2. **Set up the equation for helium and methane**:
We need to equate the rms velocities of helium and methane:
\[ v_{rms, \text{He}} = v_{rms, \text{CH}_4} \]
This gives us:
\[ \sqrt{\frac{3RT_{He}}{M_{He}}} = \sqrt{\frac{3RT_{CH_4}}{M_{CH_4}}} \]

3. **Substitute known values**:
- The molar mass of helium (\( M_{He} \)) is approximately 4 g/mol.
- The molar mass of methane (\( M_{CH_4} \)) is approximately 16 g/mol.
- The temperature for methane is given as 327°C, which we need to convert to Kelvin:
\[ T_{CH_4} = 327 + 273 = 600 \, K \]

4. **Rewrite the equation**:
Substituting the values into the equation, we have:
\[ \sqrt{\frac{3RT_{He}}{4}} = \sqrt{\frac{3R \cdot 600}{16}} \]

5. **Square both sides to eliminate the square root**:
\[ \frac{3RT_{He}}{4} = \frac{3R \cdot 600}{16} \]

6. **Cancel \( 3R \) from both sides**:
\[ \frac{T_{He}}{4} = \frac{600}{16} \]

7. **Solve for \( T_{He} \)**:
\[ T_{He} = 4 \cdot \frac{600}{16} \]
Simplifying this:
\[ T_{He} = 4 \cdot 37.5 = 150 \, K \]

8. **Convert Kelvin to Celsius** (if needed):
Since the question asks for the temperature in Celsius, we convert:
\[ T_{He} = 150 - 273 = -123 \, °C \]

### Final Answer:
The temperature at which the root mean square velocity of helium gas becomes the same as that of methane at 327°C is **150 K** or **-123°C**.