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 find the temperature at which the root mean square (RMS) velocity of helium gas becomes the same as that of methane (CH₄) at a given temperature of 327°C, we can follow these steps:

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

### Step 2: Convert the given temperature to Kelvin
The temperature given is 327°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
Thus,
\[ T = 327 + 273.15 = 600.15 \, K \approx 600 \, K \]

### Step 3: Set the RMS velocities equal for helium and methane
Let \( M_{He} \) be the molar mass of helium (4 g/mol) and \( M_{CH_4} \) be the molar mass of methane (16 g/mol). We set the RMS velocities equal:
\[ \sqrt{\frac{3RT_{He}}{M_{He}}} = \sqrt{\frac{3RT_{CH_4}}{M_{CH_4}}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{3RT_{He}}{M_{He}} = \frac{3RT_{CH_4}}{M_{CH_4}} \]

### Step 5: Cancel out common terms
Since \( R \) and \( 3 \) are common on both sides, they can be canceled out:
\[ \frac{T_{He}}{M_{He}} = \frac{T_{CH_4}}{M_{CH_4}} \]

### Step 6: Substitute the known values
Substituting the values of \( M_{He} \) and \( M_{CH_4} \):
\[ \frac{T_{He}}{4} = \frac{600}{16} \]

### Step 7: Solve for \( T_{He} \)
Cross-multiplying gives:
\[ 16T_{He} = 2400 \]
Now, divide both sides by 16:
\[ T_{He} = \frac{2400}{16} = 150 \, K \]

### Conclusion
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**.