The root mean square velocity of the molecules in a sample of helium is `5//7th` that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is `0^(@)C`, then the temperature of the helium sample is about

Options

A. `0^(@)C`
B. `0 K`
C. `273^(@)C`
D. `100^(@)C`

Explanation

To find the temperature of the helium sample given that the root mean square (RMS) velocity of helium is \( \frac{5}{7} \) that of hydrogen and the temperature of hydrogen is \( 0^\circ C \), we can follow these steps:

### Step 1: Understand the relationship between RMS velocity, temperature, and molar mass
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: Set up the equation for both gases
For helium (He) and hydrogen (H2), we can express the relationship between their RMS velocities as follows:
\[ \frac{V_{rms, He}}{V_{rms, H2}} = \sqrt{\frac{T_{He}}{T_{H2}} \cdot \frac{M_{H2}}{M_{He}}} \]

### Step 3: Substitute known values
Given:
- \( V_{rms, He} = \frac{5}{7} V_{rms, H2} \)
- \( T_{H2} = 0^\circ C = 273 \, K \)
- Molar mass of hydrogen \( M_{H2} = 2 \, g/mol \)
- Molar mass of helium \( M_{He} = 4 \, g/mol \)

Substituting these values into the equation:
\[ \frac{5}{7} = \sqrt{\frac{T_{He}}{273} \cdot \frac{2}{4}} \]

### Step 4: Simplify the equation
This simplifies to:
\[ \frac{5}{7} = \sqrt{\frac{T_{He}}{273} \cdot \frac{1}{2}} \]

### Step 5: Square both sides
Squaring both sides gives:
\[ \left(\frac{5}{7}\right)^2 = \frac{T_{He}}{273} \cdot \frac{1}{2} \]
\[ \frac{25}{49} = \frac{T_{He}}{546} \]

### Step 6: Solve for \( T_{He} \)
Now, we can solve for \( T_{He} \):
\[ T_{He} = \frac{25}{49} \cdot 546 \]
Calculating this gives:
\[ T_{He} = \frac{13650}{49} \approx 278.57 \, K \]

### Step 7: Convert to Celsius
To convert Kelvin to Celsius:
\[ T_{He} \approx 278.57 - 273 \approx 5.57^\circ C \]

### Conclusion
The temperature of the helium sample is approximately \( 5.57^\circ C \).