At what temperature the rms velocity of helium molecules will be equal to that of hydrogen molecules at NTP ?

Options

A. 844 k
B. 64 k
C. `273^(@)c`
D. 273 k

Explanation

To find the temperature at which the root mean square (rms) velocity of helium molecules will be equal to that of hydrogen molecules at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Understand the formula for rms velocity
The rms velocity (v_rms) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where:
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the mass of a gas molecule.

### Step 2: Identify the molecular masses
- The molar mass of hydrogen (H₂) is approximately 2 g/mol, which means the mass of one hydrogen molecule (H₂) is:
\[ m_{H_2} = \frac{2 \text{ g/mol}}{N_A} = \frac{2 \times 10^{-3} \text{ kg/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 3.32 \times 10^{-27} \text{ kg} \]
- The molar mass of helium (He) is approximately 4 g/mol, so the mass of one helium atom is:
\[ m_{He} = \frac{4 \text{ g/mol}}{N_A} = \frac{4 \times 10^{-3} \text{ kg/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 6.64 \times 10^{-27} \text{ kg} \]

### Step 3: Calculate the rms velocity of hydrogen at NTP
At NTP, the temperature \( T \) is 273 K. Therefore, the rms velocity of hydrogen at NTP is:
\[ v_{rms, H_2} = \sqrt{\frac{3k \cdot 273}{m_{H_2}}} \]

### Step 4: Set the rms velocities equal
We want to find the temperature \( T_{He} \) at which the rms velocity of helium equals that of hydrogen:
\[ v_{rms, He} = v_{rms, H_2} \]
This gives us:
\[ \sqrt{\frac{3kT_{He}}{m_{He}}} = \sqrt{\frac{3k \cdot 273}{m_{H_2}}} \]

### Step 5: Square both sides and simplify
Squaring both sides, we get:
\[ \frac{3kT_{He}}{m_{He}} = \frac{3k \cdot 273}{m_{H_2}} \]
Cancelling \( 3k \) from both sides, we have:
\[ \frac{T_{He}}{m_{He}} = \frac{273}{m_{H_2}} \]
Thus:
\[ T_{He} = 273 \cdot \frac{m_{He}}{m_{H_2}} \]

### Step 6: Substitute the masses
Substituting the values of \( m_{He} \) and \( m_{H_2} \):
\[ T_{He} = 273 \cdot \frac{6.64 \times 10^{-27}}{3.32 \times 10^{-27}} = 273 \cdot 2 = 546 \text{ K} \]

### Final Answer
The temperature at which the rms velocity of helium molecules will be equal to that of hydrogen molecules at NTP is **546 K**.

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