The temperature at which rms velocity of helium molecules is equal to the rms velocity of hydrogen molecules at `NTP` is

Options

A. `100 K`
B. `300 K`
C. `502 K`
D. `546 K`

Explanation

To find the temperature at which the root mean square (RMS) velocity of helium molecules is equal to the RMS velocity of hydrogen molecules at Normal Temperature and Pressure (NTP), we will use the formula for RMS velocity:

\[ C_{rms} = \sqrt{\frac{3RT}{M}} \]

where:
- \(C_{rms}\) is the RMS velocity,
- \(R\) is the universal gas constant,
- \(T\) is the absolute temperature in Kelvin,
- \(M\) is the molar mass of the gas.

### Step 1: Write the RMS velocity formula for Hydrogen and Helium

1. For Hydrogen (H₂):
- Molar mass \(M_H = 2 \, \text{g/mol}\)
- RMS velocity \(C_{rms, H} = \sqrt{\frac{3RT_H}{M_H}} = \sqrt{\frac{3RT_H}{2}}\)

2. For Helium (He):
- Molar mass \(M_{He} = 4 \, \text{g/mol}\)
- RMS velocity \(C_{rms, He} = \sqrt{\frac{3RT_{He}}{M_{He}}} = \sqrt{\frac{3RT_{He}}{4}}\)

### Step 2: Set the RMS velocities equal to each other

We need to find the temperature \(T_{He}\) at which the RMS velocities of both gases are equal:

\[ \sqrt{\frac{3RT_H}{2}} = \sqrt{\frac{3RT_{He}}{4}} \]

### Step 3: Square both sides to eliminate the square roots

Squaring both sides gives:

\[ \frac{3RT_H}{2} = \frac{3RT_{He}}{4} \]

### Step 4: Cancel out the common terms

We can cancel \(3R\) from both sides:

\[ \frac{T_H}{2} = \frac{T_{He}}{4} \]

### Step 5: Cross-multiply to solve for \(T_{He}\)

Cross-multiplying gives:

\[ 4T_H = 2T_{He} \]

Dividing both sides by 2:

\[ 2T_H = T_{He} \]

### Step 6: Substitute the value of \(T_H\)

At NTP, the temperature \(T_H\) is 273 K. Therefore:

\[ T_{He} = 2 \times 273 = 546 \, \text{K} \]

### Final Answer

The temperature at which the RMS velocity of helium molecules is equal to the RMS velocity of hydrogen molecules at NTP is **546 K**.