The rms speed of helium gas at `27^(@)C` and 1 atm pressure is 400` ms^(-1)`. Then the rms speed of helium molecules at temperature `27^(@)C` and 2 atm pressure is

Options

A. 100` ms^(-1)`
B. 500` ms^(-1)`
C. 400` ms^(-1)`
D. 600` ms^(-1)`

Explanation

To find the root mean square (rms) speed of helium molecules at a different pressure but the same temperature, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**: The rms speed \( v_{rms} \) 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.

2. **Identify Given Values**: From the question, we know:
- The rms speed at 1 atm and \( 27^\circ C \) is \( 400 \, \text{m/s} \).
- The temperature is constant at \( 27^\circ C \) (which is \( 300 \, K \)).
- The molar mass of helium remains constant.

3. **Determine the Effect of Pressure**: The rms speed formula does not include pressure as a variable. It only depends on temperature and molar mass. Therefore, changing the pressure from 1 atm to 2 atm does not affect the rms speed.

4. **Conclude the rms Speed**: Since the rms speed is independent of pressure, the rms speed of helium at \( 27^\circ C \) and 2 atm pressure will remain the same as at 1 atm:
\[ v_{rms} = 400 \, \text{m/s} \]

### Final Answer:
The rms speed of helium molecules at \( 27^\circ C \) and 2 atm pressure is \( 400 \, \text{m/s} \).

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