The rms speed of helium gas at `27^(@)C` and 1 atm pressure is 500` ms^(-1)`. Then the rms speed of helium molecules at temperature `27^(@)C` and 2 atm pressure is

Options

A. 800` ms^(-1)`
B. 100` ms^(-1)`
C. 700` ms^(-1)`
D. 500` ms^(-1)`

Explanation

To solve the problem, we need to understand the relationship between the root mean square (rms) speed of gas molecules and the conditions under which they are measured. The rms speed of a gas is given by the formula:

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

where:
- \( v_{rms} \) = root mean square speed
- \( R \) = universal gas constant
- \( T \) = absolute temperature in Kelvin
- \( M \) = molar mass of the gas

### Step-by-Step Solution:

1. **Identify the Given Values**:
- The rms speed of helium gas at \( 27^\circ C \) (which is \( 300 \, K \)) and 1 atm pressure is \( 500 \, m/s \).
- We need to find the rms speed at the same temperature \( 27^\circ C \) but at 2 atm pressure.

2. **Understand the Formula**:
- The rms speed formula does not include pressure. It depends on temperature and the molar mass of the gas.
- Since the temperature and the gas (helium) remain constant, the molar mass also remains constant.

3. **Analyze the Effect of Pressure**:
- The rms speed is independent of pressure. This means that changing the pressure from 1 atm to 2 atm does not affect the rms speed of the gas.

4. **Conclusion**:
- Since the rms speed does not depend on pressure and remains constant at the same temperature, the rms speed of helium gas at \( 27^\circ C \) and 2 atm pressure will still be \( 500 \, m/s \).

### Final Answer:
The rms speed of helium molecules at \( 27^\circ C \) and 2 atm pressure is \( 500 \, m/s \).