The rms speed of helium gas at `27^(@)C` and 1 atm pressure is 300` ms^(-1)`. Then the rms speed of helium molecules at temperature `27^(@)C` and 2 atm pressure is

Options

A. 300` ms^(-1)`
B. 100` ms^(-1)`
C. 400` ms^(-1)`
D. 200` ms^(-1)`

Explanation

To solve the problem, we need to understand the relationship between the root mean square (rms) speed of a gas and its properties. The formula for the rms speed (v_rms) of a gas is given by:

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

where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- The rms speed of helium gas at \( 27^\circ C \) and 1 atm is \( 300 \, m/s \).
- The temperature \( T \) is \( 27^\circ C \), which is \( 300 \, K \) (since \( 27 + 273 = 300 \)).
- The molar mass \( M \) of helium is constant.

2. **Understand the Effect of Pressure**:
- The rms speed formula does not include pressure as a variable. It depends only on temperature and molar mass.
- Since both the temperature and the molar mass remain unchanged, the rms speed will also remain unchanged regardless of the change in pressure.

3. **Conclusion**:
- Therefore, the rms speed of helium gas at \( 27^\circ C \) and 2 atm pressure will still be \( 300 \, m/s \).

### Final Answer:
The rms speed of helium molecules at \( 27^\circ C \) and 2 atm pressure is \( 300 \, m/s \).

---