The rms speed of helium gas at `27^(@)C` and 1 atm pressure is 800` ms^(-1)`. Then the rms speed of helium molecules at temperature `27^(@)C` and 2 atm pressure is

Options

A. 800` ms^(-1)`
B. 450` ms^(-1)`
C. 45` ms^(-1)`
D. 550` ms^(-1)`

Explanation

To solve the problem, we need to understand the relationship between the root mean square (rms) speed of a gas and the factors that influence it. The formula for the rms speed \( v_{rms} \) 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 Information**:
- The rms speed of helium gas at \( 27^{\circ}C \) (which is \( 300 \, K \)) and 1 atm pressure is \( 800 \, m/s \).
- We need to find the rms speed at the same temperature \( 27^{\circ}C \) but at 2 atm pressure.

2. **Understand the Variables**:
- The temperature \( T \) remains constant at \( 27^{\circ}C \) (or \( 300 \, K \)).
- The molar mass \( M \) of helium is also constant since we are dealing with the same gas.
- The only variable that changes is the pressure, which is not included in the rms speed formula.

3. **Analyze the Formula**:
- The rms speed depends on \( R \), \( T \), and \( M \) but does not depend on pressure.
- Since both temperature and the type of gas (helium) remain unchanged, the rms speed will also remain unchanged.

4. **Conclusion**:
- Therefore, the rms speed of helium molecules at \( 27^{\circ}C \) and 2 atm pressure is still \( 800 \, m/s \).

### Final Answer:
The rms speed of helium molecules at \( 27^{\circ}C \) and 2 atm pressure is \( 800 \, m/s \).

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