The rms speed of helium gas at `27^(@)C` and 1 atm pressure is 1100` ms^(-1)`. Then the rms speed of helium molecules at temperature `27^(@)C` and 2 atm pressure is

Options

A. 1800` ms^(-1)`
B. 1100` ms^(-1)`
C. 500` ms^(-1)`
D. 700` 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 rms speed (v_rms) can be calculated using 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.

### Step-by-step Solution:

1. **Identify the Given Values**:
- The rms speed of helium gas at 27°C (which is 300 K) and 1 atm pressure is given as 1100 m/s.
- We need to find the rms speed at the same temperature (27°C or 300 K) but at 2 atm pressure.

2. **Understand the Formula**:
- The rms speed formula shows that it depends on temperature (T) and molar mass (M) but does not depend on pressure (P).

3. **Analyze the Conditions**:
- Since the temperature (T) remains constant at 27°C (300 K) and the molar mass of helium does not change, the only factor that changes is the pressure. However, the rms speed is independent of pressure.

4. **Conclusion**:
- Since the rms speed does not depend on pressure, the rms speed of helium gas at 27°C and 2 atm pressure will remain the same as at 1 atm pressure.

5. **Final Answer**:
- Therefore, the rms speed of helium molecules at 27°C and 2 atm pressure is still **1100 m/s**.