The rms speed of helium gas at `27^(@)C` and 1 atm pressure is 700` ms^(-1)`. Then the rms speed of helium molecules at temperature `27^(@)C` and 2 atm pressure is

Options

A. 1800` ms^(-1)`
B. 800` ms^(-1)`
C. 700` ms^(-1)`
D. 100` ms^(-1)`

Explanation

To find the root mean square (rms) speed of helium gas at a temperature of 27°C and a pressure of 2 atm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for RMS Speed**:
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.

2. **Identify Constants**:
In this scenario, we are given that:
- The temperature \( T \) is constant at 27°C (which is 300 K).
- The molar mass \( M \) of helium is also constant.
- The gas constant \( R \) is constant.

3. **Analyze the Effect of Pressure**:
According to the ideal gas law, the rms speed of a gas does not depend on pressure when temperature and molar mass are constant. Therefore, even though the pressure changes from 1 atm to 2 atm, the rms speed remains the same.

4. **Conclusion**:
Since the rms speed at 1 atm is given as 700 m/s, and it does not change with pressure, the rms speed of helium at 2 atm and the same temperature (27°C) will also be:
\[ v_{rms} = 700 \, \text{m/s} \]

### Final Answer:
The rms speed of helium molecules at 27°C and 2 atm pressure is **700 m/s**.

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