The rms speed of helium gas at `27^(@)C` and 1 atm pressure is 900` ms^(-1)`. Then the rms speed of helium molecules at temperature `27^(@)C` and 2 atm pressure is

Options

A. 450` ms^(-1)`
B. 900` ms^(-1)`
C. 1900` ms^(-1)`
D. 1000` ms^(-1)`

Explanation

To solve the problem of finding 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 Given Values**:
- The rms speed of helium at 27°C (which is 300 K) and 1 atm is given as 900 m/s.
- The temperature remains constant at 27°C (300 K).
- The molar mass of helium (M) is constant.

3. **Consider the Effect of Pressure**:
The rms speed of a gas does not depend on the pressure. It only depends on the temperature and the molar mass of the gas. Since both the temperature and the molar mass of helium remain unchanged, the rms speed will also remain unchanged regardless of the change in pressure.

4. **Conclude the RMS Speed**:
Therefore, the rms speed of helium gas at 27°C and 2 atm pressure will still be:
\[ v_{rms} = 900 \, \text{m/s} \]

### Final Answer:
The rms speed of helium molecules at 27°C and 2 atm pressure is **900 m/s**.

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