The rms speed of helium gas at `27^(@)C` and 1 atm pressure is 1000` ms^(-1)`. Then the rms speed of helium molecules at temperature `27^(@)C` and 2 atm pressure is

Options

A. 100` ms^(-1)`
B. 1000` ms^(-1)`
C. 500` ms^(-1)`
D. 400` ms^(-1)`

Answer

To solve the problem regarding the root mean square (rms) speed of helium gas at different pressures while keeping the temperature constant, we can follow these steps:

### Step 1: Understand the formula for rms speed
The rms speed (v_rms) of a gas 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 2: Identify the constants
In this problem:
- The temperature \( T \) is constant at \( 27^\circ C \) (which is \( 300 K \)).
- The molar mass \( M \) of helium is also constant since we are dealing with the same gas.

### Step 3: Analyze the effect of pressure
The rms speed formula does not include pressure as a variable. This means that the rms speed of a gas is independent of the pressure when the temperature and the type of gas remain constant.

### Step 4: Conclusion
Since the rms speed of helium gas at \( 27^\circ C \) and 1 atm pressure is given as \( 1000 \, m/s \), and since the rms speed does not change with pressure, the rms speed of helium gas at \( 27^\circ C \) and 2 atm pressure will also be:
\[ v_{rms} = 1000 \, m/s \]

### Final Answer
The rms speed of helium molecules at \( 27^\circ C \) and 2 atm pressure is \( 1000 \, m/s \).

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