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 m/s
B. 900 m/s
C. 1800 m/s
D. 750 m/s

Answer

To find the root mean square (rms) speed of helium molecules at a temperature of \(27^\circ C\) and 2 atm pressure, we can use the formula for rms speed:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

Where:
- \(v_{rms}\) is the root mean square speed,
- \(R\) is the universal gas constant,
- \(T\) is the absolute temperature in Kelvin,
- \(M\) is the molar mass of the gas.

### Step 1: Convert the temperature to Kelvin
The temperature in Celsius is given as \(27^\circ C\). To convert this to Kelvin:

\[ T(K) = T(°C) + 273.15 = 27 + 273.15 = 300.15 \, K \]

### Step 2: Identify constants
For helium gas:
- The molar mass \(M\) is constant and approximately \(4 \, g/mol\) or \(0.004 \, kg/mol\).
- The universal gas constant \(R\) is \(8.314 \, J/(mol \cdot K)\).

### Step 3: Calculate the rms speed at 1 atm
We are given that the rms speed at \(27^\circ C\) and 1 atm pressure is \(900 \, m/s\).

### Step 4: Analyze the effect of pressure
The rms speed formula shows that the rms speed depends only on temperature and molar mass. Since the temperature and molar mass remain constant, the rms speed does not depend on the pressure.

### Step 5: Conclusion
Thus, the rms speed of helium molecules at \(27^\circ C\) and 2 atm pressure will remain the same as at 1 atm pressure, which is:

\[ v_{rms} = 900 \, m/s \]

### Final Answer
The rms speed of helium molecules at \(27^\circ C\) and 2 atm pressure is \(900 \, m/s\).

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