A mixture of 2 moles of helium gas `(atomic mass=4amu)` and 1 mole of argon gas `(atomic mass=40amu)` is kept at 300K in a container. The ratio of the rms speeds `((v_(rms)(helium))/((v_(rms)(argon))` is

Options

A. `0.32`
B. `0.45`
C. `2.24`
D. `3.16`

Explanation

To find the ratio of the root mean square (rms) speeds of helium and argon gases, we can use the formula for rms speed 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 in kg/mol.

### Step 1: Identify the molar masses
For helium:
- Molar mass \( M_{He} = 4 \, \text{amu} = 4 \times 10^{-3} \, \text{kg/mol} \)

For argon:
- Molar mass \( M_{Ar} = 40 \, \text{amu} = 40 \times 10^{-3} \, \text{kg/mol} \)

### Step 2: Write the rms speed expressions
The rms speed of helium (\( v_{rms, He} \)) is:
\[ v_{rms, He} = \sqrt{\frac{3RT}{M_{He}}} \]

The rms speed of argon (\( v_{rms, Ar} \)) is:
\[ v_{rms, Ar} = \sqrt{\frac{3RT}{M_{Ar}}} \]

### Step 3: Calculate the ratio of rms speeds
To find the ratio of the rms speeds of helium to argon, we can write:
\[ \frac{v_{rms, He}}{v_{rms, Ar}} = \frac{\sqrt{\frac{3RT}{M_{He}}}}{\sqrt{\frac{3RT}{M_{Ar}}}} = \sqrt{\frac{M_{Ar}}{M_{He}}} \]

### Step 4: Substitute the molar masses
Now, substituting the values of molar masses:
\[ \frac{v_{rms, He}}{v_{rms, Ar}} = \sqrt{\frac{40 \times 10^{-3}}{4 \times 10^{-3}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \]

### Step 5: Calculate the numerical value
Calculating \( \sqrt{10} \):
\[ \sqrt{10} \approx 3.16 \]

### Final Answer
Thus, the ratio of the rms speeds of helium to argon is approximately:
\[ \frac{v_{rms, He}}{v_{rms, Ar}} \approx 3.16 \]