A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g). The ratio of the rms speed of the argon molecules to that of the hydrogen is

Options

A. 1
B. 5
C. `1//5`
D. `1 // sqrt(5)`

Answer

To find the ratio of the root mean square (rms) speed of argon gas to that of hydrogen gas, we can use the formula for rms speed:

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

where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 1: Define the variables for both gases
Let:
- For argon (sample 1):
- Molar mass \( M_1 = 40 \, \text{g/mol} \)
- Temperature \( T_1 = 4T_2 \) (where \( T_2 \) is the temperature of hydrogen)

- For hydrogen (sample 2):
- Molar mass \( M_2 = 2 \, \text{g/mol} \)
- Temperature \( T_2 \) (absolute temperature of hydrogen)

### Step 2: Write the rms speed equations for both gases
- The rms speed of argon:
\[ v_{\text{rms}, 1} = \sqrt{\frac{3RT_1}{M_1}} = \sqrt{\frac{3R(4T_2)}{40}} = \sqrt{\frac{12RT_2}{40}} = \sqrt{\frac{3RT_2}{10}} \]

- The rms speed of hydrogen:
\[ v_{\text{rms}, 2} = \sqrt{\frac{3RT_2}{M_2}} = \sqrt{\frac{3RT_2}{2}} \]

### Step 3: Find the ratio of rms speeds
Now, we need to find the ratio \( \frac{v_{\text{rms}, 1}}{v_{\text{rms}, 2}} \):
\[ \frac{v_{\text{rms}, 1}}{v_{\text{rms}, 2}} = \frac{\sqrt{\frac{3RT_2}{10}}}{\sqrt{\frac{3RT_2}{2}}} \]

### Step 4: Simplify the ratio
The \( 3RT_2 \) terms cancel out:
\[ \frac{v_{\text{rms}, 1}}{v_{\text{rms}, 2}} = \sqrt{\frac{1/10}{1/2}} = \sqrt{\frac{2}{10}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

### Final Answer
Thus, the ratio of the rms speed of argon molecules to that of hydrogen is:
\[ \frac{v_{\text{rms}, \text{argon}}}{v_{\text{rms}, \text{hydrogen}}} = \frac{1}{\sqrt{5}} \]

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