The ratio between the root mean square speed of `H_(2)` at `50 K` and that of `O_(2)` at `800 K` is

Options

A. 4
B. 2
C. 1
D. `1//4`

Explanation

To find the ratio between the root mean square speed (Vrms) of \( H_2 \) at \( 50 \, K \) and that of \( O_2 \) at \( 800 \, K \), we can use the formula for root mean square speed:

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

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: Calculate \( V_{rms} \) for \( H_2 \)

1. **Identify the parameters for \( H_2 \)**:
- Temperature \( T_{H_2} = 50 \, K \)
- Molar mass \( M_{H_2} = 2 \, g/mol = 0.002 \, kg/mol \)

2. **Substitute into the formula**:
\[ V_{rms(H_2)} = \sqrt{\frac{3R(50)}{0.002}} \]

### Step 2: Calculate \( V_{rms} \) for \( O_2 \)

1. **Identify the parameters for \( O_2 \)**:
- Temperature \( T_{O_2} = 800 \, K \)
- Molar mass \( M_{O_2} = 32 \, g/mol = 0.032 \, kg/mol \)

2. **Substitute into the formula**:
\[ V_{rms(O_2)} = \sqrt{\frac{3R(800)}{0.032}} \]

### Step 3: Find the ratio \( \frac{V_{rms(H_2)}}{V_{rms(O_2)}} \)

1. **Set up the ratio**:
\[ \frac{V_{rms(H_2)}}{V_{rms(O_2)}} = \frac{\sqrt{\frac{3R(50)}{0.002}}}{\sqrt{\frac{3R(800)}{0.032}}} \]

2. **Simplify the ratio**:
\[ = \sqrt{\frac{3R(50)}{0.002}} \cdot \sqrt{\frac{0.032}{3R(800)}} \]
\[ = \sqrt{\frac{50 \times 0.032}{800 \times 0.002}} \]
\[ = \sqrt{\frac{1.6}{1.6}} = \sqrt{1} = 1 \]

### Conclusion
The ratio between the root mean square speed of \( H_2 \) at \( 50 \, K \) and that of \( O_2 \) at \( 800 \, K \) is \( 1 \).

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