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 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:
- \( v_{rms} \) 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 in kg/mol.

### Step 1: Calculate the root mean square speed of \( H_2 \)

For \( H_2 \) at \( 50 \, K \):
- Molar mass of \( H_2 \) = \( 2 \, g/mol = 0.002 \, kg/mol \)
- Temperature \( T_{H_2} = 50 \, K \)

Using the formula:
\[ v_{rms(H_2)} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3R \cdot 50}{0.002}} \]

### Step 2: Calculate the root mean square speed of \( O_2 \)

For \( O_2 \) at \( 800 \, K \):
- Molar mass of \( O_2 \) = \( 32 \, g/mol = 0.032 \, kg/mol \)
- Temperature \( T_{O_2} = 800 \, K \)

Using the formula:
\[ v_{rms(O_2)} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3R \cdot 800}{0.032}} \]

### Step 3: Find the ratio of the root mean square speeds

The ratio of the speeds is given by:
\[ \text{Ratio} = \frac{v_{rms(H_2)}}{v_{rms(O_2)}} = \frac{\sqrt{\frac{3R \cdot 50}{0.002}}}{\sqrt{\frac{3R \cdot 800}{0.032}}} \]

### Step 4: Simplify the ratio

This can be simplified as follows:
\[ \text{Ratio} = \sqrt{\frac{3R \cdot 50}{0.002}} \cdot \sqrt{\frac{0.032}{3R \cdot 800}} = \sqrt{\frac{50 \cdot 0.032}{0.002 \cdot 800}} \]

### Step 5: Calculate the numerical values

Calculating the values:
\[ = \sqrt{\frac{50 \cdot 0.032}{0.002 \cdot 800}} = \sqrt{\frac{1.6}{1.6}} = \sqrt{1} = 1 \]

### Final Answer

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