The ratio between the root mean square velocity 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`

Answer

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

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

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: Write the RMS velocity for \( H_2 \) and \( O_2 \)

For \( H_2 \) at 50 K:

\[ v_{rms, H_2} = \sqrt{\frac{3R \cdot T_{H_2}}{M_{H_2}}} \]

For \( O_2 \) at 800 K:

\[ v_{rms, O_2} = \sqrt{\frac{3R \cdot T_{O_2}}{M_{O_2}}} \]

### Step 2: Calculate the ratio of the RMS velocities

The ratio of the RMS velocities is given by:

\[ \frac{v_{rms, H_2}}{v_{rms, O_2}} = \frac{\sqrt{\frac{3R \cdot T_{H_2}}{M_{H_2}}}}{\sqrt{\frac{3R \cdot T_{O_2}}{M_{O_2}}}} \]

This simplifies to:

\[ \frac{v_{rms, H_2}}{v_{rms, O_2}} = \sqrt{\frac{T_{H_2} \cdot M_{O_2}}{T_{O_2} \cdot M_{H_2}}} \]

### Step 3: Substitute the values

- For \( H_2 \):
- \( T_{H_2} = 50 \, K \)
- \( M_{H_2} = 2 \, g/mol = 0.002 \, kg/mol \)

- For \( O_2 \):
- \( T_{O_2} = 800 \, K \)
- \( M_{O_2} = 32 \, g/mol = 0.032 \, kg/mol \)

Now substituting these values into the ratio:

\[ \frac{v_{rms, H_2}}{v_{rms, O_2}} = \sqrt{\frac{50 \cdot 0.032}{800 \cdot 0.002}} \]

### Step 4: Simplify the expression

Calculating the numerator and denominator:

- Numerator: \( 50 \cdot 0.032 = 1.6 \)
- Denominator: \( 800 \cdot 0.002 = 1.6 \)

Thus, we have:

\[ \frac{v_{rms, H_2}}{v_{rms, O_2}} = \sqrt{\frac{1.6}{1.6}} = \sqrt{1} = 1 \]

### Final Answer

The ratio between the root mean square velocity of \( H_2 \) at 50 K and that of \( O_2 \) at 800 K is:

\[ \boxed{1} \]