Given: rms velocity of hydrogen at 300K is `1.9 xx 10^3` m/s. The rms velocity of oxygen at 1200K will be

Options

A. `7.6 xx 10^3 m//s`
B. `3.8 xx 10^3 m//s`
C. `0.95 xx 10^3 m//s`
D. `0.475 xx 10^3 m//s `

Answer

To find the root mean square (rms) velocity of oxygen at 1200 K, we can use the formula for rms velocity:

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

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

### Step 1: Write down the known values for hydrogen
Given:
- \( V_{\text{rms}} \) of hydrogen (\( H_2 \)) at 300 K = \( 1.9 \times 10^3 \) m/s
- \( T_H = 300 \) K
- Molar mass of hydrogen (\( M_H \)) = 2 g/mol = 0.002 kg/mol

### Step 2: Calculate the rms velocity for hydrogen using the formula
Using the formula for hydrogen:

\[ V_{\text{rms, H}} = \sqrt{\frac{3RT_H}{M_H}} \]

Substituting the known values:

\[ 1.9 \times 10^3 = \sqrt{\frac{3R \times 300}{0.002}} \]

### Step 3: Write down the known values for oxygen
For oxygen (\( O_2 \)):
- \( T_O = 1200 \) K
- Molar mass of oxygen (\( M_O \)) = 32 g/mol = 0.032 kg/mol

### Step 4: Set up the equation for oxygen
Using the same formula for oxygen:

\[ V_{\text{rms, O}} = \sqrt{\frac{3RT_O}{M_O}} \]

### Step 5: Relate the two equations
Since both gases are ideal, we can relate their rms velocities using the ratio of the temperatures and molar masses:

\[ \frac{V_{\text{rms, O}}}{V_{\text{rms, H}}} = \sqrt{\frac{T_O \cdot M_H}{T_H \cdot M_O}} \]

### Step 6: Substitute the known values
Substituting the known values into the equation:

\[ \frac{V_{\text{rms, O}}}{1.9 \times 10^3} = \sqrt{\frac{1200 \cdot 0.002}{300 \cdot 0.032}} \]

### Step 7: Simplify the right-hand side
Calculating the right-hand side:

\[ \frac{1200 \cdot 0.002}{300 \cdot 0.032} = \frac{2.4}{9.6} = \frac{1}{4} \]

Thus,

\[ \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 8: Solve for \( V_{\text{rms, O}} \)
Now substituting back:

\[ \frac{V_{\text{rms, O}}}{1.9 \times 10^3} = \frac{1}{2} \]

Therefore,

\[ V_{\text{rms, O}} = 1.9 \times 10^3 \times \frac{1}{2} = 0.95 \times 10^3 \text{ m/s} \]

### Final Answer
The rms velocity of oxygen at 1200 K is:

\[ V_{\text{rms, O}} = 0.95 \times 10^3 \text{ m/s} \]

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