The average rms speed of molecules in a sample of oxygen gas at 300 K is `484ms^(-1)`., respectively. The corresponding value at 600 K is nearly (assuming ideal gas behaviour)

Options

A. `968ms^(-1)`
B. `986ms^(-1)`
C. `648ms^(-1)`
D. `684ms^(-1)`

Answer

To solve the problem of finding the average root mean square speed of oxygen gas at 600 K, given its speed at 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: The root mean square (rms) speed of gas molecules is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas. For an ideal gas, the rms speed is directly proportional to the square root of the temperature.

2. **Set Up the Proportionality**: Since \( v_{rms} \) is proportional to \( \sqrt{T} \), we can express the relationship between the rms speeds at two different temperatures as:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]
where \( v_1 \) is the rms speed at temperature \( T_1 \) and \( v_2 \) is the rms speed at temperature \( T_2 \).

3. **Plug in the Known Values**: From the problem, we have:
- \( v_1 = 484 \, \text{m/s} \) (at \( T_1 = 300 \, \text{K} \))
- \( T_2 = 600 \, \text{K} \)

We need to find \( v_2 \).

4. **Calculate the Ratio**:
\[ \frac{484}{v_2} = \sqrt{\frac{300}{600}} \]
Simplifying the right side:
\[ \sqrt{\frac{300}{600}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

5. **Rearranging the Equation**: Now we can rearrange the equation to solve for \( v_2 \):
\[ v_2 = 484 \cdot \sqrt{2} \]

6. **Calculate \( v_2 \)**: Now we need to calculate \( 484 \cdot \sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \]
Therefore:
\[ v_2 \approx 484 \cdot 1.414 \approx 684 \, \text{m/s} \]

### Final Answer:
The average rms speed of the molecules in the sample of oxygen gas at 600 K is approximately **684 m/s**.