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 (RMS) speed of oxygen gas at 600 K, given its RMS speed at 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: The RMS speed (\( V_{rms} \)) of gas molecules is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

2. **Identify Proportionality**: From the formula, we can see that \( V_{rms} \) is directly proportional to the square root of the temperature:
\[ V_{rms} \propto \sqrt{T} \]

3. **Set Up the Ratio**: We can set up the ratio of the RMS speeds at two different temperatures:
\[ \frac{V_{rms1}}{V_{rms2}} = \sqrt{\frac{T_1}{T_2}} \]
Here, \( V_{rms1} = 484 \, m/s \) at \( T_1 = 300 \, K \) and we need to find \( V_{rms2} \) at \( T_2 = 600 \, K \).

4. **Plug in the Values**: Substitute the known values into the ratio:
\[ \frac{484}{V_{rms2}} = \sqrt{\frac{300}{600}} \]

5. **Simplify the Square Root**: Simplifying the square root gives:
\[ \sqrt{\frac{300}{600}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

6. **Rearranging the Equation**: Now, rearranging the equation gives:
\[ V_{rms2} = 484 \cdot \sqrt{2} \]

7. **Calculate \( V_{rms2} \)**: Calculate \( V_{rms2} \):
\[ V_{rms2} \approx 484 \cdot 1.414 \approx 684 \, m/s \]

### Final Answer:
The corresponding value of the average RMS speed at 600 K is approximately **684 m/s**.