At `300K`,the `rms` speed of oxygen molecules is `sqrt((a+5)/(a))` times to that of its average speed in the gas. Then,the value of a will be (used `pi=(22)/(7))`

Options

A. `27`
B. `24`
C. `32`
D. `28`

Answer

To solve the problem, we need to establish the relationship between the root mean square (RMS) speed and the average speed of oxygen molecules at a temperature of 300K.

### Step-by-Step Solution:

1. **Understand the formulas**:
- The formula for RMS speed (\(V_{rms}\)) is given by:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
- The formula for average speed (\(V_{avg}\)) is given by:
\[ V_{avg} = \sqrt{\frac{8RT}{\pi M}} \]

2. **Set up the relationship**:
According to the problem, the RMS speed is given as:
\[ V_{rms} = \sqrt{\frac{a+5}{a}} \times V_{avg} \]

3. **Substituting the formulas**:
Substitute the expressions for \(V_{rms}\) and \(V_{avg}\) into the equation:
\[ \sqrt{\frac{3RT}{M}} = \sqrt{\frac{a+5}{a}} \times \sqrt{\frac{8RT}{\pi M}} \]

4. **Square both sides**:
Squaring both sides to eliminate the square roots gives:
\[ \frac{3RT}{M} = \frac{a+5}{a} \times \frac{8RT}{\pi M} \]

5. **Cancel out common terms**:
The \(RT/M\) terms can be canceled from both sides:
\[ 3 = \frac{a+5}{a} \times \frac{8}{\pi} \]

6. **Rearranging the equation**:
Rearranging gives:
\[ 3 = \frac{8(a+5)}{\pi a} \]

7. **Substituting the value of \(\pi\)**:
Using \(\pi = \frac{22}{7}\), we have:
\[ 3 = \frac{8(a+5)}{\frac{22}{7} a} \]
Multiplying both sides by \(\frac{22}{7} a\) gives:
\[ 3 \times \frac{22}{7} a = 8(a + 5) \]

8. **Simplifying**:
This simplifies to:
\[ \frac{66}{7} a = 8a + 40 \]
Multiplying through by 7 to eliminate the fraction:
\[ 66a = 56a + 280 \]

9. **Solving for \(a\)**:
Rearranging gives:
\[ 66a - 56a = 280 \]
\[ 10a = 280 \]
\[ a = 28 \]

### Final Answer:
The value of \(a\) is \(28\).