The R.M.S. velocity of the molecules in a gas at 27 °C is 300 m/s. The R.M.S. velocity of the molecules in the same gas at 927 °C is

Options

A. 1200 m/s
B. 600 m/s
C. 150 m/s
D. 75 m/s

Answer

To solve the problem, we need to find the root mean square (R.M.S.) velocity of the gas molecules at a temperature of 927 °C, given that the R.M.S. velocity at 27 °C is 300 m/s.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- The temperature in Kelvin is calculated by adding 273 to the Celsius temperature.
- For 27 °C:
\[ T_1 = 27 + 273 = 300 \, \text{K} \]
- For 927 °C:
\[ T_2 = 927 + 273 = 1200 \, \text{K} \]

2. **Use the R.M.S. Velocity Formula**:
- The R.M.S. velocity (\(V_{rms}\)) of gas molecules is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
- Where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas.

3. **Set Up the Ratio of R.M.S. Velocities**:
- Since we are dealing with the same gas, the molar mass \(M\) remains constant. Therefore, we can set up the ratio of the R.M.S. velocities at two different temperatures:
\[ \frac{V_{rms2}}{V_{rms1}} = \sqrt{\frac{T_2}{T_1}} \]

4. **Substitute Known Values**:
- We know \(V_{rms1} = 300 \, \text{m/s}\), \(T_1 = 300 \, \text{K}\), and \(T_2 = 1200 \, \text{K}\).
- Plugging these values into the equation gives:
\[ V_{rms2} = V_{rms1} \times \sqrt{\frac{T_2}{T_1}} = 300 \times \sqrt{\frac{1200}{300}} \]

5. **Calculate the Square Root**:
- Simplifying \(\frac{1200}{300}\) gives:
\[ \frac{1200}{300} = 4 \]
- Therefore:
\[ \sqrt{4} = 2 \]

6. **Final Calculation**:
- Now substituting back:
\[ V_{rms2} = 300 \times 2 = 600 \, \text{m/s} \]

### Conclusion:
The R.M.S. velocity of the molecules in the gas at 927 °C is **600 m/s**.