The rms velocity of an ideal gas at `27^@C` is `0.3 ms^(-1)`. Its rms velocity at `927^@C` (in `ms^(-1)`) is:

Options

A. `3.0`
B. `2.4`
C. `0.9`
D. `0.6`

Explanation

To find the root mean square (rms) velocity of an ideal gas at a temperature of 927°C, given that its rms velocity at 27°C is 0.3 m/s, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- First, we need to convert the given temperatures from Celsius to Kelvin.
- For 27°C:
\[ T_1 = 27 + 273 = 300 \, K \]
- For 927°C:
\[ T_2 = 927 + 273 = 1200 \, K \]

2. **Understand the Relationship:**
- The rms velocity (\(v_{rms}\)) of an ideal gas is directly proportional to the square root of the absolute temperature (in Kelvin). This can be expressed as:
\[ v_{rms} \propto \sqrt{T} \]
- Therefore, we can write the relationship between the rms velocities at two different temperatures:
\[ \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{T_1}{T_2}} \]

3. **Substitute Known Values:**
- Substitute the known values into the equation:
\[ \frac{0.3}{v_{rms2}} = \sqrt{\frac{300}{1200}} \]

4. **Calculate the Square Root:**
- Simplify the fraction inside the square root:
\[ \frac{300}{1200} = \frac{1}{4} \]
- Now, take the square root:
\[ \sqrt{\frac{1}{4}} = \frac{1}{2} \]

5. **Set Up the Equation:**
- Now we can set up the equation:
\[ \frac{0.3}{v_{rms2}} = \frac{1}{2} \]

6. **Solve for \(v_{rms2}\):**
- Cross-multiply to solve for \(v_{rms2}\):
\[ 0.3 = \frac{1}{2} v_{rms2} \]
- Multiply both sides by 2:
\[ v_{rms2} = 0.3 \times 2 = 0.6 \, m/s \]

### Final Answer:
The rms velocity of the ideal gas at 927°C is \(0.6 \, m/s\).

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