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 new temperature, we can use the relationship between rms velocity and temperature. The formula for rms velocity (\(v_{rms}\)) is given by:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

where:
- \(R\) is the gas constant,
- \(T\) is the absolute temperature in Kelvin,
- \(M\) is the molar mass of the gas.

### Step 1: Convert temperatures from Celsius to Kelvin
We need to convert the given temperatures from Celsius to Kelvin using the formula:

\[ T(K) = T(°C) + 273 \]

- For \(27°C\):
\[ T_1 = 27 + 273 = 300 \, K \]

- For \(927°C\):
\[ T_2 = 927 + 273 = 1200 \, K \]

### Step 2: Write the rms velocity equations for both temperatures
Using the rms velocity formula, we can write:

1. For \(27°C\):
\[ v_{rms1} = \sqrt{\frac{3R \cdot 300}{M}} \]

2. For \(927°C\):
\[ v_{rms2} = \sqrt{\frac{3R \cdot 1200}{M}} \]

### Step 3: Set up the ratio of the rms velocities
To find the new rms velocity, we can set up the ratio of the two rms velocities:

\[ \frac{v_{rms2}}{v_{rms1}} = \frac{\sqrt{\frac{3R \cdot 1200}{M}}}{\sqrt{\frac{3R \cdot 300}{M}}} \]

### Step 4: Simplify the ratio
Since \(3R\) and \(M\) are common in both the numerator and denominator, they cancel out:

\[ \frac{v_{rms2}}{v_{rms1}} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2 \]

### Step 5: Calculate \(v_{rms2}\)
Now, we can find \(v_{rms2}\) using the known value of \(v_{rms1}\):

\[ v_{rms2} = 2 \cdot v_{rms1} = 2 \cdot 0.3 \, m/s = 0.6 \, m/s \]

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

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