The average velocity of an ideal gas at `0^(@)C` is `0.4 ms^(-1)`. If the temperature of gas is increased to `546^(@)C` its average velocity will be

Options

A. `0.8 m s^(-1)`
B. `1.6 ms^(-1)`
C. `0.346 ms^(-1)`
D. `0.69 ms^(-1)`

Explanation

To solve the problem of finding the average velocity of an ideal gas when its temperature is increased, we can follow these steps:

### Step 1: Understand the relationship between average velocity and temperature
The average velocity (\( \bar{v} \)) of an ideal gas is related to its temperature (T) through the equation:
\[ \bar{v} \propto \sqrt{T} \]
This means that the average velocity is proportional to the square root of the absolute temperature.

### Step 2: Convert the given temperatures to Kelvin
- The initial temperature is given as \(0^\circ C\):
\[ T_1 = 0 + 273 = 273 \, K \]
- The final temperature is given as \(546^\circ C\):
\[ T_2 = 546 + 273 = 819 \, K \]

### Step 3: Set up the ratio of average velocities
Using the proportionality established in Step 1, we can write:
\[ \frac{\bar{v_1}}{\bar{v_2}} = \sqrt{\frac{T_1}{T_2}} \]
Where:
- \( \bar{v_1} = 0.4 \, m/s \) (initial average velocity)
- \( \bar{v_2} \) is the final average velocity we need to find.

### Step 4: Substitute the values into the equation
Substituting the known values into the equation gives:
\[ \frac{0.4}{\bar{v_2}} = \sqrt{\frac{273}{819}} \]

### Step 5: Solve for \( \bar{v_2} \)
First, calculate the right side:
\[ \sqrt{\frac{273}{819}} \approx \sqrt{0.333} \approx 0.577 \]
Now, we can rearrange the equation:
\[ \bar{v_2} = \frac{0.4}{0.577} \approx 0.693 \, m/s \]

### Step 6: Final answer
Thus, the average velocity of the gas when the temperature is increased to \(546^\circ C\) is approximately:
\[ \bar{v_2} \approx 0.69 \, m/s \]

### Summary
The average velocity of the ideal gas at \(546^\circ C\) is approximately \(0.69 \, m/s\).

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