The r.m.s sped of the molecules of a gas at a pressure `10^5` Pa and temperature `0^@C` is 0.5km `sec^(-1)`. If the pressure is kept constant but temperature is raised to `819^@C`, the velocity will become

Options

A. 1.5 `kms^(-1)`
B. 2 `kms^(-1)`
C. 5 `kms^(-1)`
D. 1 `kms^(-1)`

Explanation

To solve the problem, we need to determine the new root mean square (r.m.s) speed of the gas molecules when the temperature is raised while keeping the pressure constant. Here's a step-by-step solution:

### Step 1: Understand the relationship between r.m.s speed and temperature
The r.m.s speed (V_rms) of gas molecules is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 2: Convert temperatures from Celsius to Kelvin
We need to convert the given temperatures from Celsius to Kelvin:
- For the initial temperature \( T_1 = 0^\circ C \):
\[ T_1 = 0 + 273 = 273 \, K \]
- For the final temperature \( T_2 = 819^\circ C \):
\[ T_2 = 819 + 273 = 1092 \, K \]

### Step 3: Write the relationship between the initial and final r.m.s speeds
Since the pressure is constant, we can use the relationship that the r.m.s speed is proportional to the square root of the temperature:
\[ \frac{V_{1}}{V_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} \]
This implies:
\[ V_{2} = V_{1} \cdot \sqrt{\frac{T_{2}}{T_{1}}} \]

### Step 4: Substitute the known values
We know:
- \( V_{1} = 0.5 \, km/s = 500 \, m/s \)
- \( T_{1} = 273 \, K \)
- \( T_{2} = 1092 \, K \)

Now substituting these values into the equation:
\[ V_{2} = 500 \cdot \sqrt{\frac{1092}{273}} \]

### Step 5: Calculate the square root and the final speed
First, calculate the ratio:
\[ \frac{1092}{273} \approx 4 \]
Then, take the square root:
\[ \sqrt{4} = 2 \]
Now substitute back to find \( V_{2} \):
\[ V_{2} = 500 \cdot 2 = 1000 \, m/s \]

### Step 6: Convert to kilometers per second
Finally, convert \( V_{2} \) back to kilometers per second:
\[ V_{2} = \frac{1000 \, m/s}{1000} = 1 \, km/s \]

### Final Answer
Thus, the new r.m.s speed of the gas molecules when the temperature is raised to \( 819^\circ C \) is:
\[ \boxed{1 \, km/s} \]