The molecules of a given mass of gas have an rms velocity of `200 ms(-1)` at `27^@C` and pressure 1 atm. When the temperature is `127^@C` and pressure is 2 atm, the rms velocity in `m s^(-1)` will be ?

Options

A. `(100 sqrt(2))/(3)`
B. `100sqrt(2)`
C. `(400)/(sqrt3)`
D. None of these

Answer

To find the root mean square (rms) velocity of the gas at the new conditions, we can use the relationship between rms velocity and temperature. The formula for rms velocity (Vrms) is given by:

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

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

### Step 1: Convert the temperatures from Celsius to Kelvin
- Initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- Final temperature \( T_2 = 127^\circ C = 127 + 273 = 400 \, K \)

### Step 2: Write the ratio of the rms velocities
Since the rms velocity is directly proportional to the square root of the absolute temperature, we can write:

\[ \frac{V_{\text{rms},1}}{V_{\text{rms},2}} = \sqrt{\frac{T_1}{T_2}} \]

### Step 3: Substitute the known values
We know:
- \( V_{\text{rms},1} = 200 \, m/s \)
- \( T_1 = 300 \, K \)
- \( T_2 = 400 \, K \)

Substituting these values into the equation gives:

\[ \frac{200}{V_{\text{rms},2}} = \sqrt{\frac{300}{400}} \]

### Step 4: Simplify the right-hand side
Calculating the square root:

\[ \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Step 5: Rearranging to find \( V_{\text{rms},2} \)
Now we can rearrange the equation to solve for \( V_{\text{rms},2} \):

\[ V_{\text{rms},2} = 200 \cdot \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m/s \]

### Step 6: Calculate the numerical value
To get a numerical value, we can approximate \( \sqrt{3} \approx 1.732 \):

\[ V_{\text{rms},2} \approx \frac{400}{1.732} \approx 230.94 \, m/s \]

Thus, the rms velocity at \( 127^\circ C \) and \( 2 \, atm \) is approximately \( 230.94 \, m/s \).

### Final Answer
The rms velocity in \( m/s \) will be approximately \( 230.94 \, m/s \).