The molecules of a given mass of gas have an rms velocity of `200m s^(-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 solve the problem, we will use the relationship between the root mean square (rms) velocity of gas molecules, temperature, and pressure. The rms velocity is given by the formula:

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

Where:
- \( V_{rms} \) is the root mean square velocity,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( M \) is the molar mass of the gas.

However, since we are comparing two states of the same gas, we can use the proportionality of rms velocity to the square root of temperature, independent of pressure.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Given:
- Initial rms velocity, \( V_1 = 200 \, \text{m/s} \)
- Initial temperature, \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- Initial pressure, \( P_1 = 1 \, \text{atm} \)

2. **Identify Final Conditions**:
- Given:
- Final temperature, \( T_2 = 127^\circ C = 127 + 273 = 400 \, K \)
- Final pressure, \( P_2 = 2 \, \text{atm} \)

3. **Use the Proportionality of rms Velocity**:
- The relationship between the initial and final rms velocities can be expressed as:
\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \]
- Rearranging gives:
\[ V_2 = V_1 \cdot \sqrt{\frac{T_2}{T_1}} \]

4. **Substituting Values**:
- Substitute the known values into the equation:
\[ V_2 = 200 \cdot \sqrt{\frac{400}{300}} \]

5. **Calculate the Square Root**:
- Simplifying the fraction:
\[ \frac{400}{300} = \frac{4}{3} \]
- Therefore:
\[ V_2 = 200 \cdot \sqrt{\frac{4}{3}} = 200 \cdot \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, \text{m/s} \]

6. **Final Result**:
- The final rms velocity \( V_2 \) is:
\[ V_2 = \frac{400}{\sqrt{3}} \, \text{m/s} \]