The molecules of a given mass of a gas have rms velocity of `200 m//s at 27^(@)C and 1.0 xx 10^(5) N//m_(2)` pressure. When the temperature and pressure of the gas are respectively `127^(@)C and 0.05 xx 10^(5) Nm^(-2)`, the rms velocity of its molecules in `ms^(-1)` is

Options

A. `200 ms^(-1)`
B. `400 ms^(-1)`
C. `200sqrt2ms^(-1)`
D. `400/sqrt3ms^(-1)`

Explanation

To find the RMS velocity of the gas molecules at the new temperature and pressure, we can use the relationship between RMS velocity, temperature, and molar mass. The formula for RMS velocity (Vrms) is given by:

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

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
- 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 RMS velocity equations for both states
- For the initial state:
\[ V_{\text{rms1}} = \sqrt{\frac{3RT_1}{M}} = 200 \, m/s \]
- For the final state:
\[ V_{\text{rms2}} = \sqrt{\frac{3RT_2}{M}} \]

### Step 3: Set up the ratio of the two RMS velocities
Taking the ratio of the two equations:
\[ \frac{V_{\text{rms2}}}{V_{\text{rms1}}} = \sqrt{\frac{T_2}{T_1}} \]

### Step 4: Substitute the known values
Substituting \( T_1 = 300 \, K \) and \( T_2 = 400 \, K \):
\[ \frac{V_{\text{rms2}}}{200} = \sqrt{\frac{400}{300}} \]

### Step 5: Simplify the ratio
Calculating the right side:
\[ \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \]

### Step 6: Solve for \( V_{\text{rms2}} \)
Now, we can find \( V_{\text{rms2}} \):
\[ V_{\text{rms2}} = 200 \cdot \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m/s \]

### Final Answer
Thus, the RMS velocity of the gas molecules at the new temperature and pressure is:
\[ V_{\text{rms2}} = \frac{400}{\sqrt{3}} \, m/s \]