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
(a) `(400)/(sqrt(3))` (b) `(100sqrt(2))/(3)` (c) `(100)/(3)` (d) `100sqrt(2)`

Options

A. `(400)/(sqrt(3))`
B. `(100sqrt(2))/(3)`
C. `(100)/(3)`
D. `100sqrt(2)`

Explanation

To solve the problem, we will use the relationship between the root mean square (RMS) velocity of gas molecules and temperature. The RMS velocity 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.

Since \( R \) and \( M \) are constants for a given gas, we can say that the RMS velocity is directly proportional to the square root of the temperature:

\[ V_{rms} \propto \sqrt{T} \]

### Step 1: Convert temperatures to Kelvin
1. **Initial Temperature (T1)**:
- Given: \( 27^\circ C \)
- Convert to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

2. **Final Temperature (T2)**:
- Given: \( 127^\circ C \)
- Convert to Kelvin:
\[ T_2 = 127 + 273 = 400 \, K \]

### Step 2: Use the relationship of RMS velocities
Using the proportionality of RMS velocities to the square root of temperature, we have:

\[ \frac{V_{rms2}}{V_{rms1}} = \sqrt{\frac{T_2}{T_1}} \]

Where:
- \( V_{rms1} = 200 \, m/s \) (initial RMS velocity)
- \( V_{rms2} \) is the final RMS velocity we need to find.

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:

\[ \frac{V_{rms2}}{200} = \sqrt{\frac{400}{300}} \]

### Step 4: Simplify the right side
Calculate the square root:

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

### Step 5: Solve for \( V_{rms2} \)
Now, we can solve for \( V_{rms2} \):

\[ V_{rms2} = 200 \cdot \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m/s \]

### Final Answer
Thus, the RMS velocity of the gas molecules at the new temperature and pressure is:

\[ V_{rms2} = \frac{400}{\sqrt{3}} \, m/s \]

### Conclusion
The correct option is (a) \( \frac{400}{\sqrt{3}} \).

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