The molecules of a given mass of a gas have root mean square speeds of `100 ms^(-1) "at " 27^(@)C` and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at `127^(@)C` and 2.0 atmospheric pressure?

Options

A. `(200)/(sqrt3) ms^(-1)`
B. `(100)/(sqrt3)ms^(-1)`
C. `(400)/(3)ms^(-1)`
D. `(200)/(3)ms^(-1)`

Explanation

To solve the problem, we will use the relationship between the root mean square speed of gas molecules and temperature, as well as the ideal gas law. The root mean square speed (\(v_{rms}\)) 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 we are comparing two different states of the same gas, the molar mass \(M\) and the gas constant \(R\) remain constant. Therefore, we can express the relationship between the root mean square speeds at two different temperatures as follows:

\[ \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{T1}{T2}} \]

### Step-by-Step Solution:

1. **Convert the temperatures from Celsius to Kelvin:**
- For \(T1 = 27^\circ C\):
\[ T1 = 27 + 273 = 300 \, K \]
- For \(T2 = 127^\circ C\):
\[ T2 = 127 + 273 = 400 \, K \]

2. **Identify the given values:**
- \(v_{rms1} = 100 \, m/s\)
- \(T1 = 300 \, K\)
- \(T2 = 400 \, K\)

3. **Set up the ratio of root mean square speeds:**
\[ \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{T1}{T2}} \]

4. **Substitute the known values into the equation:**
\[ \frac{100}{v_{rms2}} = \sqrt{\frac{300}{400}} \]

5. **Simplify the right side of the equation:**
\[ \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

6. **Rearrange the equation to solve for \(v_{rms2}\):**
\[ 100 = v_{rms2} \cdot \frac{\sqrt{3}}{2} \]
\[ v_{rms2} = \frac{100 \cdot 2}{\sqrt{3}} = \frac{200}{\sqrt{3}} \, m/s \]

### Final Answer:
\[ v_{rms2} = \frac{200}{\sqrt{3}} \, m/s \]