The root mean square speed of gas molecules at a temperature `27K` and pressure 1.5 bar is `1 xx 10^(4) cm//sec` If both temperature and pressure are raised three times calculate the new rms speed of gas molecules .

Options

A. `9 xx 10^4 cm//s^2`
B. `3 xx 10^4 cm//s `
C. `sqrt(3) xx 10^4 cm//s `
D. `1 xx 10^4 cm//s `

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while providing a clear and structured approach.

### Step 1: Identify Initial Conditions
The initial conditions given are:
- Initial temperature \( T_1 = 27 \, \text{K} \)
- Initial pressure \( P_1 = 1.5 \, \text{bar} \)
- Initial root mean square speed \( U_{\text{rms1}} = 1 \times 10^4 \, \text{cm/s} \)

### Step 2: Determine New Conditions
Both temperature and pressure are raised three times:
- New temperature \( T_2 = 3 \times T_1 = 3 \times 27 = 81 \, \text{K} \)
- New pressure \( P_2 = 3 \times P_1 = 3 \times 1.5 = 4.5 \, \text{bar} \)

### Step 3: Understand the Relationship of RMS Speed
The root mean square speed of gas molecules is given by the formula:
\[ U_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molar mass of the gas.

From this formula, we can see that the RMS speed depends on the temperature but not on the pressure.

### Step 4: Set Up the Equations
For the initial and final conditions, we can write:
1. \( U_{\text{rms1}} = \sqrt{\frac{3RT_1}{M}} \)
2. \( U_{\text{rms2}} = \sqrt{\frac{3RT_2}{M}} \)

### Step 5: Divide the Equations
Dividing the first equation by the second gives:
\[ \frac{U_{\text{rms1}}}{U_{\text{rms2}}} = \sqrt{\frac{T_1}{T_2}} \]

### Step 6: Substitute Known Values
Now substituting the known values:
- \( U_{\text{rms1}} = 1 \times 10^4 \, \text{cm/s} \)
- \( T_1 = 27 \, \text{K} \)
- \( T_2 = 81 \, \text{K} \)

We have:
\[ \frac{1 \times 10^4}{U_{\text{rms2}}} = \sqrt{\frac{27}{81}} \]

### Step 7: Simplify the Square Root
Calculating the square root:
\[ \sqrt{\frac{27}{81}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \]

### Step 8: Solve for \( U_{\text{rms2}} \)
Rearranging the equation gives:
\[ U_{\text{rms2}} = 1 \times 10^4 \times \sqrt{3} \]

### Step 9: Final Calculation
Thus, the new RMS speed is:
\[ U_{\text{rms2}} = \sqrt{3} \times 10^4 \, \text{cm/s} \]

### Final Answer
The new root mean square speed of gas molecules is:
\[ U_{\text{rms2}} = \sqrt{3} \times 10^4 \, \text{cm/s} \]