The root mean square of gas molecules at 25 K and 1.5 bar is `"100 m s"^(-1)`. If the temperature is raised to 100 K and the pressure to 6.0 bar, the root mean square speed becomes

Options

A. `200ms^(-1)`
B. `100ms^(-1)`
C. `400ms^(-1)`
D. `1600ms^(-1)`

Answer

To solve the problem of finding the new root mean square (RMS) speed of gas molecules when the temperature and pressure are changed, we can use the formula for the root mean square speed of gas molecules:

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

where:
- \( v_{rms} \) = root mean square speed
- \( R \) = universal gas constant
- \( T \) = absolute temperature in Kelvin
- \( M \) = molar mass of the gas

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial temperature \( T_1 = 25 \, K \)
- Initial pressure \( P_1 = 1.5 \, bar \)
- Initial \( v_{rms1} = 100 \, m/s \)

2. **Identify Final Conditions:**
- Final temperature \( T_2 = 100 \, K \)
- Final pressure \( P_2 = 6.0 \, bar \)

3. **Use the RMS Speed Formula:**
The formula for the root mean square speed can be expressed as:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
Since the gas is the same, the molar mass \( M \) remains constant.

4. **Set Up the Ratio of RMS Speeds:**
Using the RMS speed formula for both conditions, we can set up the following ratio:
\[ \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{T_1}{T_2}} \]
Rearranging gives:
\[ v_{rms2} = v_{rms1} \cdot \sqrt{\frac{T_2}{T_1}} \]

5. **Substitute Known Values:**
Substitute \( v_{rms1} = 100 \, m/s \), \( T_1 = 25 \, K \), and \( T_2 = 100 \, K \):
\[ v_{rms2} = 100 \cdot \sqrt{\frac{100}{25}} = 100 \cdot \sqrt{4} = 100 \cdot 2 = 200 \, m/s \]

6. **Final Answer:**
The new root mean square speed \( v_{rms2} \) is \( 200 \, m/s \).