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 follow these steps:

### Step 1: Understand the formula for RMS speed
The root mean square speed (v_rms) of gas molecules 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.

### Step 2: Identify the initial conditions
From the problem, we have:
- Initial temperature \( T_1 = 25 \, K \)
- Initial pressure \( P_1 = 1.5 \, bar \)
- Initial RMS speed \( v_{rms1} = 100 \, m/s \)

### Step 3: Identify the new conditions
The new conditions are:
- New temperature \( T_2 = 100 \, K \)
- New pressure \( P_2 = 6.0 \, bar \)

### Step 4: Relate the initial and new RMS speeds
Since the molar mass \( M \) and the gas constant \( R \) remain constant, we can express the ratio of the RMS speeds at the two different temperatures:
\[ \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{T_1}{T_2}} \]

### Step 5: Substitute the known values
Substituting the known values into the equation:
\[ \frac{100}{v_{rms2}} = \sqrt{\frac{25}{100}} \]
This simplifies to:
\[ \frac{100}{v_{rms2}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 6: Solve for \( v_{rms2} \)
Cross-multiplying gives:
\[ 100 = \frac{1}{2} v_{rms2} \]
Thus,
\[ v_{rms2} = 200 \, m/s \]

### Final Answer
The new root mean square speed of the gas molecules at 100 K and 6.0 bar is \( 200 \, m/s \).

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