The rms velocity of an ideal gas at 300 K is 12240 `cms^(-1)`, what is the most probable velocity at that temperature?

Options

A. `10000`
B. `11280`
C. 1000
D. 12240

Answer

To find the most probable velocity of an ideal gas given its root mean square (rms) velocity, we can use the relationship between these two velocities. The formulas for rms velocity (v_rms) and most probable velocity (v_mp) are as follows:

1. **RMS Velocity Formula**:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin
- \( M \) = molar mass of the gas

2. **Most Probable Velocity Formula**:
\[ v_{mp} = \sqrt{\frac{2RT}{M}} \]

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Given \( v_{rms} = 12240 \, \text{cm/s} \)
- Temperature \( T = 300 \, \text{K} \)

**Step 2: Write the ratio of the two velocities.**
The relationship between rms velocity and most probable velocity can be expressed as:
\[ \frac{v_{rms}}{v_{mp}} = \sqrt{\frac{3}{2}} \]

**Step 3: Rearrange the equation to find the most probable velocity.**
From the ratio, we can express \( v_{mp} \) in terms of \( v_{rms} \):
\[ v_{mp} = v_{rms} \cdot \sqrt{\frac{2}{3}} \]

**Step 4: Substitute the known value of \( v_{rms} \).**
Now, substituting \( v_{rms} = 12240 \, \text{cm/s} \):
\[ v_{mp} = 12240 \cdot \sqrt{\frac{2}{3}} \]

**Step 5: Calculate \( \sqrt{\frac{2}{3}} \).**
Calculating \( \sqrt{\frac{2}{3}} \):
\[ \sqrt{\frac{2}{3}} \approx 0.8165 \]

**Step 6: Calculate \( v_{mp} \).**
Now, substituting this value back:
\[ v_{mp} \approx 12240 \cdot 0.8165 \approx 10000 \, \text{cm/s} \]

### Final Answer:
The most probable velocity at 300 K is approximately \( 10000 \, \text{cm/s} \).

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