At certain temperature, the r.m.s. velocity for `CH_4` gas molecules is 100 m/sec. This velocity for `SO_2` gas molecules at same temperature will be

Options

A. 400 m/sec
B. 200 m/sec
C. 50 m/sec
D. 25 m/sec

Answer

To find the root mean square (r.m.s.) velocity of \( SO_2 \) gas molecules at the same temperature where the r.m.s. velocity of \( CH_4 \) gas molecules is given as 100 m/s, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for r.m.s. velocity**:
The r.m.s. velocity (\( v_{rms} \)) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molar mass of the gas.

2. **Write the r.m.s. velocity for \( CH_4 \)**:
For methane (\( CH_4 \)), we can express the r.m.s. velocity as:
\[ v_{rms, CH_4} = \sqrt{\frac{3RT}{M_{CH_4}}} \]
Given that \( v_{rms, CH_4} = 100 \, \text{m/s} \), we can denote this as Equation (1).

3. **Write the r.m.s. velocity for \( SO_2 \)**:
For sulfur dioxide (\( SO_2 \)), the r.m.s. velocity can be expressed as:
\[ v_{rms, SO_2} = \sqrt{\frac{3RT}{M_{SO_2}}} \]
We will denote this as Equation (2).

4. **Divide the two equations**:
To find the relationship between the r.m.s. velocities of the two gases, we can divide Equation (1) by Equation (2):
\[ \frac{v_{rms, CH_4}}{v_{rms, SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{CH_4}}} \]

5. **Substitute known values**:
The molar mass of \( CH_4 \) is 16 g/mol and the molar mass of \( SO_2 \) is 64 g/mol. Substituting these values gives:
\[ \frac{100}{v_{rms, SO_2}} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2 \]

6. **Solve for \( v_{rms, SO_2} \)**:
Rearranging the equation gives:
\[ v_{rms, SO_2} = \frac{100}{2} = 50 \, \text{m/s} \]

### Final Answer:
The r.m.s. velocity for \( SO_2 \) gas molecules at the same temperature is **50 m/s**.