The root mean square velocity of a gas molecule at 100 K and 0.5 atm pressure is `106.4 m s^(-1)`. If the temperature is rasied to 400 K and the pressure is raised to 2 atm, then root mean square velocity becomes

Options

A. `106.4 ms^(-1)`
B. `425.6 ms^(-1)`
C. `212.8 ms^(-1)`
D. `851.2ms^(-1)`

Explanation

To solve the problem of finding the root mean square (RMS) velocity of a gas molecule when the temperature and pressure are changed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for RMS Velocity**:
The root mean square velocity (v_rms) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) = universal gas constant
- \( T \) = absolute temperature in Kelvin
- \( M \) = molar mass of the gas (in kg/mol)

2. **Identify Initial Conditions**:
From the problem, we have:
- Initial temperature \( T_1 = 100 \, K \)
- Initial pressure \( P_1 = 0.5 \, atm \)
- Initial RMS velocity \( v_1 = 106.4 \, m/s \)

3. **Identify Final Conditions**:
The final conditions are:
- Final temperature \( T_2 = 400 \, K \)
- Final pressure \( P_2 = 2 \, atm \)
- Final RMS velocity \( v_2 \) (which we need to find)

4. **Use the Relationship of RMS Velocity with Temperature**:
The RMS velocity is directly proportional to the square root of the temperature, so we can write:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]

5. **Substitute Known Values**:
Plugging in the known values:
\[ \frac{106.4}{v_2} = \sqrt{\frac{100}{400}} \]

6. **Simplify the Right Side**:
Calculate the square root:
\[ \sqrt{\frac{100}{400}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

7. **Set Up the Equation**:
Now we have:
\[ \frac{106.4}{v_2} = \frac{1}{2} \]

8. **Cross Multiply to Solve for \( v_2 \)**:
Cross multiplying gives:
\[ 106.4 = \frac{1}{2} v_2 \implies v_2 = 2 \times 106.4 \]

9. **Calculate \( v_2 \)**:
\[ v_2 = 212.8 \, m/s \]

### Final Answer:
The root mean square velocity at 400 K and 2 atm pressure is \( 212.8 \, m/s \).