A gas is filled in a container at pressure `P_(0)`. If the mass of molecules is halved and their rms speed is doubled, then the resultant pressure would be

Options

A. `2 P_(0)`
B. `4 P_(0)`
C. `(P_(0))/(4)`
D. `(P_(0))/(2)`

Explanation

To solve the problem, we will use the relationship between pressure, mass of gas molecules, and their root mean square (rms) speed according to the kinetic theory of gases.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Let the initial pressure of the gas be \( P_0 \).
- Let the mass of one molecule of the gas be \( m \).
- Let the initial rms speed of the gas be \( v_{\text{rms}} \).

2. **Changes in Conditions**:
- The mass of the molecules is halved:
\[ m' = \frac{m}{2} \]
- The rms speed is doubled:
\[ v'_{\text{rms}} = 2 v_{\text{rms}} \]

3. **Using the Kinetic Theory of Gases**:
- According to the kinetic theory, the pressure \( P \) of a gas is given by:
\[ P \propto \frac{m \cdot v_{\text{rms}}^2}{V} \]
- Since the volume \( V \) of the container is constant, we can express the relationship for the new pressure \( P' \) as:
\[ P' \propto m' \cdot (v'_{\text{rms}})^2 \]

4. **Substituting the New Values**:
- Substitute \( m' \) and \( v'_{\text{rms}} \) into the equation:
\[ P' = k \cdot m' \cdot (v'_{\text{rms}})^2 \]
- Where \( k \) is a constant of proportionality. Thus:
\[ P' = k \cdot \left(\frac{m}{2}\right) \cdot (2 v_{\text{rms}})^2 \]

5. **Calculating \( P' \)**:
- Expanding the equation:
\[ P' = k \cdot \left(\frac{m}{2}\right) \cdot (4 v_{\text{rms}}^2) \]
- Simplifying:
\[ P' = k \cdot \frac{m \cdot 4 v_{\text{rms}}^2}{2} = k \cdot 2 m v_{\text{rms}}^2 \]
- Since \( k \cdot m v_{\text{rms}}^2 = P_0 \), we have:
\[ P' = 2 P_0 \]

6. **Final Result**:
- Therefore, the resultant pressure \( P' \) is:
\[ P' = 2 P_0 \]

### Conclusion:
The resultant pressure after halving the mass of the molecules and doubling their rms speed is \( 2 P_0 \).