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

Explanation

To solve the problem, we need to understand the relationship between pressure, mass of the gas molecules, and their root mean square (rms) speed.

### Step-by-Step Solution:

1. **Understanding the Pressure Formula**:
The pressure \( P \) of a gas can be expressed using the formula:
\[ P = \frac{1}{3} \rho v_{rms}^2 \]
where \( \rho \) is the density of the gas and \( v_{rms} \) is the root mean square speed of the gas molecules.

2. **Relating Density to Mass**:
The density \( \rho \) can be expressed in terms of mass \( m \) and volume \( V \):
\[ \rho = \frac{m}{V} \]
Substituting this into the pressure formula gives:
\[ P = \frac{1}{3} \left(\frac{m}{V}\right) v_{rms}^2 \]

3. **Identifying Changes in Mass and Speed**:
According to the problem, the mass of the molecules is halved:
\[ m' = \frac{m}{2} \]
and the rms speed is doubled:
\[ v_{rms}' = 2 v_{rms} \]

4. **Substituting New Values into the Pressure Formula**:
The new pressure \( P' \) can be calculated using the new mass and new rms speed:
\[ P' = \frac{1}{3} \left(\frac{m'}{V}\right) (v_{rms}')^2 \]
Substituting the new values:
\[ P' = \frac{1}{3} \left(\frac{\frac{m}{2}}{V}\right) (2 v_{rms})^2 \]

5. **Simplifying the Expression**:
Now simplify the expression:
\[ P' = \frac{1}{3} \left(\frac{\frac{m}{2}}{V}\right) (4 v_{rms}^2) \]
\[ P' = \frac{1}{3} \left(\frac{m}{2V}\right) (4 v_{rms}^2) \]
\[ P' = \frac{4}{2} \cdot \frac{1}{3} \left(\frac{m}{V}\right) v_{rms}^2 \]
\[ P' = 2 \cdot P \]

6. **Final Result**:
Since the initial pressure was \( P_0 \), we have:
\[ P' = 2 P_0 \]

Thus, the resultant pressure would be \( 2 P_0 \).