Gas at a pressure `P_(0)` in contained as a vessel. If the masses of all the molecules are halved and their speeds are doubles. The resulting pressure P will be equal to

Options

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

Explanation

To solve the problem, we need to understand how pressure is related to the mass of gas molecules and their speeds. The pressure of a gas can be expressed using the kinetic theory of gases, which states:

\[ P = \frac{1}{3} \frac{mN \bar{v}^2}{V} \]

where:
- \( P \) is the pressure,
- \( m \) is the mass of a single molecule,
- \( N \) is the number of molecules,
- \( \bar{v} \) is the root mean square speed of the molecules,
- \( V \) is the volume of the gas.

### Step-by-step Solution:

1. **Initial Pressure Calculation**:
The initial pressure \( P_0 \) is given by:
\[ P_0 = \frac{1}{3} \frac{m N v^2}{V} \]
where \( m \) is the initial mass of the molecules, \( N \) is the number of molecules, \( v \) is the initial speed of the molecules, and \( V \) is the volume.

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

3. **New Pressure Calculation**:
We now calculate the new pressure \( P \) with the modified mass and speed:
\[ P = \frac{1}{3} \frac{m' N (v')^2}{V} \]
Substituting the new values:
\[ P = \frac{1}{3} \frac{\left(\frac{m}{2}\right) N (2v)^2}{V} \]
Simplifying this:
\[ P = \frac{1}{3} \frac{\left(\frac{m}{2}\right) N (4v^2)}{V} \]
\[ P = \frac{1}{3} \frac{2m N v^2}{V} \]
\[ P = 2 \left(\frac{1}{3} \frac{m N v^2}{V}\right) \]
\[ P = 2 P_0 \]

4. **Final Result**:
Therefore, the resulting pressure \( P \) is:
\[ P = 2 P_0 \]

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