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

Answer

To solve the problem, we will analyze the relationship between pressure, mass, and the root mean square (rms) speed of gas molecules. We will use the formula for pressure in terms of density and rms speed.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Let the initial mass of the gas molecules be \( m_i \).
- The initial rms speed is \( v_{rms, initial} = v_i \).
- The initial pressure is given as \( P_0 \).

2. **Determine New Conditions:**
- The mass of the molecules is halved:
\[ m_f = \frac{m_i}{2} \]
- The rms speed is doubled:
\[ v_{rms, final} = 2v_i \]

3. **Use the Pressure Formula:**
The pressure of the gas can be expressed as:
\[ P = \frac{1}{3} \rho v_{rms}^2 \]
where \( \rho \) (density) can be expressed as:
\[ \rho = \frac{m}{V} = \frac{n \cdot m}{V} \]
Here, \( n \) is the number of molecules, \( m \) is the mass of one molecule, and \( V \) is the volume.

4. **Relate Initial and Final Pressures:**
The relationship between initial and final pressures can be expressed as:
\[ \frac{P_{initial}}{P_{final}} = \frac{m_{initial}}{m_{final}} \cdot \frac{v_{initial}^2}{v_{final}^2} \]

5. **Substituting Values:**
Substitute the known values into the equation:
\[ \frac{P_0}{P_f} = \frac{m_i}{\frac{m_i}{2}} \cdot \frac{v_i^2}{(2v_i)^2} \]
Simplifying this gives:
\[ \frac{P_0}{P_f} = \frac{m_i}{\frac{m_i}{2}} \cdot \frac{v_i^2}{4v_i^2} \]
\[ = 2 \cdot \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]

6. **Finding the Final Pressure:**
Rearranging gives:
\[ P_f = 2P_0 \]

### Final Result:
The resultant pressure \( P_f \) is:
\[ P_f = 2P_0 \]