If masses of all molecule of a gas are halved and their speed doubled then the ratio of initial and final pressure will be

Options

A. `2 :1`
B. `1 : 2 `
C. `4 : 1`
D. `1 : 4`

Explanation

To solve the problem, we need to analyze how the pressure of a gas changes when the mass of its molecules is halved and their speed is doubled. We will use the kinetic gas equation to derive the ratio of the initial and final pressures.

### Step-by-Step Solution:

1. **Understand the Kinetic Gas Equation**:
The pressure \( P \) of a gas can be expressed using the kinetic gas equation:
\[ P = \frac{1}{3} \frac{m n}{V} v_{\text{rms}}^2 \]
where:
- \( P \) = pressure
- \( m \) = mass of each molecule
- \( n \) = total number of molecules
- \( V \) = volume
- \( v_{\text{rms}} \) = root mean square speed of the molecules.

2. **Initial Conditions**:
Let the initial mass of each molecule be \( m \) and the initial root mean square speed be \( v_{\text{rms}} \). The initial pressure \( P \) can be written as:
\[ P = \frac{1}{3} \frac{m n}{V} v_{\text{rms}}^2 \]

3. **New Conditions**:
According to the problem:
- The mass of each molecule is halved: \( m' = \frac{m}{2} \)
- The speed of the molecules is doubled: \( v'_{\text{rms}} = 2 v_{\text{rms}} \)

4. **Calculate Final Pressure**:
Substitute the new values into the kinetic gas equation to find the final pressure \( P' \):
\[ P' = \frac{1}{3} \frac{m' n}{V} (v'_{\text{rms}})^2 \]
Substituting \( m' \) and \( v'_{\text{rms}} \):
\[ P' = \frac{1}{3} \frac{\left(\frac{m}{2}\right) n}{V} (2 v_{\text{rms}})^2 \]
Simplifying this:
\[ P' = \frac{1}{3} \frac{\left(\frac{m}{2}\right) n}{V} (4 v_{\text{rms}}^2) = \frac{1}{3} \frac{m n}{V} \cdot \frac{4}{2} v_{\text{rms}}^2 = \frac{2}{3} \frac{m n}{V} v_{\text{rms}}^2 \]
Therefore:
\[ P' = 2P \]

5. **Calculate the Ratio of Initial to Final Pressure**:
The ratio of the initial pressure \( P \) to the final pressure \( P' \) is:
\[ \frac{P}{P'} = \frac{P}{2P} = \frac{1}{2} \]

### Conclusion:
The ratio of the initial pressure to the final pressure is \( \frac{1}{2} \).