If the mass of each molecule of a gas is halved and speed is doubled. Find the ratio of initial and final pressure.

Options

A. `1/3`
B. `1/2`
C. `1/5`
D. `1/4`

Answer

To solve the problem, we need to find the ratio of the initial pressure (P1) to the final pressure (P2) when the mass of each molecule of a gas is halved and the speed is doubled.

### Step-by-Step Solution:

1. **Understand the given parameters**:
- Let the mass of each molecule of gas initially be \( m_1 \).
- The final mass of each molecule is \( m_2 = \frac{m_1}{2} \) (halved).
- Let the initial speed of the molecules be \( c_1 \).
- The final speed of the molecules is \( c_2 = 2c_1 \) (doubled).
- The volume \( V \) remains constant.

2. **Use the formula for pressure**:
The pressure of an ideal gas can be expressed as:
\[ P = \frac{1}{3} \cdot \frac{N \cdot m \cdot c^2}{V} \]
where \( N \) is the number of molecules, \( m \) is the mass of each molecule, \( c \) is the speed of the molecules, and \( V \) is the volume.

3. **Calculate the initial pressure \( P_1 \)**:
Using the initial parameters:
\[ P_1 = \frac{1}{3} \cdot \frac{N \cdot m_1 \cdot c_1^2}{V} \]

4. **Calculate the final pressure \( P_2 \)**:
Using the final parameters:
\[ P_2 = \frac{1}{3} \cdot \frac{N \cdot m_2 \cdot c_2^2}{V} \]
Substituting \( m_2 \) and \( c_2 \):
\[ P_2 = \frac{1}{3} \cdot \frac{N \cdot \left(\frac{m_1}{2}\right) \cdot (2c_1)^2}{V} \]
Simplifying \( P_2 \):
\[ P_2 = \frac{1}{3} \cdot \frac{N \cdot \left(\frac{m_1}{2}\right) \cdot 4c_1^2}{V} \]
\[ P_2 = \frac{2}{3} \cdot \frac{N \cdot m_1 \cdot c_1^2}{V} \]

5. **Find the ratio of initial and final pressure**:
Now, we can find the ratio \( \frac{P_1}{P_2} \):
\[ \frac{P_1}{P_2} = \frac{\frac{1}{3} \cdot \frac{N \cdot m_1 \cdot c_1^2}{V}}{\frac{2}{3} \cdot \frac{N \cdot m_1 \cdot c_1^2}{V}} \]
Simplifying the ratio:
\[ \frac{P_1}{P_2} = \frac{1}{2} \]

### Final Answer:
The ratio of initial pressure to final pressure is:
\[ \frac{P_1}{P_2} = \frac{1}{2} \]