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

Answer

To solve the problem, we need to analyze how the changes in the mass of the gas molecules and their speeds affect the pressure of the gas.

### Step-by-Step Solution:

1. **Understand the relationship of pressure with mass and speed**:
The pressure \( P \) of a gas can be expressed as:
\[ P = \frac{1}{3} \times \text{(number of molecules)} \times \text{(mass of each molecule)} \times \text{(vrms)}^2 \div \text{(volume)} \]
Here, \( \text{vrms} \) is the root mean square speed of the gas molecules.

2. **Identify the initial conditions**:
Let the initial pressure be \( P_0 \), the mass of each molecule be \( m_1 \), and the initial speed be \( v_1 \). Thus, we can write:
\[ P_0 = k \cdot m_1 \cdot v_1^2 \]
where \( k \) is a constant that includes the number of molecules and volume.

3. **Identify the final conditions**:
According to the problem, the mass of each molecule is halved, so:
\[ m_2 = \frac{m_1}{2} \]
The speed of each molecule is doubled, so:
\[ v_2 = 2v_1 \]

4. **Express the final pressure**:
The final pressure \( P_2 \) can be expressed as:
\[ P_2 = k \cdot m_2 \cdot v_2^2 \]
Substituting the values of \( m_2 \) and \( v_2 \):
\[ P_2 = k \cdot \left(\frac{m_1}{2}\right) \cdot (2v_1)^2 \]

5. **Simplify the expression for \( P_2 \)**:
\[ P_2 = k \cdot \left(\frac{m_1}{2}\right) \cdot (4v_1^2) \]
\[ P_2 = k \cdot m_1 \cdot v_1^2 \cdot \frac{4}{2} \]
\[ P_2 = k \cdot m_1 \cdot v_1^2 \cdot 2 \]
Since \( k \cdot m_1 \cdot v_1^2 = P_0 \), we can substitute:
\[ P_2 = 2P_0 \]

6. **Conclusion**:
The resulting pressure \( P \) after halving the mass and doubling the speed of the gas molecules is:
\[ P = 2P_0 \]

### Final Answer:
The resulting pressure \( P \) will be equal to \( 2P_0 \).

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