If the pressure of a gas is reduced to half and temperature is doubled, its volume becomes

Options

A. `(V)/(4)`
B. `2V^(3)`
C. 6V
D. 4 V

Explanation

To solve the problem, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of gas
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
Let the initial pressure be \( P \), the initial volume be \( V \), and the initial temperature be \( T \).

2. **Apply Changes**:
According to the problem:
- The pressure is reduced to half: \( P' = \frac{P}{2} \)
- The temperature is doubled: \( T' = 2T \)

3. **Use the Ideal Gas Law**:
The Ideal Gas Law for the initial state is:
\[ PV = nRT \]
For the new state, the Ideal Gas Law becomes:
\[ P'V' = nRT' \]

4. **Substitute the New Conditions**:
Substitute \( P' \) and \( T' \) into the new state equation:
\[ \left(\frac{P}{2}\right)V' = nR(2T) \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ V' = \frac{nR(2T)}{\frac{P}{2}} = \frac{2nRT}{\frac{P}{2}} = \frac{2nRT \cdot 2}{P} = \frac{4nRT}{P} \]

6. **Relate to the Initial Volume**:
From the initial state, we know:
\[ V = \frac{nRT}{P} \]
So we can express \( V' \) in terms of \( V \):
\[ V' = 4 \left(\frac{nRT}{P}\right) = 4V \]

### Conclusion:
Thus, the new volume \( V' \) becomes:
\[ V' = 4V \]

### Final Answer:
The volume becomes 4 times the original volume, or \( V' = 4V \).

---