For a given mass of a gas, if pressure is reduced to half and temperature is doubled, then volume V will become

Options

A. `4V`
B. `2V^2`
C. `V/4`
D. `8V`.

Explanation

To solve the problem, we will use the Ideal Gas Law, which states that for a given mass of gas, the relationship between pressure (P), volume (V), and temperature (T) can be expressed as:

\[ PV = nRT \]

Where:
- \( P \) = Pressure of the gas
- \( V \) = Volume of the gas
- \( n \) = Number of moles of the gas (constant for a given mass)
- \( R \) = Universal gas constant
- \( T \) = Temperature of the gas in Kelvin

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Let the initial pressure be \( P_1 = P \)
- Let the initial volume be \( V_1 = V \)
- Let the initial temperature be \( T_1 = T \)

2. **Identify Final Conditions**:
- The pressure is reduced to half:
\[ P_2 = \frac{P}{2} \]
- The temperature is doubled:
\[ T_2 = 2T \]

3. **Use the Ideal Gas Law**:
- According to the Ideal Gas Law, we can write:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

4. **Substitute Known Values**:
- Substitute \( P_1, V_1, T_1, P_2, T_2 \) into the equation:
\[ \frac{P \cdot V}{T} = \frac{\left(\frac{P}{2}\right) V_2}{2T} \]

5. **Simplify the Equation**:
- Rearranging gives:
\[ \frac{PV}{T} = \frac{P V_2}{4T} \]
- Cancel \( P \) and \( T \) from both sides:
\[ V = \frac{V_2}{4} \]

6. **Solve for \( V_2 \)**:
- Rearranging gives:
\[ V_2 = 4V \]

### Conclusion:
The final volume \( V_2 \) will be \( 4V \).