If the aboslute temperature of a gas having volume `V cm^(3)` is doubled and the pressure is reduced to half, the final volume will be

Options

A. 0.25 V
B. 0.50 V
C. 40 V
D. 4 V

Explanation

To solve the problem, we will use the Ideal Gas Law, which states that \( PV = nRT \). We will analyze the situation step by step.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Let the initial volume \( V_1 = V \) cm³.
- Let the initial temperature \( T_1 = T \) (in absolute temperature).
- Let the initial pressure \( P_1 = P \).

2. **Identify Final Conditions**:
- The final temperature \( T_2 \) is given as double the initial temperature:
\[ T_2 = 2T \]
- The final pressure \( P_2 \) is given as half the initial pressure:
\[ P_2 = \frac{P}{2} \]

3. **Apply the Ideal Gas Law**:
- For the initial state, we have:
\[ P_1 V_1 = nRT_1 \quad \text{(1)} \]
- For the final state, we have:
\[ P_2 V_2 = nRT_2 \quad \text{(2)} \]

4. **Divide the Two Equations**:
- Dividing equation (1) by equation (2):
\[ \frac{P_1 V_1}{P_2 V_2} = \frac{nRT_1}{nRT_2} \]
- Simplifying gives:
\[ \frac{P_1}{P_2} \cdot \frac{V_1}{V_2} = \frac{T_1}{T_2} \]

5. **Substitute Known Values**:
- Substitute \( P_1 = P \), \( P_2 = \frac{P}{2} \), \( V_1 = V \), \( T_1 = T \), and \( T_2 = 2T \):
\[ \frac{P}{\frac{P}{2}} \cdot \frac{V}{V_2} = \frac{T}{2T} \]
- This simplifies to:
\[ 2 \cdot \frac{V}{V_2} = \frac{1}{2} \]

6. **Solve for \( V_2 \)**:
- Rearranging gives:
\[ \frac{V}{V_2} = \frac{1}{4} \]
- Therefore:
\[ V_2 = 4V \]

### Final Answer:
The final volume \( V_2 \) is \( 4V \) cm³.

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