A Gas is enclosed in a metallic container of volume Vand its initial pressure is p. It is slowly compressed to a volume V/2 and then suddenly compressed to V/4. Find the final pressure of the gas. From the initial state the gas is suddenly compressed to V/2 and then slowly compressed to V/4, what will be the final pressure now.

Answer

To solve the problem, we will analyze the two scenarios described in the question step by step.

### Part 1: Slow Compression to V/2 and Sudden Compression to V/4

1. **Initial Conditions**:
- Initial volume \( V \)
- Initial pressure \( P \)

2. **First Step - Slow Compression to V/2**:
- Since the compression is slow, it is an isothermal process (temperature remains constant).
- According to Boyle's Law for isothermal processes, \( P_1 V_1 = P_2 V_2 \).
- Here, \( P_1 = P \), \( V_1 = V \), \( V_2 = \frac{V}{2} \).
- Therefore, we can write:
\[ P \cdot V = P_2 \cdot \frac{V}{2} \]
- Rearranging gives:
\[ P_2 = 2P \]

3. **Second Step - Sudden Compression to V/4**:
- This compression is sudden, so it is an adiabatic process.
- For adiabatic processes, we use the relation \( P V^\gamma = \text{constant} \), where \( \gamma \) is the heat capacity ratio \( \frac{C_p}{C_v} \).
- Now, we have \( P_2 = 2P \) and \( V_2 = \frac{V}{2} \), and we want to find the final pressure \( P_f \) when the volume is suddenly compressed to \( V_f = \frac{V}{4} \).
- Using the adiabatic relation:
\[ P_2 \left( \frac{V}{2} \right)^\gamma = P_f \left( \frac{V}{4} \right)^\gamma \]
- Substituting \( P_2 = 2P \):
\[ 2P \left( \frac{V}{2} \right)^\gamma = P_f \left( \frac{V}{4} \right)^\gamma \]
- Simplifying gives:
\[ 2P \cdot \frac{V^\gamma}{2^\gamma} = P_f \cdot \frac{V^\gamma}{4^\gamma} \]
- Canceling \( V^\gamma \) from both sides:
\[ 2P \cdot \frac{1}{2^\gamma} = P_f \cdot \frac{1}{4^\gamma} \]
- Rearranging gives:
\[ P_f = 2P \cdot \frac{4^\gamma}{2^\gamma} = 2P \cdot 2^{2\gamma} = 2^{2\gamma + 1} P \]

### Part 2: Sudden Compression to V/2 and Then Slow Compression to V/4

1. **Initial Conditions**:
- Initial volume \( V \)
- Initial pressure \( P \)

2. **First Step - Sudden Compression to V/2**:
- This is an adiabatic process.
- Using the same adiabatic relation:
\[ P \cdot V^\gamma = P_1 \cdot \left( \frac{V}{2} \right)^\gamma \]
- Rearranging gives:
\[ P_1 = P \cdot 2^\gamma \]

3. **Second Step - Slow Compression to V/4**:
- This is an isothermal process.
- Using Boyle's Law:
\[ P_1 \cdot \frac{V}{2} = P_f \cdot \frac{V}{4} \]
- Substituting \( P_1 = P \cdot 2^\gamma \):
\[ P \cdot 2^\gamma \cdot \frac{V}{2} = P_f \cdot \frac{V}{4} \]
- Canceling \( V \) from both sides:
\[ P \cdot 2^\gamma \cdot \frac{1}{2} = P_f \cdot \frac{1}{4} \]
- Rearranging gives:
\[ P_f = 2P \cdot 2^\gamma = 2^{\gamma + 1} P \]

### Final Answers:
1. For the first scenario (slow to V/2, sudden to V/4):
\[ P_f = 2^{2\gamma + 1} P \]
2. For the second scenario (sudden to V/2, slow to V/4):
\[ P_f = 2^{\gamma + 1} P \]