Initial pressure and volume of a gas are P and V respectively. First its volume is expanded to 4V by isothermal process and then again its volume makes to be V by adiabatic process then its final pressure is `(gamma = 1.5)` -

Options

A. 8P
B. 4P
C. P
D. 2P

Explanation

To solve the problem step-by-step, we will analyze the two processes the gas undergoes: an isothermal expansion followed by an adiabatic compression.

### Step 1: Understand the Initial Conditions
The initial pressure and volume of the gas are given as \( P \) and \( V \) respectively.

### Step 2: Isothermal Expansion
The gas expands isothermally from volume \( V \) to \( 4V \). In an isothermal process, the product of pressure and volume remains constant, which can be expressed as:
\[ PV = nRT \]
Since the temperature \( T \) remains constant, we can write:
\[ P \cdot V = P' \cdot (4V) \]
Where \( P' \) is the pressure after the isothermal expansion.

Rearranging gives:
\[ P' = \frac{P \cdot V}{4V} = \frac{P}{4} \]
Thus, after the isothermal expansion, the pressure \( P' \) is \( \frac{P}{4} \).

### Step 3: Adiabatic Compression
Next, the gas undergoes an adiabatic process where its volume changes from \( 4V \) back to \( V \). For an adiabatic process, we use the relation:
\[ PV^\gamma = \text{constant} \]
Applying this to the initial and final states of the adiabatic process, we have:
\[ P' \cdot (4V)^\gamma = P_C \cdot V^\gamma \]
Where \( P_C \) is the final pressure after the adiabatic process.

Substituting \( P' = \frac{P}{4} \):
\[ \frac{P}{4} \cdot (4V)^\gamma = P_C \cdot V^\gamma \]
This simplifies to:
\[ \frac{P}{4} \cdot 4^\gamma \cdot V^\gamma = P_C \cdot V^\gamma \]
Dividing both sides by \( V^\gamma \):
\[ \frac{P \cdot 4^\gamma}{4} = P_C \]
This simplifies to:
\[ P_C = \frac{P \cdot 4^\gamma}{4} = P \cdot 4^{\gamma - 1} \]

### Step 4: Substitute the Value of \( \gamma \)
Given that \( \gamma = 1.5 \):
\[ P_C = P \cdot 4^{1.5 - 1} = P \cdot 4^{0.5} = P \cdot 2 \]
Thus, the final pressure \( P_C \) is:
\[ P_C = 2P \]

### Final Answer
The final pressure of the gas after the two processes is \( \boxed{2P} \).

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