A mass of ideal gas at pressure P is expanded isothermally to four times the originl volume and then slowly compressed adiabatically to its original volume. Assuming `gamma` to be 1.5 , the new pressure of the gas is

Options

A. 0.58333333333333
B. P
C. 0.66666666666667
D. P/2

Explanation

To solve the problem of an ideal gas that is expanded isothermally to four times its original volume and then slowly compressed adiabatically to its original volume, we will follow these steps:

### Step 1: Understand the Isothermal Expansion
In an isothermal process, the temperature remains constant. For an ideal gas, the relationship between pressure (P), volume (V), and temperature (T) is given by the ideal gas law:
\[ PV = nRT \]
where \( n \) is the number of moles and \( R \) is the universal gas constant.

Given that the gas expands isothermally to four times its original volume, we can denote the initial volume as \( V_1 \) and the final volume after expansion as \( V_2 = 4V_1 \).

### Step 2: Apply the Ideal Gas Law
Since the process is isothermal, we can express the initial and final states using the ideal gas law:
\[ P_1 V_1 = nRT \]
\[ P_2 V_2 = nRT \]

From the second equation, substituting \( V_2 = 4V_1 \):
\[ P_2 (4V_1) = nRT \]

### Step 3: Relate the Pressures
From the two equations, we can set them equal to each other since both equal \( nRT \):
\[ P_1 V_1 = P_2 (4V_1) \]

Dividing both sides by \( V_1 \):
\[ P_1 = 4P_2 \]

Thus, we can express \( P_2 \):
\[ P_2 = \frac{P_1}{4} \]

### Step 4: Understand the Adiabatic Compression
Next, we need to consider the adiabatic compression back to the original volume \( V_1 \). For an adiabatic process, the relationship between pressure and volume is given by:
\[ P V^\gamma = \text{constant} \]
where \( \gamma \) is the heat capacity ratio (given as 1.5).

### Step 5: Apply the Adiabatic Condition
Let \( P_3 \) be the pressure after the adiabatic compression. We can write:
\[ P_2 V_2^\gamma = P_3 V_1^\gamma \]

Substituting \( V_2 = 4V_1 \):
\[ P_2 (4V_1)^\gamma = P_3 V_1^\gamma \]

### Step 6: Solve for \( P_3 \)
Now, substituting \( P_2 = \frac{P_1}{4} \):
\[ \frac{P_1}{4} (4V_1)^\gamma = P_3 V_1^\gamma \]

This simplifies to:
\[ \frac{P_1}{4} \cdot 4^\gamma V_1^\gamma = P_3 V_1^\gamma \]

Dividing both sides by \( V_1^\gamma \):
\[ \frac{P_1 \cdot 4^\gamma}{4} = P_3 \]

### Step 7: Calculate \( P_3 \)
Since \( 4^\gamma = 4^{1.5} = 8 \):
\[ P_3 = \frac{P_1 \cdot 8}{4} = 2P_1 \]

### Final Result
Thus, the new pressure \( P_3 \) of the gas after the adiabatic compression is:
\[ P_3 = 2P_1 \]