One litre gas at `400K` and `300atm` pressure is compressed to a pressure of `600 atm` and `200K`. The compressibility factor is changed from `1.2 to 1.6` respectively. Calculate the final volume of the gas.

Explanation

To solve the problem step by step, we will use the compressibility factor and the ideal gas law.

### Step 1: Understand the given data
We have the following initial conditions:
- Initial volume (V1) = 1 L
- Initial temperature (T1) = 400 K
- Initial pressure (P1) = 300 atm
- Initial compressibility factor (Z1) = 1.2

Final conditions:
- Final temperature (T2) = 200 K
- Final pressure (P2) = 600 atm
- Final compressibility factor (Z2) = 1.6

### Step 2: Write the equations for compressibility factors
The compressibility factor (Z) is defined as:
\[ Z = \frac{PV}{RT} \]
Where:
- P = Pressure
- V = Volume
- R = Ideal gas constant
- T = Temperature

For the initial state:
\[ Z_1 = \frac{P_1 V_1}{RT_1} \]
For the final state:
\[ Z_2 = \frac{P_2 V_2}{RT_2} \]

### Step 3: Set up the equation using the compressibility factors
From the above equations, we can express the volumes in terms of the compressibility factors:
\[ \frac{Z_1}{Z_2} = \frac{P_1 V_1 / RT_1}{P_2 V_2 / RT_2} \]

This simplifies to:
\[ \frac{Z_1}{Z_2} = \frac{P_1 V_1 T_2}{P_2 V_2 T_1} \]

### Step 4: Rearranging the equation to find V2
Rearranging gives us:
\[ V_2 = \frac{P_2 V_1 T_1 Z_2}{P_1 T_2 Z_1} \]

### Step 5: Substitute the known values
Now we can substitute the known values into the equation:
- \( P_1 = 300 \, \text{atm} \)
- \( V_1 = 1 \, \text{L} \)
- \( T_1 = 400 \, \text{K} \)
- \( Z_1 = 1.2 \)
- \( P_2 = 600 \, \text{atm} \)
- \( T_2 = 200 \, \text{K} \)
- \( Z_2 = 1.6 \)

Substituting these values:
\[ V_2 = \frac{600 \times 1 \times 400 \times 1.6}{300 \times 200 \times 1.2} \]

### Step 6: Calculate V2
Calculating the right-hand side:
- Numerator: \( 600 \times 1 \times 400 \times 1.6 = 384000 \)
- Denominator: \( 300 \times 200 \times 1.2 = 72000 \)

Now, divide the numerator by the denominator:
\[ V_2 = \frac{384000}{72000} = 5.33 \, \text{L} \]

### Final Answer
The final volume of the gas (V2) is approximately **5.33 L**.

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