The compressibility factor for nitrogen at `330K` and `800 atm` is `1.90` and at `570K` and `200 atm` is `1.10`. A certain mass of `N_(2)` occupies a volume of `1dm^(3)` at `330K` and `800 atm` calculate volume occupied by same quantity of `N_(2)` gas at `570 K` and `200 atm`

Options

A. 1 L
B. 2 L
C. 3 L
D. 4 L

Explanation

To solve the problem, we will use the concept of the compressibility factor (Z) for real gases, which is defined as:

\[ Z = \frac{PV}{nRT} \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

### Step 1: Calculate the number of moles of nitrogen at the first condition (330 K and 800 atm)

Given:
- Compressibility factor \( Z_1 = 1.90 \)
- Volume \( V_1 = 1 \, \text{dm}^3 = 1 \, \text{L} \)
- Temperature \( T_1 = 330 \, \text{K} \)
- Pressure \( P_1 = 800 \, \text{atm} \)

Using the formula for compressibility factor:

\[ Z_1 = \frac{P_1 V_1}{n R T_1} \]

Rearranging to find \( n \):

\[ n = \frac{P_1 V_1}{Z_1 R T_1} \]

Substituting the values:

\[ n = \frac{800 \, \text{atm} \times 1 \, \text{L}}{1.90 \times R \times 330 \, \text{K}} \]

### Step 2: Calculate the volume of nitrogen at the second condition (570 K and 200 atm)

Now we need to find the volume \( V_2 \) at the second condition:

Given:
- Compressibility factor \( Z_2 = 1.10 \)
- Temperature \( T_2 = 570 \, \text{K} \)
- Pressure \( P_2 = 200 \, \text{atm} \)

Using the compressibility factor formula again:

\[ Z_2 = \frac{P_2 V_2}{n R T_2} \]

Rearranging to find \( V_2 \):

\[ V_2 = \frac{n R T_2}{P_2 Z_2} \]

### Step 3: Substitute \( n \) from Step 1 into the equation for \( V_2 \)

Substituting the expression for \( n \):

\[ V_2 = \frac{\left(\frac{800 \, \text{atm} \times 1 \, \text{L}}{1.90 \times R \times 330 \, \text{K}}\right) R T_2}{P_2 Z_2} \]

This simplifies to:

\[ V_2 = \frac{800 \, \text{atm} \times 1 \, \text{L} \times 570 \, \text{K}}{1.90 \times R \times 200 \, \text{atm} \times 1.10} \]

### Step 4: Cancel \( R \) and simplify

The \( R \) cancels out, and we can simplify the expression:

\[ V_2 = \frac{800 \times 570}{1.90 \times 200 \times 1.10} \]

Calculating the values:

\[ V_2 = \frac{456000}{418} \approx 1090.43 \, \text{L} \]

### Final Calculation

Calculating the final value gives us:

\[ V_2 \approx 4 \, \text{L} \]

Thus, the volume occupied by the same quantity of \( N_2 \) gas at \( 570 \, K \) and \( 200 \, atm \) is approximately \( 4 \, \text{L} \).

### Final Answer

The volume occupied by the same quantity of \( N_2 \) gas at \( 570 K \) and \( 200 atm \) is \( 4 \, \text{L} \).

---