Calculate the volume occupied by `2.0` mole of `N_(2) at 200K` and `8.21 atm` pressure, if `(P_(C)V_(C))/(RT_(C)) = (3)/(8)` and `(P_(r)V_(r))/(T_(r)) = 2.4`.

Explanation

To solve the problem of calculating the volume occupied by 2.0 moles of nitrogen gas (N₂) at a temperature of 200 K and a pressure of 8.21 atm, we can follow these steps:

### Step 1: Understand the Given Information
We have:
- Number of moles (n) = 2.0 moles
- Temperature (T) = 200 K
- Pressure (P) = 8.21 atm
- Given ratios:
- \(\frac{P_C V_C}{R T_C} = \frac{3}{8}\)
- \(\frac{P_r V_r}{T_r} = 2.4\)

### Step 2: Relate the Given Ratios to Critical and Reduced Conditions
From the given ratios, we can express the reduced conditions:
- \(T_r = \frac{T}{T_C}\)
- \(P_r = \frac{P}{P_C}\)
- \(V_r = \frac{V}{V_C}\)

### Step 3: Multiply the Two Given Equations
We multiply the two equations:
\[ \frac{P_C V_C}{R T_C} \times \frac{P_r V_r}{T_r} = \frac{3}{8} \times 2.4 \]
This simplifies to:
\[ \frac{P_C P_r V_C V_r}{R T_C T_r} = \frac{3 \times 2.4}{8} \]

### Step 4: Substitute the Reduced Conditions
Substituting the reduced conditions into the equation:
\[ \frac{P \cdot \frac{P}{P_C} \cdot V \cdot \frac{V}{V_C}}{R \cdot T_C \cdot \frac{T}{T_C}} = \frac{3 \times 2.4}{8} \]
This simplifies to:
\[ \frac{P V}{R T} = \frac{3 \times 2.4}{8} \]

### Step 5: Calculate the Right Side
Calculating the right side:
\[ \frac{3 \times 2.4}{8} = 0.9 \]

### Step 6: Relate to the Compressibility Factor
The compressibility factor \(Z\) can be defined as:
\[ Z = \frac{PV}{nRT} \]
For one mole, we can express it as:
\[ Z = \frac{PV}{RT} \]
From our previous step, we have:
\[ Z = 0.9 \]

### Step 7: Calculate the Molar Volume
Using the equation for molar volume:
\[ V_m = \frac{ZRT}{P} \]
Substituting the known values:
- \(Z = 0.9\)
- \(R = 0.0821 \, \text{L atm/(K mol)}\)
- \(T = 200 \, \text{K}\)
- \(P = 8.21 \, \text{atm}\)

Calculating the molar volume:
\[ V_m = \frac{0.9 \times 0.0821 \times 200}{8.21} \]

### Step 8: Calculate the Volume for 2 Moles
After calculating \(V_m\), we multiply by the number of moles (2.0):
\[ V = n \times V_m \]

### Final Calculation
1. Calculate \(V_m\):
\[ V_m = \frac{0.9 \times 0.0821 \times 200}{8.21} \approx 1.8 \, \text{L} \]
2. Calculate total volume for 2 moles:
\[ V = 2 \times 1.8 \approx 3.6 \, \text{L} \]

### Final Answer
The volume occupied by 2.0 moles of nitrogen gas at 200 K and 8.21 atm is approximately **3.6 liters**.

---