Reduced temperature for benzene is `0.7277` and its reduced volume is `0.40`. Calculated the reduced pressure of benzene.

Answer

To calculate the reduced pressure (Pr) of benzene using the given reduced temperature (Tr) and reduced volume (Vr), we can follow these steps:

### Step 1: Write down the formula
We will use the law of corresponding states, which gives us the relationship:
\[ \frac{P_r - 3}{V_r^3 (3V_r - 1)} = 8RT_r \]
Where:
- \(P_r\) = reduced pressure
- \(V_r\) = reduced volume
- \(T_r\) = reduced temperature

### Step 2: Substitute the known values
From the question, we have:
- \(T_r = 0.7277\)
- \(V_r = 0.40\)

Substituting these values into the equation:
\[ \frac{P_r - 3}{(0.40)^3 (3 \times 0.40 - 1)} = 8 \times 0.7277 \]

### Step 3: Calculate \(V_r^3\) and \(3V_r - 1\)
First, calculate \(V_r^3\):
\[ (0.40)^3 = 0.064 \]
Next, calculate \(3V_r - 1\):
\[ 3 \times 0.40 - 1 = 1.2 - 1 = 0.2 \]

### Step 4: Substitute these values back into the equation
Now, substituting these values back into the equation:
\[ \frac{P_r - 3}{0.064 \times 0.2} = 8 \times 0.7277 \]
Calculating the right side:
\[ 8 \times 0.7277 = 5.8216 \]
So we have:
\[ \frac{P_r - 3}{0.0128} = 5.8216 \]

### Step 5: Solve for \(P_r\)
Now, multiply both sides by \(0.0128\):
\[ P_r - 3 = 5.8216 \times 0.0128 \]
Calculating the right side:
\[ 5.8216 \times 0.0128 = 0.0745 \]
Now, add 3 to both sides:
\[ P_r = 0.0745 + 3 = 3.0745 \]

### Step 6: Final calculation
The reduced pressure of benzene is approximately:
\[ P_r \approx 3.0745 \text{ atm} \]

### Final Answer
The reduced pressure of benzene is approximately **3.0745 atm**.

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