The density of steam at `27^@C` and `8.314xx10^4` pascal is `0.8 kg m^(-3)`.The compressibility factor would be

Options

A. 0.75
B. 1
C. 0.88
D. 1.1

Explanation

To calculate the compressibility factor \( Z \) for steam at the given conditions, we can follow these steps:

### Step 1: Understand the formula for the compressibility factor
The compressibility factor \( Z \) is defined as:
\[ Z = \frac{PV}{RT} \]
where:
- \( P \) = pressure in pascals
- \( V \) = volume in cubic meters
- \( R \) = universal gas constant (approximately \( 8.314 \, \text{J/(mol K)} \))
- \( T \) = temperature in Kelvin

### Step 2: Convert the temperature to Kelvin
Given temperature is \( 27^\circ C \). To convert this to Kelvin:
\[ T = 27 + 273.15 = 300.15 \, K \approx 300 \, K \]

### Step 3: Calculate the volume of one mole of steam
We know that the mass of one mole of steam (water) is \( 18 \, g \) or \( 18 \times 10^{-3} \, kg \). The density of steam is given as \( 0.8 \, kg/m^3 \). The volume \( V \) can be calculated using the formula:
\[ V = \frac{\text{mass}}{\text{density}} = \frac{18 \times 10^{-3} \, kg}{0.8 \, kg/m^3} \]
Calculating this gives:
\[ V = \frac{18 \times 10^{-3}}{0.8} = 22.5 \times 10^{-3} \, m^3 = 0.0225 \, m^3 \]

### Step 4: Substitute the values into the compressibility factor formula
Now we can substitute the values into the formula for \( Z \):
- \( P = 8.314 \times 10^4 \, Pa \)
- \( V = 0.0225 \, m^3 \)
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( T = 300 \, K \)

Substituting these values into the formula:
\[ Z = \frac{(8.314 \times 10^4) \times (0.0225)}{(8.314) \times (300)} \]

### Step 5: Simplify the equation
Calculating the numerator:
\[ 8.314 \times 10^4 \times 0.0225 = 1864.65 \]
Calculating the denominator:
\[ 8.314 \times 300 = 2494.2 \]
Now substituting these values back into the equation for \( Z \):
\[ Z = \frac{1864.65}{2494.2} \approx 0.748 \]

### Step 6: Final result
The compressibility factor \( Z \) is approximately:
\[ Z \approx 0.75 \]

### Conclusion
Thus, the compressibility factor \( Z \) for steam at \( 27^\circ C \) and \( 8.314 \times 10^4 \, Pa \) is approximately \( 0.75 \).

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