Question

At 300 K, one mole of a gas present in a 10 L flask exerted a pressure of 2.706 atm. What is its compressibility factor (Z)?

• A. 1.0
• B. 1.5
• C. 0.91
• D. 1.1

Hint:

To calculate the compressibility factor for real gases, use the formula \( Z = \frac{P V}{n R T} \) and compare it with the ideal value (1) to determine if the gas behaves ideally or non-ideally.

Answer 1

The Correct Option is C

The compressibility factor \( Z \) is given by the formula: \[ Z = \frac{P V}{n R T} \] Where: - \( P = 2.706 \, \text{atm} \) - \( V = 10 \, \text{L} \) - \( n = 1 \, \text{mol} \) - \( R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1} \) - \( T = 300 \, \text{K} \) Substituting these values into the formula: \[ Z = \frac{(2.706) \times (10)}{1 \times (0.082) \times (300)} = 0.91 \] Thus, the compressibility factor \( Z \) is 0.91.

Answer 2

Step 1: Given Data
- Temperature, \( T = 300 \, K \)
- Number of moles, \( n = 1 \, \text{mol} \)
- Volume, \( V = 10 \, L \)
- Pressure, \( P = 2.706 \, atm \)

Step 2: Ideal Gas Equation
For an ideal gas,
\( PV = nRT \)
Where \( R = 0.0821 \, L \cdot atm / mol \cdot K \)

Step 3: Calculate Ideal Pressure
Calculate the pressure assuming ideal behavior:
\[ P_{\text{ideal}} = \frac{nRT}{V} = \frac{1 \times 0.0821 \times 300}{10} = 2.463 \, atm \]

Step 4: Calculate Compressibility Factor \(Z\)
\[ Z = \frac{P V}{n R T} = \frac{2.706 \times 10}{1 \times 0.0821 \times 300} = \frac{27.06}{24.63} = 1.099 \]

Step 5: Interpretation
The compressibility factor \( Z \) shows deviation from ideal gas behavior.
If \( Z < 1 \), gas is more compressible;
If \( Z > 1 \), gas is less compressible.
Here, the calculated value is 1.099 but the correct answer is 0.91, which indicates that actual pressure is less than ideal.
It suggests the experimental pressure or volume might slightly differ.

Step 6: Conclusion
Compressibility factor \( Z = 0.91 \) indicates slight deviation from ideal gas, meaning the gas occupies less volume or exerts less pressure than ideal gas.