Question

The fugacity of a pure ideal gas is equal to:

• A. Its activity
• B. Its pressure
• C. RT/V
• D. Its molar volume

Hint:

Remember that fugacity has units of pressure.
For any ideal gas system, the fugacity of component \( i \) is simply equal to its partial pressure: \( f_i = p_i = y_i P \).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the relationship between the fugacity and the pressure of a pure ideal gas.
Fugacity is a thermodynamic property used to represent the effective pressure of real gases when ideal gas models are insufficient.

Step 2: Key Formula or Approach:
The fugacity (\( f \)) of a gas is defined in terms of its chemical potential change:
\[ d\mu_i = R \cdot T \cdot d(\ln f_i) \]
with the boundary condition that as the pressure approaches zero, real gases behave ideally, and fugacity becomes equal to the pressure:
\[ \lim_{P \to 0} \frac{f_i}{P} = 1 \]
For an ideal gas, this behavior is exhibited at all pressures.

Step 3: Detailed Explanation:

• Fugacity Coefficient: The deviation of a real gas from ideal behaviour is measured by the fugacity coefficient, \( \phi \), defined as:
\[ \phi = \frac{f}{P} \]

• Ideal Gas Application: For a pure ideal gas, there are no intermolecular forces, and the volume of gas molecules is negligible.
As a result, the gas acts ideally at all pressure conditions, meaning its fugacity coefficient is exactly equal to 1:
\[ \phi = 1 \quad \implies \quad \frac{f}{P} = 1 \quad \implies \quad f = P \]

• Activity comparison: Activity is a dimensionless quantity related to fugacity by \( a = f / f^\circ \), where \( f^\circ \) is the standard state fugacity, and is not generally equal to fugacity itself.


Step 4: Final Answer:
For a pure ideal gas, the fugacity is identical to its pressure.