Difficulty: Easy
Correct Answer: a/V^2 (pressure correction due to attractions)
Explanation:
Introduction / Context:
The van der Waals (vdW) equation modifies the ideal-gas law to capture real-gas behavior by introducing two empirical parameters, a and b. These corrections embody intermolecular attractions and finite molecular size, respectively. Identifying which term corresponds to which physical effect is vital in high-pressure gas processing and critical-phenomena studies.
Given Data / Assumptions:
Concept / Approach:
Attractive forces effectively reduce the measured pressure relative to an ideal gas at the same state because molecules pull each other inward. The vdW model compensates by adding a/V^2 to the measured pressure P. Conversely, the finite size of molecules reduces the accessible volume, represented by (V − b). Thus a/V^2 is directly tied to attractions, while b is tied to repulsive core/excluded-volume effects.
Step-by-Step Solution:
Recall physical roles: a → attractions; b → finite size.Identify the position of a: appears as a pressure correction a/V^2.Therefore, the term accounting for intermolecular forces (attractions) is a/V^2.Select option “a/V^2 (pressure correction due to attractions)”.
Verification / Alternative check:
In the limit a → 0 and b → 0, the equation reduces to PV = R T, confirming that a and b are corrections beyond ideal behavior.
Why Other Options Are Wrong:
R T is the thermal energy driving expansion, not a force correction.(V − b) represents excluded volume (repulsive size effect), not attractions.1/(R T) is not a term in the vdW equation; it has no direct physical meaning here.
Common Pitfalls:
Mixing up roles of a and b; forgetting that attractive forces reduce pressure whereas excluded volume reduces effective V.
Final Answer:
a/V^2 (pressure correction due to attractions)
Discussion & Comments