---
title: "The graph below displays the compressibility factor, \\(\\dfrac{PV}{nRT}\\), as a function of pressure for \\(1.00 \\text{ mol}\\) samples of three different real gases (\\(\\text{Gas 1}\\), \\(\\text{Gas 2}\\), and \\(\\text{Gas 3}\\)) and an ideal gas at a constant temperature of \\(300 \\text{ K}\\).  Which of the following correctly ranks the strength of the attractive intermolecular forces among the particles of each gas and provides the correct reasoning for why the value of \\(\\dfrac{PV}{nRT}\\) drops below \\(1.0\\) at moderate pressures?"
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url: "https://nerd-notes.com/ubq/123694/"
date_modified: "2026-09-28T12:02:00+00:00"
---

# The graph below displays the compressibility factor, \(\dfrac{PV}{nRT}\), as a function of pressure for \(1.00 \text{ mol}\) samples of three different real gases (\(\text{Gas 1}\), \(\text{Gas 2}\), and \(\text{Gas 3}\)) and an ideal gas at a constant temperature of \(300 \text{ K}\).

Which of the following correctly ranks the strength of the attractive intermolecular forces among the particles of each gas and provides the correct reasoning for why the value of \(\dfrac{PV}{nRT}\) drops below \(1.0\) at moderate pressures?

The graph below displays the compressibility factor, \(\dfrac{PV}{nRT}\), as a function of pressure for \(1.00 \text{ mol}\) samples of three different real gases (\(\text{Gas 1}\), \(\text{Gas 2}\), and \(\text{Gas 3}\)) and an ideal gas at a constant temperature of \(300 \text{ K}\).

Which of the following correctly ranks the strength of the attractive intermolecular forces among the particles of each gas and provides the correct reasoning for why the value of \(\dfrac{PV}{nRT}\) drops below \(1.0\) at moderate pressures?

![A Cartesian coordinate graph showing compressibility factor on the vertical axis versus pressure on the horizontal axis. The horizontal axis is labeled 'Pressure (atm)' with tick marks at 0, 100, 200, 300, 400, 500, and 600. The vertical axis is labeled 'PV / nRT' with tick marks at 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, and 1.8. A horizontal dashed line at PV / nRT = 1.0 spans the entire width and is labeled 'Ideal gas'. Light gray rectangular gridlines extend across the plot. Three continuous curves start at the point (0, 1.0). The solid curve, labeled 'Gas 1', curves downward to a minimum value of 0.50 near 150 atm, intersects the 1.0 line near 380 atm, and rises to 1.55 at 600 atm. The dashed curve, labeled 'Gas 2', curves downward to a minimum value of 0.78 near 150 atm, intersects the 1.0 line near 320 atm, and rises to 1.45 at 600 atm. The dotted curve, labeled 'Gas 3', dips imperceptibly to 0.98 near 50 atm, crosses 1.0 near 100 atm, and rises steadily to 1.35 at 600 atm. No other curves, labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790596920-TelkpH.jpg)

- **A.** The strength of attractive intermolecular forces is \(\text{Gas 1} > \text{Gas 2} > \text{Gas 3}\), because attractive forces between gas particles reduce the force of collisions with the container walls, causing the measured pressure to be lower than ideal.
- **B.** The strength of attractive intermolecular forces is \(\text{Gas 1} > \text{Gas 2} > \text{Gas 3}\), because the finite volume of the gas particles reduces the volume available for movement, causing the measured volume to be lower than ideal.
- **C.** The strength of attractive intermolecular forces is \(\text{Gas 3} > \text{Gas 2} > \text{Gas 1}\), because attractive forces between gas particles reduce the force of collisions with the container walls, causing the measured pressure to be lower than ideal.
- **D.** The strength of attractive intermolecular forces is \(\text{Gas 3} > \text{Gas 2} > \text{Gas 1}\), because inelastic collisions between gas particles cause a decrease in average kinetic energy at moderate pressures.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123694/*