DAT General Chemistry: The Ideal Gas Law

DAT High-Yield Concept

The ideal gas equation is $PV = nRT$ , relating pressure, volume, moles, and temperature.
Temperature must always be expressed in Kelvin ( $K = \pu{degC} + 273.15$ ).
Match the universal gas constant $R$ to the units of pressure given in the problem.
Ideal behavior assumes gas particles have zero volume and experience no intermolecular forces.

1. The Ideal Gas Law Equation

The ideal gas law unifies the state variables of a gas sample into a single mathematical relationship:

PV = nRT

Where $P$ is absolute pressure, $V$ is volume, $n$ is the number of moles of gas, $R$ is the universal gas constant, and $T$ is absolute temperature in Kelvin.

Molecular collisions within a closed container (the propane tank) are shown (right). The arrows represent the random motions and collisions of these molecules. The pressure and temperature of the gas are directly proportional: as the temperature is increased, the pressure of the propane increases by the same factor. A simple consequence of this proportionality is that on a hot summer day, the propane tank pressure will be elevated, and thus propane tanks must be rated to withstand such increases in pressure. (Source: BlyumJ · CC BY-SA 4.0 · Wikimedia Commons)

P-V diagram for an ideal gas (Source: T. Wolverson · CC BY-SA 3.0 · Wikimedia Commons)

Ideal Gas Assumptions: An ideal gas consists of point-mass particles in continuous random motion that undergo perfectly elastic collisions, exert no intermolecular attractive or repulsive forces, and occupy negligible physical volume compared to the container.

2. Selecting the Universal Gas Constant R

The universal gas constant $R$ serves as a proportionality factor. You must select the numerical value of $R$ that matches the pressure units in the question:

Pressure Unit	Gas Constant Value ( $R$ )	Standard Units ( $V, n, T$ )
Atmospheres ( $\pu{atm}$ )	$\pu{0.08206 Latm//molK}$	$\pu{L}$ , $\pu{mol}$ , $\pu{K}$
Torr / Millimeters of Mercury ( $\pu{torr}$ / $\pu{mmHg}$ )	$\pu{62.36 Ltorr//molK}$	$\pu{L}$ , $\pu{mol}$ , $\pu{K}$
Kilopascals / Joules ( $\pu{kPa}$ / $\pu{J}$ )	$\pu{8.314 LkPa//molK}$ (or $\pu{8.314 J//mol*K}$ )	$\pu{L}$ (or $\pu{m^3}$ ), $\pu{mol}$ , $\pu{K}$

Note: Empirical gas relationships integrate into the ideal gas law (the combined gas law is covered in depth in its own lesson).

3. Solving for Gas Variables

Rearrange $PV = nRT$ to isolate the target unknown:

Pressure: $P = \frac{nRT}{V}$
Volume: $V = \frac{nRT}{P}$
Moles: $n = \frac{PV}{RT}$
Temperature: $T = \frac{PV}{nR}$

Worked Example 1: Solving for Volume

Problem: Calculate the volume occupied by $\pu{0.500 mol}$ of $\ce{O2}$ gas at $\pu{1.50 atm}$ pressure and $\pu{27 degC}$ .

Step 1: Convert temperature to Kelvin.

T = 27 + 273.15 = \pu{300.15 K}

Step 2: Solve for volume.

V = \frac{nRT}{P} = \frac{(\pu{0.500 mol})(\pu{0.08206 L*atm//mol*K})(\pu{300.15 K})}{\pu{1.50 atm}} = \pu{8.21 L}

Exam Trap: Plugging temperature in Celsius directly into $PV = nRT$ without converting to Kelvin is a top DAT distractor error!

Worked Example 2: Solving for Pressure

Problem: A $\pu{5.00 L}$ container holds $\pu{0.250 mol}$ of helium ( $\ce{He}$ ) gas at $\pu{298 K}$ . Calculate the gas pressure in $\pu{kPa}$ .

Step 1: Select R matching kilopascals. Use $R = \pu{8.314 L*kPa//mol*K}$ .

Step 2: Solve for pressure.

P = \frac{nRT}{V} = \frac{(\pu{0.250 mol})(\pu{8.314 L*kPa//mol*K})(\pu{298 K})}{\pu{5.00 L}} = \pu{124 kPa}

Exam Trap: Using $R = \pu{0.08206}$ when the target answer demands $\pu{kPa}$ yields an incorrect pressure in atmospheres instead of kilopascals.

Memory Aid: "Perky Vipers Needs Really Tough Training" →

P \cdot V = n \cdot R \cdot T

. Remember: Kelvin is Key, R must Match P.

Critical Thinking Questions

If the absolute temperature of a flexible container holding $\ce{N2}$ gas doubles while internal pressure is held constant, how does the container volume change?
Why does a gas depart from ideal gas law predictions at very low temperatures or extremely high pressures?
If two identical rigid bulbs at the same temperature contain equal masses of $\ce{He}$ and $\ce{O2}$ gas respectively, which bulb has a higher pressure?

References

Sources returned by Google Search grounding during lesson verification.

6.1 The Ideal Gas Law - UCalgary Chemistry Textbook — chem.libretexts.org
Gases for the DAT - Shemmassian Academic Consulting — shemmassianconsulting.com

The Ideal Gas Law

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