Ideal Gas Law — PV = nRT Explained with Examples

The ideal gas law packs four quantities into one equation:

where $P$ is pressure, $V$ is volume, $n$ is the amount of gas in moles, $R$ is the gas constant, and $T$ is absolute temperature in Kelvin. Know any three and you can solve for the fourth. A common chemistry value of the constant is

which is convenient when pressure is in atm and volume in liters; other units call for a matching value of $R$. Reading each symbol once: $P$ is pressure, $V$ is volume, $n$ is moles, $R$ is the gas constant tying the units together, and $T$ is temperature in Kelvin. The fast intuition behind the equation is direct: more gas means a larger $n$, hotter gas means a larger $T$, and either tends to raise pressure or volume unless one of them is held fixed.

Why the equation holds, and what it assumes

$PV = nRT$ is the model for an ideal gas, one whose particles have negligible volume and no intermolecular forces except during collisions. Those assumptions are why it combines the separate gas laws into one statement: Boyle's law lives in the $P$-$V$ trade-off at fixed $n$ and $T$, Charles's law in the $V$-$T$ link at fixed $P$ and $n$, and Avogadro's law in the $V$-$n$ link at fixed $P$ and $T$. Picture a sealed container: heat it at fixed volume and amount and pressure rises; let it expand at roughly fixed pressure and volume rises instead. The equation keeps all of those relationships in one place.

The assumptions also set the limits. The model works best at lower pressure and higher temperature, where particles are far apart, and real gases deviate more at high pressure or near condensation. And temperature must always be in Kelvin, or the answer comes out wrong.

Worked example: solving for volume

A gas sample has $n = 0.50\ \mathrm{mol}$, $T = 300\ \mathrm{K}$, and $P = 1.20\ \mathrm{atm}$. Find the volume with $R = 0.08206\ \mathrm{L \cdot atm \cdot mol^{-1} \cdot K^{-1}}$. Rearrange once:

Substitute and simplify:

So the volume is about $10.3\ \mathrm{L}$. The workflow is the lesson: pick a matching $R$, keep $T$ in Kelvin, rearrange once, then sanity-check. A half mole at room temperature near $1\ \mathrm{atm}$ taking up several liters is plausible, so it passes. The same four steps handle any of the four unknowns. To solve for pressure you would rearrange to $P = nRT/V$; for moles, $n = PV/RT$; for temperature, $T = PV/nR$. Only the algebra of the first rearrangement changes, while the unit discipline and the closing sanity check stay identical.

Try your own numbers

Change one value in the example and predict the effect before computing. Double the moles to $n = 1.0\ \mathrm{mol}$, keeping $T$ and $P$ fixed, and volume should double:

which is indeed about twice $10.3\ \mathrm{L}$. To run a different set of numbers or units, try a similar case in GPAI Solver.

Calculation pitfalls

• Using Celsius instead of Kelvin. Convert $27^\circ\mathrm{C}$ to $300\ \mathrm{K}$ before substituting.
• Mixing units without changing $R$. Pa and $\mathrm{m^3}$ need a different constant than atm and liters.
• Treating the law as exact for every gas. It is an approximation, often very good but not equally accurate for every gas in every condition.
• Forgetting what is held fixed. "Higher temperature means higher pressure" is only directly true at constant volume and amount.

The ideal gas law shows up in introductory chemistry, thermodynamics, gas-collection problems, lab calculations, and engineering approximations, and it is the bridge to molar volume and real-gas deviations once the four-variable relationship feels natural.