Combined gas law and ideal gas equation PV = nRT: Worked examples

We only use tools already built in the parent note: $P V = n R T$ , the combined law $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ , and $ρ = \frac{P M}{R T}$ . Nothing new — just every situation.

The scenario matrix

Read this as a checklist. Each row is a class of problem; the last column names the worked example that covers it.

Cell	What is fixed / what changes	Which tool	Covered by
A. Single-state solve	one fixed state, find the missing 4th quantity	$P V = n R T$	Ex 1
B. Sealed state-change	$n$ fixed, state $1 \to 2$	combined law	Ex 2
C. One variable held constant	e.g. isothermal ( $T$ fixed)	reduces to Boyle/Charles	Ex 3
D. Amount changes	$n$ not fixed (leak / add gas)	full $P V = n R T$ each state	Ex 4
E. Density / molar mass	mass and volume linked	$ρ = \frac{P M}{R T}$	Ex 5
F. Degenerate / limiting	a quantity $\to 0$ or $\to \infty$	reasoning on the equation	Ex 6
G. Real-world word problem	translate words $\to$ symbols	combined law	Ex 7
H. Exam twist (units + Celsius trap)	mismatched units, $^{\circ}$ C given	convert first, then plug	Ex 8

Worked examples

Ex 1 — Cell A: single-state solve

Ex 2 — Cell B: sealed state-change

Worked example Combined law, both P and T change

Q: A sealed cylinder of gas is at $2.0$ atm, $5.0$ L, $250$ K. It is compressed to $2.0$ L and warmed to $400$ K. Find the new pressure.

Forecast: volume shrank (pushes $P$ up) and it got hotter (pushes $P$ up) — expect $P_{2}$ clearly above $2.0$ atm.

Choose the tool: $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ . Why this step? $n$ is unknown but the cylinder is sealed, so $n$ is constant and cancels.
Solve for $P_{2}$ : $P_{2} = P_{1} \frac{V _{1}}{V _{2}} \frac{T _{2}}{T _{1}}$ . Why this step? Rearrange so the single unknown sits alone.
Plug in: $P_{2} = 2.0 \times \frac{5.0}{2.0} \times \frac{400}{250} = 2.0 \times 2.5 \times 1.6 = 8.0 atm$ . Why this step? Each ratio is a dimensionless factor; multiply them onto the old pressure.

Verify: $\frac{P _{2} V _{2}}{T _{2}} = \frac{8.0 \times 2.0}{400} = 0.04$ and $\frac{P _{1} V _{1}}{T _{1}} = \frac{2.0 \times 5.0}{250} = 0.04$ . Equal ✓, and $P_{2} > P_{1}$ as forecast.

Ex 3 — Cell C: one variable held constant (isothermal)

Figure — Combined gas law and ideal gas equation PV = nRT

Ex 4 — Cell D: amount of gas changes

Worked example Gas is added → n is NOT constant

Q: A rigid $10.0$ L tank at $300$ K holds gas at $2.0$ atm. More gas is pumped in until the pressure reads $5.0$ atm at the same temperature. How many moles were added?

Forecast: pressure more than doubled at fixed $V, T$ , so moles more than doubled — the added amount is a bit more than the starting amount.

Do NOT use the combined law: $n$ changes, so $\frac{P V}{T}$ is not constant. Use full $P V = n R T$ at each state. Why this step? The combined law's cancellation of $n$ only holds for a sealed, fixed- $n$ sample.
Moles before: $n_{1} = \frac{P _{1} V}{R T} = \frac{2.0 \times 10.0}{0.0821 \times 300} = 0.812 mol$ . Why this step? We need an absolute count, so we need $R$ back.
Moles after: $n_{2} = \frac{P _{2} V}{R T} = \frac{5.0 \times 10.0}{0.0821 \times 300} = 2.030 mol$ . Why this step? Same tank and temperature, only pressure changed.
Added: $Δ n = n_{2} - n_{1} = 2.030 - 0.812 = 1.22 mol$ . Why this step? "How many added" is the difference of the two counts.

Verify: because $V, T$ fixed, $n \propto P$ , so $\frac{n _{2}}{n _{1}} = \frac{5.0}{2.0} = 2.5$ ; indeed $\frac{2.030}{0.812} = 2.5$ ✓. More than doubled, as forecast.

Ex 5 — Cell E: density → molar mass

Ex 6 — Cell F: degenerate & limiting cases

Worked example What happens at the edges?

Q: Using $P V = n R T$ , reason out three limiting cases: (a) What is $V$ if $n \to 0$ ? (b) What happens to $P$ as $V \to 0$ at fixed $n, T$ ? (c) What volume does the equation predict as $T \to 0 K$ at fixed $n, P$ ?

Forecast: empty container → zero volume of gas; squeeze to nothing → pressure blows up; freeze toward absolute zero → volume heads to zero.

(a) Set $n = 0$ : $P V = 0 \cdot R T = 0$ , so with any real $P > 0$ , $V = 0$ . Why this step? No particles means no gas to occupy space — the maths agrees with the picture.
(b) Let $V \to 0$ : $P = \frac{n R T}{V}$ ; as the denominator shrinks toward $0$ , $P \to \infty$ . Why this step? Fewer and fewer roomier walls means hits become infinitely frequent — pressure diverges. (Real gases resist this; see Real Gases and van der Waals Equation.)
(c) Let $T \to 0$ : $V = \frac{n R T}{P} \to 0$ . Why this step? This is the ideal prediction — and it is exactly why $T$ must be measured from absolute zero. In reality the gas liquefies first, so the equation breaks before $T = 0$ .

Verify: all three limits agree with the balloon-of-balls intuition. (a) $V = 0$ when $n = 0$ ✓; (b) $P$ unbounded ✓; (c) $V \to 0$ ✓. These are the boundaries that make Kinetic Theory of Gases and real-gas corrections necessary.

Ex 7 — Cell G: real-world word problem

Worked example Weather balloon rising

Q: A weather balloon is released holding $50.0$ L of gas at ground level: $1.00$ atm, $290 K$ . High up the pressure is $0.40$ atm and the temperature $220 K$ . The balloon is not sealed rigidly — it stretches, but no gas escapes. Find the new volume.

Forecast: lower pressure lets it expand a lot; lower temperature shrinks it a little. Pressure dropped $2.5 \times$ , temperature only fell modestly, so net expansion — expect well above $50$ L.

Translate words to symbols: $P_{1} = 1.00, V_{1} = 50.0, T_{1} = 290$ ; $P_{2} = 0.40, T_{2} = 220$ , find $V_{2}$ . No gas escapes → $n$ constant → combined law. Why this step? "No gas escapes" is the signal that $n$ is fixed, so we may use the combined law.
Solve for $V_{2}$ : $V_{2} = V_{1} \frac{P _{1}}{P _{2}} \frac{T _{2}}{T _{1}}$ . Why this step? Rearrange $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ for the one unknown.
Plug in: $V_{2} = 50.0 \times \frac{1.00}{0.40} \times \frac{220}{290} = 50.0 \times 2.5 \times 0.7586 = 94.8 L$ . Why this step? Two competing dimensionless factors: pressure ratio $2.5$ (expand) beats temperature ratio $0.76$ (shrink).

Verify: $\frac{P _{2} V _{2}}{T _{2}} = \frac{0.40 \times 94.8}{220} = 0.1724$ ; $\frac{P _{1} V _{1}}{T _{1}} = \frac{1.00 \times 50.0}{290} = 0.1724$ . Equal ✓. Expanded as forecast.

Ex 8 — Cell H: exam twist (units + Celsius trap)

Worked example Everything is in the "wrong" units

Q: $8.00$ g of oxygen ( $O_{2}$ , $M = 32.0 g/mol$ ) is held in a $5.00$ L vessel at $27^{\circ} C$ . Find the pressure in atm.

Forecast: $0.25$ mol in $5$ L is dilute, so expect a modest pressure, order $1$ atm.

Convert the temperature FIRST: $T = 27 + 273.15 = 300.15 K$ (use $300 K$ ). Why this step? The laws require absolute temperature; plugging $27$ would be the classic Celsius mistake.
Find moles: $n = \frac{m}{M} = \frac{8.00}{32.0} = 0.250 mol$ . Why this step? $P V = n R T$ needs moles, not grams.
Isolate $P$ : $P = \frac{n R T}{V}$ , with $R = 0.0821$ (atm & L wanted). Why this step? Choose $R$ to match the units we were given and asked for.
Plug in: $P = \frac{0.250 \times 0.0821 \times 300}{5.00} = 1.23 atm$ . Why this step? Substitute the converted, consistent quantities.

Verify: dilute gas gives $\approx 1.2$ atm as forecast ✓. Had we wrongly used $27$ K, we'd have gotten $0.11$ atm — absurdly low, a red flag for the Celsius trap.

Recall

Recall Test yourself on the matrix

When $n$ changes, may I use the combined law? ::: No — use full $P V = n R T$ at each state (Ex 4). Isothermal state-change collapses to which single law? ::: Boyle's law, $P_{1} V_{1} = P_{2} V_{2}$ (Ex 3). As $V \to 0$ at fixed $n, T$ , what does $P$ do? ::: $P \to \infty$ (Ex 6b). First thing to do when temperature is given in $^{\circ}$ C? ::: Convert to kelvin, $T (K) = T (^{\circ} C) + 273.15$ (Ex 8). Which $R$ with pressure in atm and volume in litres? ::: $0.0821 L atm K^{- 1} mol^{- 1}$ .

Connections

Parent topic — the core equation these scenarios exercise
Boyle's Law — the isothermal cell (Ex 3)
Charles's Law — the isobaric limit
Avogadro's Law — the $V \propto n$ view behind Ex 4
Kinetic Theory of Gases — why the limits in Ex 6 behave as they do
Real Gases and van der Waals Equation — where Ex 6's $V \to 0$ prediction breaks
Dalton's Law of Partial Pressures — the multi-gas extension