Combined gas law and ideal gas equation PV = nRT: Exercises

Here is the whole toolbox on one card, so you never have to hunt:

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

The picture above is the decision map we will use again and again: read the question, ask "is $n$ fixed and changing states?" → combined law; "is density or molar mass involved?" → $ρ = P M / R T$ ; otherwise plug straight into $P V = n R T$ .

Level 1 — Recognition

Here you only need to spot the right relation and rearrange it. No traps yet.

Recall Solution L1·1

WHAT we want: $V$ . WHY $P V = n R T$ : we know $P$ , $n$ , $T$ and want $V$ — the master equation contains all four. Rearrange to isolate $V$ : $V = \frac{n R T}{P}$ . Units are atm and litres, so use $R = 0.0821$ . $V = \frac{3.0 \times 0.0821 \times 250}{1.5} = \frac{61.575}{1.5} = 41.05 L$ Answer: $V \approx 41.1 L$ .

Recall Solution L1·2

WHAT changes: $V$ and $P$ only; $n$ sealed, $T$ constant. That is exactly Boyle's Law, the $P V = const$ special case (see Boyle's Law). $P_{1} V_{1} = P_{2} V_{2} \Rightarrow P_{2} = P_{1} \frac{V _{1}}{V _{2}} = 2.0 \times \frac{1.0}{0.40} = 5.0 atm$ WHY bigger: you squeezed the gas into $2/5$ of its room, so hits on the wall are more frequent → pressure up. Answer: $P_{2} = 5.0 atm$ .

Level 2 — Application

Now you must convert units and keep them consistent.

Recall Solution L2·1

WHAT first: convert temperature. $T = 27 + 273.15 = 300.15 K$ (use $300 K$ ). WHY $P V = n R T$ : all of $n, V, T$ known, want $P$ . $P = \frac{n R T}{V} = \frac{0.50 \times 0.0821 \times 300}{12.0} = \frac{12.315}{12.0} = 1.026 atm$ Answer: $P \approx 1.03 atm$ .

Recall Solution L2·2

WHY $R = 8.314$ : the volume is in m³ and we want pascals — that is SI, so use the SI value of $R$ . $P = \frac{n R T}{V} = \frac{1.20 \times 8.314 \times 350}{0.0100} = \frac{3491.88}{0.0100} = 349188 Pa$ Answer: $P \approx 3.49 \times 1 0^{5} Pa$ (about $3.45 atm$ ).

Level 3 — Analysis

Here the "obvious" formula is a trap; you must reason about what is held fixed.

Recall Solution L3·1

WHY not $P V = n R T$ : we never learn $n$ , but the balloon is sealed so $n$ is constant → it cancels. Use the combined law. $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}} \Rightarrow V_{2} = V_{1} \frac{P _{1}}{P _{2}} \frac{T _{2}}{T _{1}}$ $V_{2} = 8.0 \times \frac{1.0}{0.40} \times \frac{250}{290} = 8.0 \times 2.5 \times 0.8621 = 17.24 L$ WHY it expands: the drop in outside pressure ( $\times 2.5$ swelling) beats the cooling ( $\times 0.862$ shrinking), so net it grows. Answer: $V_{2} \approx 17.2 L$ .

Recall Solution L3·2

WHY the combined law is illegal here: $n$ is not constant — you added gas. So $\frac{P V}{T} = n R$ is a different constant before and after. Go back to $P V = n R T$ for each state. With $V, T$ (and $R$ ) identical in both states, $P \propto n$ : $\frac{n _{2}}{n _{1}} = \frac{P _{2}}{P _{1}} = \frac{3.2}{2.0} = 1.6$ Answer: moles increased by a factor of $1.6$ (a $60%$ increase).

Level 4 — Synthesis

Two ideas at once: density/molar-mass forms combined with the ideal gas law.

Recall Solution L4·1

WHY $ρ = P M / R T$ : it is the ideal gas law rewritten with $n = m / M$ and $ρ = m / V$ — the only tool that links density to molar mass. Rearrange for $M$ : $M = \frac{ρR T}{P}$ . $M = \frac{0.714 \times 0.0821 \times 273}{1} = 16.00 g/mol$ $16 g/mol$ → this is $CH_{4}$ (methane). Answer: $M \approx 16.0 g/mol$ , methane.

Recall Solution L4·2

Plan: first find moles with $P V = n R T$ , then mass with $m = n M$ . $n = \frac{P V}{R T} = \frac{10.0 \times 50.0}{0.0821 \times 300} = \frac{500}{24.63} = 20.30 mol$ $m = n M = 20.30 \times 32.0 = 649.6 g$ Answer: $m \approx 650 g$ (about $0.65 kg$ ).

Level 5 — Mastery

Multi-step, messy, and you must decide which tool at each stage.

Recall Solution L5·1

WHY per-component $P V = n R T$ : each gas fills the whole $10.0 L$ on its own and behaves ideally, so its partial pressure obeys $P_{i} V = n_{i} R T$ (this is Dalton's Law of Partial Pressures). $P_{N_{2}} = \frac{n R T}{V} = \frac{0.20 \times 0.0821 \times 300}{10.0} = \frac{4.926}{10.0} = 0.4926 atm$ $P_{O_{2}} = \frac{0.30 \times 0.0821 \times 300}{10.0} = \frac{7.389}{10.0} = 0.7389 atm$ (b) Total pressure = sum of partials (Dalton): $P_{total} = 0.4926 + 0.7389 = 1.2315 atm$ (c) Mole fraction of $O_{2}$ = its moles ÷ total moles: $x_{O_{2}} = \frac{0.30}{0.20 + 0.30} = \frac{0.30}{0.50} = 0.60$ Answers: $P_{N_{2}} \approx 0.493 atm$ , $P_{O_{2}} \approx 0.739 atm$ , $P_{total} \approx 1.23 atm$ , $x_{O_{2}} = 0.60$ .

Recall Solution L5·2

Stage 1 (leak): $V$ and $T$ fixed, but $n$ dropped — the pressure fall from $6.0$ to $2.0 atm$ already tells us the new amount. We don't even need $n$ explicitly; the state right after the valve closes is: $P = 2.0 atm$ , $V = 4.0 L$ , $T = 400 K$ . WHY we cannot use combined law across the leak: $n$ changed while gas escaped, so bridge only after the valve is shut. Stage 2 (heat, sealed): now $n$ is constant, $V$ constant. Apply combined law between the closed state and the hot state: $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}, V_{1} = V_{2} \Rightarrow \frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$ $P_{2} = P_{1} \frac{T _{2}}{T _{1}} = 2.0 \times \frac{500}{400} = 2.0 \times 1.25 = 2.5 atm$ Answer: final pressure $= 2.5 atm$ .

Recall Solution L5·3

WHY $ρ = P M / R T$ : we want density from $P, M, T$ directly. $ρ = \frac{P M}{R T} = \frac{2.0 \times 44.0}{0.0821 \times 350} = \frac{88.0}{28.735} = 3.063 g/L$ Answer: $ρ \approx 3.06 g/L$ . Reality note: at high pressure / low temperature $CO_{2}$ deviates from ideal behaviour — see Real Gases and van der Waals Equation. Here conditions are mild, so the ideal estimate is trustworthy.

Wrap-up recall

Recall Which tool for which clue? (hide and test yourself)

"Find $P / V / T / n$ , one state given" ::: $P V = n R T$ , isolate the unknown "Before → after, sealed sample" ::: combined law $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ "Density or molar mass appears" ::: $ρ = \frac{P M}{R T}$ "Gas added, leaked, or reacted" ::: $n$ changed — use $P V = n R T$ per state, never combined law "Two gases share a container" ::: partial pressures $P_{i} V = n_{i} R T$ , then sum (Dalton) "Temperature in $^{\circ}$ C" ::: add $273.15$ to get kelvin first

Connections

Boyle's Law — used in L1·2
Charles's Law — the $P$ / $T$ move in L5·2
Avogadro's Law — the $P \propto n$ logic in L3·2
Dalton's Law of Partial Pressures — the whole of L5·1
Real Gases and van der Waals Equation — the caveat in L5·3
Kinetic Theory of Gases — why pressure rises when you add molecules