Real gases — deviations from ideality, compressibility factor Z: Worked examples

This page is a drill floor. The parent note built the tools; here we throw every kind of question at them, one at a time, and never leave a case unshown. Before each example, you'll try to guess the answer first — that's the whole point.

Every symbol we use was defined in the parent: $Z = \frac{P V _{m}}{R T}$ is the ratio of "what we measured" to "what the ideal law predicted"; $a$ measures attraction; $b$ measures the space molecules physically occupy; $V_{m} = V / n$ is volume per mole; $T_{B} = \frac{a}{R b}$ is the Boyle temperature. If any of these feel shaky, re-read the parent — this page assumes you can state them but shows you how to use them.

The scenario matrix

Every question this topic can throw at you falls into one of these cells. The examples below hit every single one.

Cell	What makes it distinct	Example
A. Plain $Z$ from data	just plug $P, V, n, T$ into $Z = P V / n R T$	Ex 1
B. $Z < 1$ sign case	attraction wins → interpret direction	Ex 2
C. $Z > 1$ sign case	size/repulsion wins ( $H_{2}$ , high P)	Ex 3
D. Degenerate: $Z = 1$ exactly	ideal behaviour — is it luck or Boyle T?	Ex 4
E. Limiting case: $V_{m} \to \infty$ (low $P$ )	slope formula $Z \approx 1 + \frac{1}{V _{m}} (b - \frac{a}{R T})$	Ex 5
F. Limiting case: $P \to \infty$ (huge $P$ )	size term blows up, $Z \to \infty$	Ex 6
G. Boyle temperature	solve $T_{B} = a / R b$ , or "at what $T$ ideal?"	Ex 7
H. Real-world word problem	scuba/industrial, translate words to numbers	Ex 8
I. Exam twist	"find the real volume" — inverting $Z$	Ex 9

Example 1 — Cell A: plain $Z$ from raw data

Worked example The most basic question

$0.75$ mol of a gas occupies $V = 15 L$ at $P = 1.2 atm$ and $T = 320 K$ . Find $Z$ . Use $R = 0.0821 atm\cdotL\cdotmol^{- 1} K^{- 1}$ .

Forecast: Is $Z$ going to be near 1, well above, or well below? (These are mild conditions — low pressure — so guess slightly below 1 if attractions matter, near 1 otherwise.)

Step 1 — Write the definition using measured quantities. $Z = \frac{P V}{n R T}$ Why this step? $Z$ is literally "real over ideal". The numerator $P V$ is what the world gave us; the denominator $n R T$ is what the ideal law would give for the same $n, T$ .

Step 2 — Substitute. $Z = \frac{1.2 \times 15}{0.75 \times 0.0821 \times 320} = \frac{18}{19.704} = 0.9135$ Why this step? Pure arithmetic — no physics judgement yet, just compute the ratio.

Step 3 — Interpret. $Z = 0.914 < 1$ → the gas is more compressible than ideal → attractions are winning at this low pressure. Why this step? A number alone isn't an answer in chemistry; the direction of deviation tells the story.

Verify: Units cancel: $\frac{atm \cdot L}{mol \cdot ( atm\cdotL\cdotmol ^{- 1} K ^{- 1} ) \cdot K}$ = dimensionless ✔. And $0.91$ is close to 1 (mild conditions), matching our forecast. ✔

Example 2 — Cell B: $Z < 1$ , attraction dominates

Worked example Reading a

$Z < 1$ result At $50 atm$ and $273 K$ , $CO_{2}$ shows $Z \approx 0.35$ . What is physically happening?

Forecast: $Z$ is far below 1 — this is a strong signal. Guess: attractions are dominating hard, and the real volume is much smaller than ideal.

Step 1 — Compare to 1. $Z = 0.35$ is way below 1. Why this step? The sign of $(Z - 1)$ names the winning effect; here $(Z - 1) < 0$ → attraction.

Step 2 — Translate to volume. Since $Z = \frac{V _{real}}{V _{ideal}}$ (rearranging $Z = P V / n R T$ with $V_{ideal} = n R T / P$ ), we get $V_{real} = 0.35 V_{ideal}$ . Why this step? At fixed $P, T, n$ , $Z$ is the ratio of real volume to ideal volume. This makes "more compressible" concrete: the real gas shrank to 35% of the predicted volume.

Step 3 — Physical cause. $CO_{2}$ has a large $a$ (big, polarizable molecule), and we're near its critical region (high $P$ , moderate $T$ ). Attraction pulls molecules together → volume collapses → $Z ≪ 1$ . Why this step? $CO_{2}$ 's large $a$ (see Intermolecular Forces) is exactly why the attraction term $- a / (R T V_{m})$ is large and negative here.

Verify: $V_{real} = 0.35 V_{ideal}$ means the gas occupies far less than ideal — consistent with $Z < 1$ meaning "more compressible". ✔ (Sanity: $Z = 0.35$ near critical conditions is realistic for $CO_{2}$ .)

Example 3 — Cell C: $Z > 1$ , size/repulsion dominates

Worked example The

$H_{2}$ / high-pressure case Hydrogen at $r oo m t e m p er a t u r e$ and $200 atm$ has $Z \approx 1.13$ . Explain, and find $V_{real} / V_{ideal}$ .

Forecast: $Z > 1$ — the opposite of Ex 2. Guess: the real gas takes up more room than ideal predicts, because the molecules' own volume resists squeezing.

Step 1 — Sign check: $Z - 1 = + 0.13 > 0$ → repulsion/finite-size wins. Why this step? Positive deviation always means the size term $\frac{V _{m}}{V _{m} - b}$ is overpowering the attraction term.

Step 2 — Volume ratio: $V_{real} = Z V_{ideal} = 1.13 V_{ideal}$ . The gas refuses to be squeezed to the ideal volume — it stays 13% bigger. Why this step? Same identity as Ex 2, but now $Z > 1$ so real volume exceeds ideal.

Step 3 — Why $H_{2}$ specifically? $H_{2}$ has a tiny $a$ (weak attraction — see Intermolecular Forces). At room $T$ , $b > a / R T$ , so the size effect wins even at moderate pressure. Its Boyle Temperature is far below room temperature. Why this step? This connects the number to the parent's low- $P$ slope rule: $Z$ starts above 1 whenever $b > a / R T$ .

Verify: $He$ and $H_{2}$ are the textbook "always $Z > 1$ at room $T$ " gases; $1.13$ is a realistic value. ✔ Volume ratio $> 1$ matches $Z > 1$ . ✔

Example 4 — Cell D: the degenerate case $Z = 1$

Worked example When a

real gas fakes ideal $N_{2}$ ( $a = 1.35$ , $b = 0.0387$ , $R = 0.0821$ ) is at low pressure and $T = 425 K$ . Compute its low- $P$ slope term $(b - \frac{a}{R T})$ . What is $Z$ ?

Forecast: $425 K$ is suspiciously close to $N_{2}$ 's Boyle temperature (parent Ex 3). Guess: the slope term is $\approx 0$ , so $Z \approx 1$ even though $N_{2}$ is a real gas.

Step 1 — Compute $\frac{a}{R T}$ . $\frac{a}{R T} = \frac{1.35}{0.0821 \times 425} = \frac{1.35}{34.8925} = 0.03869 L\cdotmol^{- 1}$ Why this step? This is the "attraction pulls $Z$ down" contribution to the slope. We must weigh it against $b$ .

Step 2 — Subtract. $b - \frac{a}{R T} = 0.0387 - 0.03869 = 0.00001 \approx 0$ Why this step? The parent's low- $P$ formula $Z \approx 1 + \frac{1}{V _{m}} (b - \frac{a}{R T})$ says: if this bracket is zero, $Z = 1$ regardless of $V_{m}$ .

Step 3 — Conclude: the two lies exactly cancel. $N_{2}$ at $425 K$ behaves ideally over a range of low pressures — this is the definition of the Boyle temperature. Why this step? $Z = 1$ here is not luck and not because forces vanished; the size effect and attraction effect are equal and opposite.

Verify: $425 K$ is exactly $T_{B} = a / (R b) = 1.35/ (0.0821 \times 0.0387) = 425 K$ from parent Ex 3, so the bracket must be zero. ✔

Example 5 — Cell E: low-pressure limit, the slope of $Z$ vs $P$

Worked example Predicting the initial direction

For a gas with $a = 3.6 atm\cdotL^{2} mol^{- 2}$ , $b = 0.043 L\cdotmol^{- 1}$ , at $T = 300 K$ : does $Z$ start above or below 1 as pressure rises from zero? ( $R = 0.0821$ )

Forecast: Big $a$ (strong attraction), room-ish $T$ → guess attractions win → $Z$ starts below 1.

Step 1 — Recall the low- $P$ slope formula. $Z \approx 1 + \frac{1}{V _{m}} (b - \frac{a}{R T})$ Why this step? At low pressure $1/ V_{m} \propto P$ , so the sign of the bracket $(b - a / R T)$ is the sign of the initial slope of $Z$ vs $P$ .

Step 2 — Compute $a / R T$ . $\frac{a}{R T} = \frac{3.6}{0.0821 \times 300} = \frac{3.6}{24.63} = 0.1462$ Why this step? We need to compare the attraction size $a / R T$ against $b$ .

Step 3 — Compare. $a / R T = 0.1462$ vs $b = 0.043$ . Since $\frac{a}{R T} > b$ , the bracket is negative ( $0.043 - 0.1462 = - 0.1032$ ), so $Z$ starts below 1. Why this step? Negative slope from zero pressure means the curve dips first — the classic $CO_{2}$ -like shape.

Verify: Look at the figure below — the curve with $a / R T > b$ (magenta) dips below 1 first. Bracket $= - 0.103 < 0$ ✔ matches forecast. ✔

Example 6 — Cell F: the high-pressure limit

Worked example What happens as we crush the gas

Using $Z = \frac{V _{m}}{V _{m} - b} - \frac{a}{R T V _{m}}$ , what does $Z$ do as $V_{m} \to b^{+}$ (molecules packed as tight as possible)?

Forecast: At extreme squeezing the molecules' own volume dominates everything. Guess: $Z \to + \infty$ (shoots way above 1).

Step 1 — Examine the first term as $V_{m} \to b$ . $\frac{V _{m}}{V _{m} - b} ⟶ \frac{b}{0 ^{+}} = + \infty$ Why this step? The denominator $(V_{m} - b)$ is the free space; as it goes to zero, the gas becomes essentially incompressible and this term explodes.

Step 2 — Examine the second term. $\frac{a}{R T V _{m}} \to \frac{a}{R T b}$ — a finite number. Why this step? The attraction correction stays bounded; a finite thing cannot cancel an infinite thing.

Step 3 — Combine: $Z \to + \infty$ . Every real gas eventually has $Z > 1$ at high enough pressure. Why this step? This confirms the parent's mistake-buster: the dip below 1 is only a low-to-moderate $P$ feature; the size term always wins in the end.

Verify: Numerically, take $b = 0.04$ , $V_{m} = 0.041$ : first term $= 0.041/0.001 = 41$ , huge. Second term (say $a = 1.4, R = 0.0821, T = 300$ ) $= 1.4/ (0.0821 \cdot 300 \cdot 0.041) = 1.386$ , small. $Z = 41 - 1.39 = 39.6 ≫ 1$ ✔. See the figure — all curves rise above 1 at high $P$ . ✔

Example 7 — Cell G: Boyle temperature two ways

Worked example Solve for

$T_{B}$ and use it A gas has $a = 2.28 atm\cdotL^{2} mol^{- 2}$ and $b = 0.0428 L\cdotmol^{- 1}$ ( $R = 0.0821$ ). (i) Find $T_{B}$ . (ii) At $T = 300 K$ (below $T_{B}$ ), does $Z$ start above or below 1?

Forecast: $T_{B}$ likely a few hundred K. Below $T_{B}$ , attractions win → $Z$ dips below 1 first.

Step 1 — Apply the Boyle temperature formula. $T_{B} = \frac{a}{R b} = \frac{2.28}{0.0821 \times 0.0428} = \frac{2.28}{0.0035139} = 648.8 K$ Why this step? $T_{B}$ is where the low- $P$ slope bracket $(b - a / R T)$ vanishes; setting it to zero gives $T_{B} = a / (R b)$ .

Step 2 — Compare $300 K$ to $T_{B} = 649 K$ . We're below $T_{B}$ . Why this step? Below $T_{B}$ the attraction term $a / R T$ is larger (colder → less thermal energy to overcome pulls) → attractions dominate.

Step 3 — Conclude: at $300 K$ , $Z$ starts below 1, then rises above 1 at high $P$ (Cell F). Why this step? Below Boyle temperature is exactly the "dip then rise" regime.

Verify: Check bracket at $300 K$ : $a / R T = 2.28/ (0.0821 \cdot 300) = 0.09257$ ; $b - a / R T = 0.0428 - 0.09257 = - 0.0498 < 0$ ✔ (below 1, matches forecast). At $T_{B} = 648.8$ : $a / R T_{B} = 2.28/ (0.0821 \cdot 648.8) = 0.0428 = b$ ✔ bracket zero.

Example 8 — Cell H: real-world word problem (scuba tank)

Worked example Filling a scuba cylinder

A $12 L$ scuba cylinder holds $O_{2}$ at $P = 150 atm$ and $T = 300 K$ . Ideal law predicts $n_{ideal} = P V / R T$ moles. If the real $Z$ under these conditions is $0.95$ , how many moles are actually in the tank?

Forecast: High pressure but $Z$ still below 1 means attractions still slightly winning — real tank holds slightly more moles than ideal predicts (gas packed a bit tighter than ideal).

Step 1 — Compute the ideal-law moles. $n_{ideal} = \frac{P V}{R T} = \frac{150 \times 12}{0.0821 \times 300} = \frac{1800}{24.63} = 73.08 mol$ Why this step? This is the naive answer a student using $P V = n R T$ would give — our baseline.

Step 2 — Correct with $Z$ . From $Z = \frac{P V}{n _{real} R T}$ , solve for $n_{real}$ : $n_{real} = \frac{P V}{Z R T} = \frac{n _{ideal}}{Z} = \frac{73.08}{0.95} = 76.93 mol$ Why this step? $Z < 1$ means the real gas is more compressible, so more molecules fit than ideal predicts — we divide by $Z$ ( $< 1$ ), which increases the count.

Step 3 — Interpret. About $77$ mol are really in the tank, ~ $3.9$ mol more than the ideal estimate — a $5%$ error if you'd trusted $P V = n R T$ . Why this step? Real-gas corrections matter for engineering: dosing, buoyancy and gas duration all depend on the true amount.

Verify: $n_{real} = 73.08/0.95 = 76.93$ ; check: $Z = P V / (n_{real} R T) = 1800/ (76.93 \cdot 24.63) = 1800/1894.8 = 0.95$ ✔. $Z < 1 \Rightarrow n_{real} > n_{ideal}$ ✔.

Example 9 — Cell I: exam twist (invert $Z$ to get real volume)

Worked example "Find the real molar volume"

$1 mol$ of ammonia at $P = 10 atm$ , $T = 300 K$ has $Z = 0.88$ . (i) What molar volume does the ideal law predict? (ii) What is the real molar volume?

Forecast: $Z < 1$ → real volume is smaller than ideal. Ideal $V_{m}$ is a few litres; real should be ~ $0.88 \times$ that.

Step 1 — Ideal molar volume. $V_{m, ideal} = \frac{R T}{P} = \frac{0.0821 \times 300}{10} = \frac{24.63}{10} = 2.463 L\cdotmol^{- 1}$ Why this step? Set $Z = 1$ in $Z = P V_{m} / R T$ and solve — this is the "pretend" volume.

Step 2 — Real molar volume via $Z$ . $V_{m, real} = Z \cdot V_{m, ideal} = 0.88 \times 2.463 = 2.167 L\cdotmol^{- 1}$ Why this step? Rearranging $Z = P V_{m} / R T$ gives $V_{m, real} = Z R T / P$ ; since $Z < 1$ , the real gas shrinks below the ideal prediction — attractions pulling it in.

Step 3 — Sanity. Ammonia has strong hydrogen bonding (large $a$ , see Intermolecular Forces), so $Z < 1$ at moderate $P$ is expected. Why this step? The physics must match the number: strong attraction ⇒ $Z < 1$ ⇒ smaller real volume. ✔

Verify: $V_{m, real} = 0.88 \times 2.463 = 2.167$ ; back-check $Z = P V_{m, real} / R T = 10 \cdot 2.167/24.63 = 21.67/24.63 = 0.88$ ✔.

Recall Quick self-test across the matrix

Which cell does each belong to? "Given $P, V, n, T$ , find $Z$ ." ::: Cell A (plain plug-in). " $Z = 1.13$ for $H_{2}$ at room T — explain." ::: Cell C (size dominates). " $b - a / R T = 0$ , so $Z = 1$ ." ::: Cell D (degenerate / Boyle condition). "As $V_{m} \to b$ , what is $Z$ ?" ::: Cell F ( $Z \to + \infty$ ). "Scuba tank: real moles?" ::: Cell H (word problem, $n_{real} = n_{ideal} / Z$ ).