van der Waals equation (P + a - V²)(V − b) = RT — physical meaning of a, b: Worked examples

Before anything, the two tools we lean on the whole page:

Read the pressure formula as a tug-of-war between two terms:

Repulsion / size term: $\frac{n R T}{V - nb}$ . Because we shrank the available volume from $V$ to $V - nb$ , this fraction is bigger than the ideal $\frac{n R T}{V}$ . This term pushes $P$ up.
Attraction term: $- \frac{a n ^{2}}{V ^{2}}$ . A straight subtraction — attractions pull $P$ down.

Every scenario below is just: which term wins? The figure below plots exactly these two terms for CO₂ so you can see the competition before we run any numbers.

Figure — van der Waals equation (P + a - V²)(V − b) = RT — physical meaning of a, b

Look at the plot: the green push-up curve and the orange pull-down curve. Where the orange sits above the gap between the blue real- $P$ curve and the gray ideal line, attraction is winning (blue dips below gray, $Z < 1$ ). At small $V$ on the left, the green curve rockets up as $V \to nb$ — that is the finite-size wall of Ex 2/Ex 3. The red marker is Ex 1: at $V = 1$ L attraction wins, so blue is below gray. Every worked example below is one vertical slice of this picture.

The scenario matrix

Every case this topic can pose falls into one of these cells. The right column names the worked example that covers it.

#	Case class	What is happening physically	Sign of $(P_{real} - P_{ideal})$ / value of $Z$	Example
C1	Attraction dominates (moderate P)	$a$ -term $>$ size-term	$P_{real} < P_{ideal}$ , $Z < 1$	Ex 1
C2	Size/repulsion dominates (very high P)	size-term $>$ $a$ -term	$P_{real} > P_{ideal}$ , $Z > 1$	Ex 2
C3	Degenerate: $a = 0$ (only size matters)	hard spheres, no stickiness	always $Z > 1$	Ex 3
C4	Degenerate: $b = 0$ (only attraction)	point particles that attract	always $Z < 1$	Ex 4
C5	Limiting: $V \to \infty$ / low P	molecules far apart	$Z \to 1$ (ideal)	Ex 5
C6	The exact cancel ( $Z = 1$ but not ideal)	two errors happen to balance — the Boyle idea	$Z = 1$	Ex 6
C7	Word problem (compare two gases)	read $a$ , $b$ off a table	which liquefies / which is bigger	Ex 7
C8	Exam twist: back-solve for $a$	given $P$ , find a constant	numeric	Ex 8

We use CO₂ data throughout where handy: $a = 3.59 atm L^{2} mol^{- 2}$ , $b = 0.0427 L mol^{- 1}$ , $R = 0.0821 atm L mol^{- 1} K^{- 1}$ .

Ex 1 — Cell C1: attraction dominates

Worked example 1 mol CO₂ in

$1.00 L$ at $300 K$ Find the real pressure and compare to the ideal-gas prediction.

Forecast: guess before computing — is real pressure higher or lower than ideal? (Moderate density, CO₂ is fairly sticky… lean toward lower.)

Step 1 — ideal pressure first, for a baseline. $P_{ideal} = \frac{n R T}{V} = \frac{1 \times 0.0821 \times 300}{1.00} = 24.63 atm$ Why this step? We need something to compare against; "real vs ideal" is meaningless without the ideal number.

Step 2 — the size term. $\frac{n R T}{V - nb} = \frac{24.63}{1.00 - 0.0427} = \frac{24.63}{0.9573} = 25.73 atm$ Why this step? Available volume is $V - nb$ , not $V$ ; this raises the fraction from $24.63$ to $25.73$ .

Step 3 — the attraction term. $\frac{a n ^{2}}{V ^{2}} = \frac{3.59 \times 1 ^{2}}{1.0 0 ^{2}} = 3.59 atm$ Why this step? This is the amount pulled off the pressure by attractions.

Step 4 — combine. $P = 25.73 - 3.59 = 22.14 atm$ Why this step? The pressure formula is literally (size push-up) minus (attraction pull-down), so we subtract Step 3 from Step 2. The $3.59$ pull-down beats the $1.10$ push-up ( $25.73 - 24.63$ ), so real $<$ ideal — forecast confirmed.

Verify: $Z = \frac{P V _{m}}{R T} = \frac{22.14 \times 1.00}{0.0821 \times 300} = \frac{22.14}{24.63} = 0.899 < 1$ . Attractions dominate ✓, matching cell C1.

Ex 2 — Cell C2: size/repulsion dominates ( $Z > 1$ )

Worked example 1 mol H₂ squeezed into

$V = 0.0500 L$ at $300 K$ H₂ barely attracts ( $a = 0.244 atm L^{2} mol^{- 2}$ ) but has co-volume $b = 0.0266 L mol^{- 1}$ . Find $P$ and $Z$ , and compare to ideal.

Forecast: H₂ is tiny and nonpolar — attraction is weak. At this squeeze the finite-size term should win. Guess $P_{real} > P_{ideal}$ , $Z > 1$ .

Step 1 — size term. $\frac{n R T}{V - nb} = \frac{1 \times 0.0821 \times 300}{0.0500 - 0.0266} = \frac{24.63}{0.0234} = 1052.6 atm$ Why this step? $V - nb = 0.0234$ is very small, so the push-up fraction becomes enormous — the finite-size wall.

Step 2 — attraction term. $\frac{a n ^{2}}{V ^{2}} = \frac{0.244}{0.050 0 ^{2}} = \frac{0.244}{0.00250} = 97.6 atm$ Why this step? H₂'s tiny $a$ keeps the pull-down small even at this density; this is the amount subtracted.

Step 3 — combine. $P = 1052.6 - 97.6 = 955.0 atm$ Why this step? Same subtraction — real $P$ = size push-up (Step 1) minus attraction pull-down (Step 2). The push-up dwarfs the pull-down, so $P$ ends up large.

Step 4 — ideal comparison. $P_{ideal} = 24.63/0.0500 = 492.6 atm$ , so $P_{real} = 955.0 > 492.6$ . Why this step? Comparing real to ideal is the whole point of the cell — it tells us which correction won. Here real exceeds ideal, so the size ( $b$ ) term is dominant — forecast confirmed, this is a true $Z > 1$ case.

Verify: $Z = \frac{955.0 \times 0.0500}{0.0821 \times 300} = \frac{47.75}{24.63} = 1.939 > 1$ . Size/repulsion dominates ✓, matching cell C2. (Contrast Ex 1: CO₂'s large $a$ keeps it below ideal at comparable density — that is why we picked weakly-attracting H₂ here.)

Ex 3 — Cell C3: degenerate $a = 0$ (only size)

Worked example A "hard-sphere gas": set

$a = 0$ , keep $b = 0.0427$ . 1 mol, $V = 0.200 L$ , $T = 300 K$ . This is the pretend gas with size but no stickiness.

Forecast: with the pull-down term gone, must $Z$ be above, below, or exactly 1?

Step 1 — the equation collapses. $P = \frac{n R T}{V - nb} - 0 = \frac{24.63}{0.200 - 0.0427} = \frac{24.63}{0.1573} = 156.6 atm$ Why this step? With $a = 0$ only the size term survives, and it is always larger than the ideal fraction.

Step 2 — the $Z$ formula shortcut. With $a = 0$ , $Z = \frac{P V _{m}}{R T} = \frac{V _{m}}{V _{m} - b} = \frac{0.200}{0.1573} = 1.271$ Why this step? Substituting $P = R T / (V_{m} - b)$ into $Z$ cancels $R T$ and leaves a pure geometric ratio — a clean formula worth memorising: hard spheres always give $Z = \frac{V _{m}}{V _{m} - b} > 1$ .

Verify: $Z = 1.271 > 1$ ✓ (cell C3). And numerically $P V_{m} / R T = 156.6 \times 0.200/24.63 = 1.271$ ✓ — the two routes agree.

Ex 4 — Cell C4: degenerate $b = 0$ (only attraction)

Ex 5 — Cell C5: limiting behaviour $V \to \infty$ (low pressure)

Worked example 1 mol CO₂ in a

huge $V = 100.0 L$ at $300 K$ . Show it acts ideal.

Forecast: both corrections should shrink toward nothing. Guess $Z \approx 1$ .

Step 1 — size term barely differs. $\frac{n R T}{V - nb} = \frac{24.63}{100.0 - 0.0427} = \frac{24.63}{99.957} = 0.24641 atm$ Why this step? $nb = 0.0427$ is negligible against $100$ , so $V - nb \approx V$ . The size term is $0.24641$ atm — almost exactly the ideal $0.24630$ atm, so the correction it adds is only $\sim 0.0001$ atm, i.e. negligible.

Step 2 — attraction term vanishes even faster. $\frac{a n ^{2}}{V ^{2}} = \frac{3.59}{100. 0 ^{2}} = 0.000359 atm$ Why this step? The $1/ V^{2}$ shape makes attraction die off faster than the $1/ V$ -ish size effect as $V$ grows.

Step 3 — combine. $P = 0.24641 - 0.000359 = 0.24605 atm, P_{ideal} = 24.63/100 = 0.24630$ Why this step? Same subtraction (push-up minus pull-down); both corrections are now tiny, so the result sits almost exactly on the ideal value.

Verify: $Z = \frac{0.24605 \times 100.0}{24.63} = \frac{24.605}{24.63} = 0.99898 \approx 1$ ✓ (cell C5). Every real gas $\to$ ideal as $P \to 0$ .

Ex 6 — Cell C6: the exact cancel ( $Z = 1$ but NOT ideal)

Worked example Two-part twist. (a) At

300 K, find the molar volume $V_{m}$ at which CO₂ has $Z = 1$ (yet is a real gas). (b) Show there is a special temperature — the Boyle temperature — at which $Z \to 1$ for all large $V$ . This is the essence of the Boyle Temperature idea.

Forecast: for part (a) guess whether such a $V_{m}$ exists at 300 K (the volume where "size push-up" exactly cancels "attraction pull-down"). For part (b) guess whether the cancel can be made to hold for every volume at once.

Step 1 — write the exact $Z$ for van der Waals. Combine $P$ with $Z = P V_{m} / R T$ (per mole, $n = 1$ ): $Z = \frac{P V _{m}}{R T} = \frac{V _{m}}{V _{m} - b} - \frac{a}{R T V _{m}}$ Why this step? Plug $P = \frac{R T}{V _{m} - b} - \frac{a}{V _{m}^{2}}$ into $Z = P V_{m} / R T$ . The first term becomes $\frac{V _{m}}{V _{m} - b}$ (the exact size ratio from Ex 3) and the second becomes $\frac{a}{R T V _{m}}$ (the exact attraction drop from Ex 4). This form is exact — no approximation yet.

Step 2 (part a) — set $Z = 1$ at 300 K and solve for $V_{m}$ . $\frac{V _{m}}{V _{m} - b} - \frac{a}{R T V _{m}} = 1$ Multiply through by $V_{m} (V_{m} - b)$ and use $V_{m} / (V_{m} - b) - 1 = b / (V_{m} - b)$ : $\frac{b}{V _{m} - b} = \frac{a}{R T V _{m}} ⟹ b R T V_{m} = a (V_{m} - b)$ $V_{m} (b R T - a) = - ab ⟹ V_{m} = \frac{ab}{a - b R T}$ Why this step? This is the exact condition for the two errors to cancel — we solved a genuine algebraic equation, no expansion. Plug numbers ( $a = 3.59, b = 0.0427, R = 0.0821, T = 300$ ): $a - b R T = 3.59 - 0.0427 \times 0.0821 \times 300 = 3.59 - 1.0517 = 2.538$ $V_{m} = \frac{3.59 \times 0.0427}{2.538} = \frac{0.15329}{2.538} = 0.0604 L mol^{- 1}$ So at 300 K there is a real-gas volume ( $V_{m} \approx 0.0604$ L) where $Z = 1$ exactly, even though the gas is far from ideal.

Step 3 (part b) — the temperature that cancels for all large $V_{m}$ . Take the exact $Z$ and expand only the size ratio for the low-density regime $b / V_{m} ≪ 1$ (molecules far apart): $\frac{V _{m}}{V _{m} - b} = \frac{1}{1 - b / V _{m}} = 1 + \frac{b}{V _{m}} + \frac{b ^{2}}{V _{m}^{2}} + \dots$ Why this step? The geometric series $\frac{1}{1 - x} = 1 + x + x^{2} + \dots$ is valid only when $x = b / V_{m} < 1$ — i.e. when the molar volume is much larger than the co-volume. We keep terms to first order in $b / V_{m}$ ; the dropped terms are $O (b^{2} / V_{m}^{2})$ , which are negligible precisely because $b / V_{m} ≪ 1$ . So: $Z \approx 1 + \frac{b}{V _{m}} - \frac{a}{R T V _{m}} = 1 + \frac{1}{V _{m}} (b - \frac{a}{R T}) + O (\frac{b ^{2}}{V _{m}^{2}})$ Why each term? The $+ b / V_{m}$ is the size push-up (from the expansion); the $- a / (R T V_{m})$ is the attraction pull-down (exact). Both share a $1/ V_{m}$ , so they collect into one bracket.

Step 4 (part b) — kill the bracket. For $Z \approx 1$ regardless of (large) $V_{m}$ , the bracket must vanish: $b - \frac{a}{R T} = 0 ⟹ T = \frac{a}{R b} = \frac{3.59}{0.0821 \times 0.0427} = 1024 K$ Why this step? Only at this Boyle temperature do the first-order size and attraction corrections cancel simultaneously for every large volume, so the gas mimics ideal over a wide range.

Verify (a): put $V_{m} = 0.0604$ back into the exact $Z$ : $\frac{0.0604}{0.0604 - 0.0427} - \frac{3.59}{0.0821 \times 300 \times 0.0604} = \frac{0.0604}{0.0177} - \frac{3.59}{1.4877} = 3.413 - 2.413 = 1.000$ ✓. Verify (b): $\frac{a}{R b} = \frac{3.59}{0.0821 \times 0.0427} = 1024 K$ ✓. Since 300 K is far below 1024 K, at ordinary volumes CO₂ has $Z < 1$ (attraction wins) — exactly what Ex 1 found.

Ex 7 — Cell C7: word problem (compare two gases)

Worked example Two cylinders, same

$T$ and $P$ . Gas X: $a = 6.49, b = 0.0562$ . Gas Y: $a = 0.211, b = 0.0171$ (SI-ish, same units). (a) Which liquefies more easily? (b) Which has larger molecules?

Forecast: guess before reading the rule — big $a$ or big $b$ controls liquefaction?

Step 1 — liquefaction is about attraction, so compare $a$ . $a_{X} = 6.49 > a_{Y} = 0.211$ . Why this step? $a$ measures Intermolecular Forces; stronger pull ⇒ molecules can be held together ⇒ higher critical temperature ⇒ easier to liquefy. $\Rightarrow$ Gas X liquefies more easily.

Step 2 — molecular size is about excluded volume, so compare $b$ . $b_{X} = 0.0562 > b_{Y} = 0.0171$ . Why this step? $b = 4 N_{A} \cdot \frac{4}{3} π r^{3}$ grows with molecular radius; bigger $b$ ⇒ bigger molecule. $\Rightarrow$ Gas X molecules are larger. (Here X is like CO₂/a hydrocarbon; Y like H₂/He.)

Verify: ratio check — $a_{X} / a_{Y} = 6.49/0.211 = 30.8$ and $b_{X} / b_{Y} = 0.0562/0.0171 = 3.29$ . Why this step? Both ratios exceeding 1 confirms X beats Y on both counts (stickier and bulkier), so there is no contradiction in our two conclusions — the checks cross-validate Steps 1 and 2 ✓.

Ex 8 — Cell C8: exam twist (back-solve for a constant)

Worked example A gas has

$b = 0.0500 L mol^{- 1}$ . For 1 mol in $V = 2.00 L$ at $T = 350 K$ , the measured pressure is $P = 13.80 atm$ . Find $a$ .

Forecast: ideal pressure here is $n R T / V = 0.0821 \times 350/2 = 14.37$ . Measured is lower ⇒ attractions present ⇒ expect a positive $a$ .

Step 1 — rearrange the vdW pressure formula for $a$ . $\frac{a n ^{2}}{V ^{2}} = \frac{n R T}{V - nb} - P$ Why this step? We know every quantity on the right, so isolate the only unknown, $a$ .

Step 2 — plug the size term. $\frac{n R T}{V - nb} = \frac{0.0821 \times 350}{2.00 - 0.0500} = \frac{28.735}{1.95} = 14.736 atm$ Why this step? This is the push-up pressure; subtract the measured $P$ from it to get the attraction pull-down.

Step 3 — solve for $a$ . $\frac{a}{V ^{2}} = 14.736 - 13.80 = 0.936 atm \Rightarrow a = 0.936 \times V^{2} = 0.936 \times 4.00 = 3.74 atm L^{2} mol^{- 2}$ Why this step? Multiply through by $V^{2} = 4.00$ (with $n = 1$ ) to free $a$ .

Verify: rebuild $P$ with $a = 3.74$ : $P = 14.736 - \frac{3.74}{4.00} = 14.736 - 0.935 = 13.80 atm$ ✓. Positive $a$ , matching the forecast that attractions lowered the pressure.

Recall Which cell am I in? (self-test)

$Z < 1$ , moderate P ::: attraction dominates (C1) $Z > 1$ , extreme squeeze of a weakly-attracting gas ::: size/repulsion dominates (C2) Set $a = 0$ : $Z$ is always ::: greater than 1, equal to $V_{m} / (V_{m} - b)$ (C3) Set $b = 0$ : $Z$ is always ::: less than 1, equal to $1 - a / (R T V_{m})$ (C4) $V \to \infty$ ::: $Z \to 1$ , ideal (C5) Temperature where corrections cancel to first order ::: Boyle temperature $T = a / (R b)$ (C6) Compare which gas liquefies easier ::: bigger $a$ (C7) Given $P$ , find a constant ::: rearrange $a n^{2} / V^{2} = n R T / (V - nb) - P$ (C8)

Connections

Parent: physical meaning of a and b
Compressibility Factor Z — every example ended with a $Z$ check
Boyle Temperature — the exact-cancel cell C6, $T = a / (R b)$
Critical Constants and Liquefaction — why big $a$ (Ex 7) means easy liquefaction
Intermolecular Forces — the physics behind the $a$ term
Ideal Gas Law — the $V \to \infty$ limit (Ex 5)
Kinetic Theory of Gases — the assumptions these corrections repair