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

Constants used throughout (state them so no symbol is unexplained):

The van der Waals equation we keep reusing:

Level 1 — Recognition

L1.1

Which van der Waals constant patches the pressure term, and which patches the volume term? State the units of each.

Recall Solution

$a$ patches pressure: it lives inside the bracket $(P + \frac{a n ^{2}}{V ^{2}})$ , which is a pressure being corrected. Units of $a$ : $atm L^{2} mol^{- 2}$ . $b$ patches volume: it lives inside $(V - nb)$ , a volume being corrected. Units of $b$ : $L mol^{- 1}$ . Mnemonic: "a Attracts, b Bulks."

L1.2

Two gases: $a (He) = 0.034$ , $a (NH_{3}) = 4.17$ (atm L²/mol²). Which one liquefies more easily, and why?

Recall Solution

NH₃. Larger $a$ ⇒ stronger intermolecular attractions ⇒ higher critical temperature ⇒ easier to pull molecules together into a liquid. Helium's tiny $a$ means almost no attraction, so it is one of the hardest gases to liquefy.

L1.3

For an ideal gas, what is $Z$ ? What does $Z < 1$ tell you about which correction ( $a$ or $b$ ) is winning?

Recall Solution

Ideal gas: $Z = 1$ exactly. $Z < 1$ means the measured $P V_{m}$ is smaller than $R T$ — pressure got softened — so attractions ( $a$ ) dominate.

Level 2 — Application

L2.1

1.00 mol of CO₂ occupies 2.00 L at 350 K. Given $a = 3.59$ , $b = 0.0427$ , find the real pressure $P$ .

Recall Solution

Use $P = \frac{n R T}{V - nb} - \frac{a n ^{2}}{V ^{2}}$ .

Repulsion term: $\frac{n R T}{V - nb} = \frac{1 \times 0.0821 \times 350}{2.00 - 0.0427} = \frac{28.735}{1.9573} = 14.68$ atm.
Attraction term: $\frac{a n ^{2}}{V ^{2}} = \frac{3.59 \times 1}{4.00} = 0.898$ atm.
$P = 14.68 - 0.898 = 13.78$ atm. Ideal law would give $n R T / V = 28.735/2 = 14.37$ atm, so the real gas is slightly lower — attractions win at this modest density.

L2.2

A gas has $b = 0.0304 L mol^{- 1}$ . Estimate the radius $r$ of one molecule (in metres). Use $b = 4 N_{A} \cdot \frac{4}{3} π r^{3}$ .

Recall Solution

WHY this formula: $b$ is $4 \times$ the actual molar volume of the molecules, and the actual molar volume is $N_{A} \times \frac{4}{3} π r^{3}$ . First convert $b$ to m³ per mole: $0.0304 L = 0.0304 \times 1 0^{- 3} m^{3} = 3.04 \times 1 0^{- 5} m^{3}$ . $\frac{4}{3} π r^{3} = \frac{b}{4 N _{A}} = \frac{3.04 \times 1 0 ^{- 5}}{4 \times 6.022 \times 1 0 ^{23}} = 1.262 \times 1 0^{- 29} m^{3} .$ $r^{3} = \frac{3}{4 π} \times 1.262 \times 1 0^{- 29} = 3.013 \times 1 0^{- 30} m^{3} \Rightarrow r = 1.44 \times 1 0^{- 10} m \approx 1.4 \overset{˚}{A} .$ A believable atomic-scale radius.

L2.3

Compute $Z$ for the CO₂ of L2.1 ( $P = 13.78$ atm, $V_{m} = 2.00$ L, $T = 350$ K). Is it $< 1$ or $> 1$ ? Interpret.

Recall Solution

$Z = \frac{P V _{m}}{R T} = \frac{13.78 \times 2.00}{0.0821 \times 350} = \frac{27.56}{28.735} = 0.959$ . $Z < 1$ ⇒ attractions dominate ⇒ gas is more compressible than an ideal gas here. Consistent with the pressure coming out below ideal.

Level 3 — Analysis

L3.1

Gas X: $a = 1.36$ , $b = 0.0319$ . Gas Y: $a = 6.49$ , $b = 0.0562$ . At the same moderate pressure and temperature, which shows the larger deviation of $Z$ below 1, and which is the bigger molecule?

Recall Solution

Deviation of $Z$ below 1 is driven by attraction ( $a$ ). Y has the larger $a$ ( $6.49 > 1.36$ ), so Y dips further below $Z = 1$ (stronger attractions) at moderate pressure.
Bigger molecule is set by $b$ (excluded volume). Y has larger $b$ ( $0.0562 > 0.0319$ ), so Y is the larger molecule too. Here both point to Y, but in general $a$ and $b$ can disagree — always separate the two questions.

L3.2

Show algebraically why the pressure-correction term must scale as $(n / V)^{2}$ and not as $(n / V)$ .

Recall Solution

The inward tug on one wall-molecule is proportional to how many molecules are near it, i.e. to the number density $\propto n / V$ . But the total pressure reduction also scales with how many wall-molecules are being tugged, again $\propto n / V$ . The pressure loss is the product of two density factors: $Δ P \propto (\frac{n}{V}) \times (\frac{n}{V}) = (\frac{n}{V})^{2} .$ Attraction is a pair phenomenon; one factor for the pulled molecule, one for the pullers. Writing the constant as $a$ : $Δ P = a n^{2} / V^{2}$ .

L3.3

For 1 mol at $V_{m} = 22.4$ L and 273 K, compute the ideal pressure and the vdW pressure for CO₂ ( $a = 3.59$ , $b = 0.0427$ ). By roughly what percent do they differ? Why is the difference tiny here?

Recall Solution

Ideal: $P_{ideal} = \frac{R T}{V _{m}} = \frac{0.0821 \times 273}{22.4} = 1.0006$ atm. vdW: $P = \frac{R T}{V _{m} - b} - \frac{a}{V _{m}^{2}} = \frac{22.4133}{22.3573} - \frac{3.59}{501.76} = 1.00251 - 0.007155 = 0.99536$ atm. Difference: $\frac{1.0006 - 0.99536}{1.0006} \times 100 \approx 0.52%$ . Why tiny: at 22.4 L the molecules are far apart (low density), so both finite-size and attraction corrections are small. Deviations grow at high pressure / low temperature.

Level 4 — Synthesis

L4.1

The Boyle temperature is the temperature at which a real gas behaves ideally over a range of low pressures (its $Z$ stays near 1). A low-pressure analysis gives $T_{B} = \frac{a}{R b}$ . Compute $T_{B}$ for N₂ ( $a = 1.36$ , $b = 0.0391$ ).

Recall Solution

$T_{B} = \frac{a}{R b} = \frac{1.36}{0.0821 \times 0.0391} = \frac{1.36}{0.0032101} = 423.7 K .$ Below $T_{B}$ the attraction term wins (curve dips, $Z < 1$ ); above $T_{B}$ the volume term wins ( $Z > 1$ ); right at $T_{B}$ the two low-pressure effects cancel and the gas mimics ideal. See Boyle Temperature.

L4.2

For a van der Waals gas the critical temperature is $T_{c} = \frac{8 a}{27 R b}$ . Using $T_{B} = a / (R b)$ from L4.1, prove $T_{B} = \frac{27}{8} T_{c}$ , then compute $T_{c}$ for N₂ and check the ratio.

Recall Solution

Ratio proof: $\frac{T _{B}}{T _{c}} = \frac{a / ( R b )}{8 a / ( 27 R b )} = \frac{27}{8} = 3.375$ . So $T_{B} = 3.375 T_{c}$ . N₂ numbers: $T_{c} = \frac{8 \times 1.36}{27 \times 0.0821 \times 0.0391} = \frac{10.88}{0.086673} = 125.5 K$ . Check: $3.375 \times 125.5 = 423.6 K$ , matching $T_{B}$ from L4.1 (rounding). See Critical Constants and Liquefaction.

L4.3

Using $V_{c} = 3 b$ and $P_{c} = \frac{a}{27 b ^{2}}$ , compute the critical compressibility $Z_{c} = \frac{P _{c} V _{c}}{R T _{c}}$ and show it is a universal constant (independent of $a, b$ ).

Recall Solution

Substitute: $Z_{c} = \frac{P _{c} V _{c}}{R T _{c}} = \frac{\frac{a}{27 b ^{2}} \cdot 3 b}{R \cdot \frac{8 a}{27 R b}} = \frac{\frac{3 a}{27 b}}{\frac{8 a}{27 b}} = \frac{3}{8} = 0.375.$ The $a$ , $b$ , $R$ all cancel — every van der Waals gas predicts $Z_{c} = 3/8 = 0.375$ , a universal number. (Real gases sit near $0.27$ – $0.30$ , which is why vdW is only approximate.)

Level 5 — Mastery

L5.1 (degenerate case: $a = 0$ )

Set $a = 0$ but keep $b > 0$ . Write the resulting equation for $P$ and predict whether $Z > 1$ or $Z < 1$ . Verify with $b = 0.0427$ , $n = 1$ , $V = 1.00$ L, $T = 300$ K.

Recall Solution

With $a = 0$ : $P = \frac{n R T}{V - nb}$ (only the size correction survives). Since $V - nb < V$ , this $P$ is larger than the ideal $n R T / V$ . So $Z = \frac{P V}{n R T} = \frac{V}{V - nb} > 1$ always. Numbers: $P = \frac{0.0821 \times 300}{1 - 0.0427} = \frac{24.63}{0.9573} = 25.73$ atm. Ideal $= 24.63$ atm. $Z = \frac{25.73 \times 1}{24.63} = 1.0446 > 1$ . Pure excluded-volume gas is harder to compress. ✓

L5.2 (degenerate case: $b = 0$ )

Now set $b = 0$ but keep $a > 0$ . Write $P$ , predict the sign of $Z - 1$ , and verify with $a = 3.59$ , same $n, V, T$ as L5.1.

Recall Solution

With $b = 0$ : $P = \frac{n R T}{V} - \frac{a n ^{2}}{V ^{2}}$ (only attraction survives). This $P$ is smaller than ideal $n R T / V$ , so $Z < 1$ always. Numbers: $P = \frac{24.63}{1} - \frac{3.59}{1} = 24.63 - 3.59 = 21.04$ atm. $Z = \frac{21.04 \times 1}{24.63} = 0.8542 < 1$ . Pure attraction gas is more compressible. ✓ L5.1 vs L5.2 together show the two corrections push $Z$ in opposite directions — the real $Z$ is a tug-of-war between them.

L5.3 (limiting behaviour + full CO₂)

For CO₂ ( $a = 3.59$ , $b = 0.0427$ ) at $T = 300$ K, $n = 1$ : compute $Z$ at $V = 1.00$ L (high density) and at $V = 50.0$ L (low density). Explain the trend and its limit as $V \to \infty$ .

Recall Solution

Both corrections act. $P = \frac{R T}{V - b} - \frac{a}{V ^{2}}$ , $Z = \frac{P V}{R T}$ .

At $V = 1.00$ L: $P = \frac{24.63}{0.9573} - \frac{3.59}{1} = 25.73 - 3.59 = 22.14$ atm. $Z = \frac{22.14}{24.63} = 0.899$ . Attraction still slightly wins ⇒ $Z < 1$ .

At $V = 50.0$ L: $P = \frac{24.63}{49.957} - \frac{3.59}{2500} = 0.49302 - 0.001436 = 0.49159$ atm. $Z = \frac{0.49159 \times 50}{24.63} = \frac{24.579}{24.63} = 0.99795 \approx 0.998$ .

Trend: as $V$ grows, molecules spread out, both corrections shrink and $Z \to 1$ . As $V \to \infty$ : $V - b \to V$ and $a / V^{2} \to 0$ , so $P \to R T / V$ , the ideal law. This is the required limit — every real gas becomes ideal at infinite dilution.

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

Recall check

L2.1 real pressure of CO₂ (1 mol, 2 L, 350 K)

13.78

atm

L2.2 molecular radius from

b = 0.0304

\approx 1.44 \times 1 0^{- 10}

m (

1.4

Å)

L4.1 Boyle temp of N₂

423.7

L4.2 critical temp of N₂

125.5

L4.3 universal critical

Z_{c}

for a vdW gas

3/8 = 0.375

L5.2

Z

for pure-attraction gas

0.854

(

< 1

)

Connections

Parent: physical meaning of a and b
Compressibility Factor Z — the $Z$ we computed all over this page
Boyle Temperature — L4.1 uses $T_{B} = a / (R b)$
Critical Constants and Liquefaction — L4.2, L4.3 critical constants
Kinetic Theory of Gases — where the ideal assumptions come from
Intermolecular Forces — the physical origin of $a$
Ideal Gas Law — the limit every problem returns to as $V \to \infty$