Chemistry 2nd Paper · Classes 11-12 · Chapter 1 · 1.6 to 1.7
Ideal and Real Gases, the van der Waals Equation.
PV = nRT has balanced every sum so far. But pack a gas into a cylinder at 500 atm, or cool it hard, and the numbers stop matching. The fault is not in the equation but in two things it takes for granted: that molecules have no volume of their own, and that they do not pull on each other. In a real gas both are wrong.
1. The ideal gas
A gas that obeys Boyle's law, Charles's law and Avogadro's law perfectly, at every temperature and pressure, is called an ideal gas.
Two tests pick out an ideal gas: (1) it obeys PV = nRT at every temperature and pressure; (2) at constant temperature its internal energy does not depend on its volume, that is (∂U/∂V)_T = 0.
2. Real gases
By whether they obey the gas laws, gases are of two kinds: ideal and real. A real (non-ideal) gas is any gas you actually meet, with attraction between its molecules and a variable volume: H₂, N₂, O₂, CO₂, CH₄.
A real gas does not obey PV = nRT exactly. At 1 atm or below, and at high temperature, it behaves roughly ideally. The lower the pressure and the higher the temperature, the closer to ideal; at high pressure and low temperature it strays a long way.
STP shows it. An ideal gas has a molar volume of 22.414 L, but real gases differ: He 22.435, H₂ 22.432, Ne 22.422, Ar 22.397, N₂ 22.396, O₂ 22.390, CO 22.388, Cl₂ 22.184, NH₃ 22.079 L. Furthest off are NH₃ and Cl₂, which condense at −33.4°C and −34.0°C, so they turn liquid easily.
3. Ideal against real
On attraction and volume: an ideal gas is imaginary, its molecules neither attract nor repel, and their total volume is negligible. A real gas exists, its molecules attract and repel, and their total volume is not negligible.
On equation, energy and Z: an ideal gas obeys PV = nRT, a real gas the van der Waals equation. An ideal gas's internal energy is independent of volume; a real gas's depends on it. Z = 1 for an ideal gas and Z ≠ 1 for a real one. Now sort seven statements into the two boxes.
Ideal and real gases
4. Amagat's curves
In 1880 the French scientist Amagat plotted PV against P for several gases at constant temperature; these are Amagat's curves. For an ideal gas PV is constant by Boyle's law, so its graph is a straight line parallel to the P axis.
Real gases give two kinds of curve. The first kind (H₂ and He at ordinary temperature): PV rises steadily with pressure and always stays above the ideal value. The second kind (O₂, N₂, CO₂): as pressure rises PV first falls below the ideal value to a minimum, then climbs back and ends above it.
Temperature changes the kind. Cool H₂ and He enough and their curves become the second kind; heat N₂, O₂ and CO₂ enough and theirs become the first kind. So Amagat drew two conclusions: the kind of deviation depends on temperature, not on the nature of the gas; and the higher the pressure, the greater the deviation from ideal.
5. The compressibility factor Z
The number that measures the deviation is the compressibility factor: Z = PV/nRT = V/V₀, where V is the real gas's molar volume and V₀ the ideal gas's at the same T and P. Z = 1 for an ideal gas; for a real gas it depends on P and T.
The further Z is from 1, above or below, the more the gas deviates. Z > 1 means the gas is harder to compress than an ideal one (H₂, He). Z < 1 means it is easier to compress (CO₂, O₂), and these gases dissolve more in water.
On a graph of Z against P, every gas has Z = 1 near zero pressure, so at very low pressure every real gas behaves ideally. Heat N₂ and its curve moves towards the ideal line, and at 700°C it nearly lies on it. The temperature at which a real gas behaves ideally is called Boyle's temperature.
Problem 1.65: if 1 mol of N₂ at 27°C and 1 atm has a volume of 24.5 L, Z = PV/RT = 1 × 24.5/(0.082 × 300) = 0.9959, so the deviation from ideal is 1 − 0.9959 = 0.0041. Problem 1.66: at what temperature does 28 g of N₂ (1 mol) in a 0.26 L cylinder at 400 atm behave ideally? Putting Z = 1, T = PV/nR = 400 × 0.26/0.082 = 1268.3 K = 995.3°C.
With Z given, the temperature is T = PV/(nRZ). For 14.0 g of N₂ (0.5 mol) at 100 kPa in 2.24 L with Z = 0.10, T = 10⁵ × 2.24×10⁻³/(0.5 × 8.314 × 0.10) = 538.8 K, about 265.7°C. The pressure is in Pa and the volume in m³ here, so R = 8.314.
6. Why real gases deviate
Cause 1, the volume of the molecules. At STP the molecules take up only 0.05% of the container, which is negligible. At 500 atm and 0°C they take up 20%, which is not. So at high pressure a real gas has a larger volume than the ideal sum predicts.
Cause 2, attraction. At low pressure the molecules are far apart and attraction barely acts. As pressure rises they come closer (attraction works from about 10 molecular diameters away) and attraction shrinks the volume. Attraction lowers the volume and molecular size raises it: two effects pulling opposite ways. Above 350 atm the molecular volume wins.
This tug of war sets Z. PV/RT < 1 comes from attraction winning, as for CO₂ and CH₄ at medium pressure. PV/RT > 1 comes from molecular volume winning, as for H₂ and He at high pressure. Drag the handle to raise the pressure.
Pressure and deviation
The drawing is not to scale. 0.05% and 20% are the book's values; 10 atm and 350 atm are the book's rough limits, not exact cut-offs.
7. The van der Waals equation
Van der Waals saw that two postulates of the kinetic theory fail for real gases: that molecular volume is negligible, and that molecules do not attract. PV = nRT fails for real gases because of exactly these two. So he made two corrections.
The volume correction. Take b as the effective volume of one mole of the molecules themselves; for n moles, nb comes off the vessel's volume V. The molecules only have the free space (V − nb) to move in.
The pressure correction. A molecule heading for the wall is pulled back by the ones behind it, so a real gas presses on the wall less than an ideal gas would. To make up the shortfall, P_a is added to P. P_a is called the internal or cohesive pressure; it is proportional to the square of the gas's density, P_a ∝ d² ∝ (n/V)², hence P_a = an²/V².
Put both corrections in and you have the van der Waals equation: (P + n²a/V²)(V − nb) = nRT, or P = nRT/(V − nb) − n²a/V². For one mole, (P + a/V²)(V − b) = RT.
8. The constants a and b
The constant a measures the strength of intermolecular attraction; its unit is atm L² mol⁻². A large a means strong attraction, usually a larger molar mass, and a gas that liquefies at a lower pressure.
The constant b is the effective volume (covolume): four times the actual volume of the molecules in one mole, b = 4 × N_A × (4/3)πr³. Bigger molecules give a bigger b; its unit is L mol⁻¹.
Table 1.6 (a in atm L² mol⁻², b in L mol⁻¹): He 0.034, 0.0237; Ne 0.211, 0.0171; Ar 1.350, 0.0322; H₂ 0.244, 0.0266; N₂ 1.350, 0.0387; O₂ 1.360, 0.0318; Cl₂ 6.490, 0.0562; CH₄ 2.253, 0.0428; CO 1.450, 0.0395; CO₂ 3.590, 0.0427; NH₃ 4.170, 0.0371. Of these, He has the smallest a (0.034), so its attraction is negligible and it is close to an ideal gas; Cl₂ has the largest a (6.490), so its attraction is the strongest. Put six of the gases in order of a.
a: a measure of attraction
9. What the equation explains
Why do the PV curves of H₂ and He sit above the ideal line? Their attraction is tiny, so (P + a/V²) ≈ P and the equation becomes P(V − b) = RT, that is PV = RT + Pb. Pb is positive, so PV > RT, and PV rises from the very start as pressure goes up.
At very low pressure and very high temperature V is huge. Then b is negligible (V − b ≈ V) and attraction is almost nothing (P + a/V² ≈ P), so the equation comes down to PV = RT. That is why a real gas behaves ideally under these conditions.
See for yourself how the two corrections push Z above or below 1. Pick a gas, raise or lower the temperature, then move the pressure point.
Z against P: which effect wins?
The curves are a model worked out from the van der Waals equation with Table 1.6's a and b, not measured values.
10. Working the numbers
To write the equation for a given mass: first find n = mass ÷ M, then put n²a in place of a, nb in place of b, and nRT on the right. 44 g of CO₂, 4 g of He or 28 g of N₂ give n = 1: (P + a/V²)(V − b) = RT. 4 g of H₂, 8 g of He or 64 g of O₂ give n = 2: (P + 4a/V²)(V − 2b) = 2RT. 96 g of O₂ gives n = 3: (P + 9a/V²)(V − 3b) = 3RT. 16 g of He gives n = 4: (P + 16a/V²)(V − 4b) = 4RT.
Exercise 1.91(a): 0.5 mol of H₂ (a = 0.244, b = 0.0266) in a 0.86 L vessel at 27°C. P = nRT/(V − nb) − an²/V² = 14.4445 atm, using R = 0.082. Exercise 1.91(b): 0.50 mol of N₂ (a = 1.35, b = 0.0387) in 0.60 L at 300 K. The ideal equation gives P = 20.515 atm and van der Waals gives P = 20.26 atm, using R = 0.08206. The real pressure is a little below the ideal one. The book's two answers use different values of R, so the calculator below lets you choose R. Numbers you type go nowhere. The notebook is yours.
P = nRT/(V − nb) − an²/V²
a and b are from Table 1.6 and the examples from the book; the working is the equation's.
11. Check yourself
Three questions in the exam's style. Start with the first.
Question 1
The constant 'a' of the van der Waals equation expresses-
Question 2
What is the basis of Amagat's curve?
Question 3
Which is the correct van der Waals equation for 16 g of He gas?
The whole note on eight cards.
Eight cards
Source: Chemistry 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Environmental Chemistry”, section 1.6 to 1.7.