S1.5 Ideal gases | IB DP Chemistry

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🎯What you need to be able to do

Recognise and state the key assumptions of the ideal gas model.
Explain the limitations of the model, and the conditions under which real gases deviate most from it.
Investigate and analyse graphs relating the pressure, volume and temperature of a fixed mass of gas.
Use the molar volume of an ideal gas at STP.
Solve problems using the ideal gas equation \( PV = nRT \) and the combined gas law.
Use the ideal gas equation to determine the molar mass of a gas from experimental data.

📚The chemistry

The ideal gas model

An ideal gas consists of moving particles for which four assumptions hold:

the particles have negligible volume compared with the volume of the container;
there are no intermolecular forces between them, either attractive or repulsive;
all collisions are perfectly elastic — no kinetic energy is lost;
the particles are in constant random motion, and their average kinetic energy is proportional to the absolute temperature.

No real gas satisfies these exactly. The model is nevertheless excellent under ordinary laboratory conditions, and it is a good illustration of the nature-of-science point that a model is judged by whether its predictions work, not by whether its assumptions are literally true.

Why real gases deviate

Real gases deviate most at low temperature and high pressure, and the reason maps directly onto the first two assumptions.

At high pressure the particles are forced close together, so the volume they themselves occupy is no longer negligible compared with the container volume — the gas is less compressible than the model predicts.
At low temperature the particles move slowly, so the intermolecular forces between them (which the model ignores) have time to act. Attractions pull particles together, reducing the force with which they strike the walls, so the measured pressure is lower than predicted — and eventually the gas condenses, which an ideal gas could never do.

The syllabus adds a comparison between gases: at the same temperature and pressure, a gas whose molecules have stronger intermolecular forces or a larger molecular volume deviates more. So ammonia (hydrogen bonding, polar) deviates far more than helium (only weak London forces, tiny atoms). That connects straight to S2.2. The guide states explicitly that no mathematical treatment of the deviation is required — there is no van der Waals equation in this course.

The relationships between P, V, T and n

You should be able to recognise and sketch these, though the names of the individual gas laws are not assessed:

\( P \) against \( V \) at constant \( T \), \( n \)
inversely proportional — a hyperbola. A plot of \( P \) against \( 1/V \) is a straight line through the origin.

\( V \) against \( T \) at constant \( P \), \( n \)
directly proportional — a straight line which, extrapolated, passes through the origin only if T is in kelvin; on a Celsius axis it cuts at −273 °C.

\( P \) against \( T \) at constant \( V \), \( n \)
directly proportional, straight line through the origin in kelvin.

\( V \) against \( n \) at constant \( P \), \( T \)
directly proportional — this is Avogadro’s law from S1.4.

That \( V \) against \( \theta/^{\circ}\mathrm{C} \) graph is one of the nicest pieces of evidence in the course: extrapolate the line back to zero volume and it meets the axis at −273 °C for every gas, which is how absolute zero can be located without ever reaching it.

Molar volume

Because equal amounts of any ideal gas occupy equal volumes under the same conditions, the molar volume is a constant at a given temperature and pressure. At STP (273 K and 100 kPa) it is \( 22.7\ \mathrm{dm^3\,mol^{-1}} \), and the value is in the data booklet — look it up rather than trusting memory, because older textbooks quote 22.4 dm³ mol⁻¹ for the obsolete 1 atm standard.

\[ n = \frac{V}{22.7} \qquad (V \text{ in } \mathrm{dm^3}, \text{ at STP only}) \]

The ideal gas equation and the combined gas law

\[ PV = nRT \qquad\qquad \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]

Both are in the data booklet, along with the gas constant \( R = 8.31\ \mathrm{J\,K^{-1}\,mol^{-1}} \). Use \( PV = nRT \) when you need an absolute quantity — an amount, a mass, a molar mass. Use the combined gas law when a fixed amount of gas is changed from one set of conditions to another, because then \( n \) and \( R \) cancel and you never need them.

Units are where this topic is won or lost. The value of \( R \) is in joules, so every quantity must be SI: \( P \) in Pa (not kPa: multiply by 1000), \( V \) in m³ (not dm³: divide by 1000; not cm³: divide by \( 10^6 \)), and \( T \) in K (never Celsius, in any gas calculation, ever). In the combined gas law the pressure and volume units merely have to match on both sides and so can be left as kPa and dm³ — but the temperature must still be in kelvin, because only kelvin makes \( T \) proportional to kinetic energy.

Finding a molar mass from gas data

Substituting \( n = m/M \) into \( PV = nRT \) and rearranging gives the standard experimental route to the molar mass of a volatile liquid or a gas:

\[ PV = \frac{m}{M}RT \qquad\Longrightarrow\qquad M = \frac{mRT}{PV} \]

Experimentally, a weighed sample of a volatile liquid is vaporised in a gas syringe or a flask of known volume in a hot water bath; the temperature and atmospheric pressure are recorded, and \( M \) follows. The obvious sources of error — vapour escaping before the volume is read, the liquid not fully vaporising, the syringe cooling as it is removed — are exactly what Paper 1B asks you to evaluate.

✏️Worked example

(a) A 0.250 g sample of a volatile liquid is completely vaporised. The vapour occupies 102 cm³ at 90.0 °C and a pressure of 101 kPa. Calculate its molar mass.
(b) A fixed mass of gas occupies 450 cm³ at 27 °C and 120 kPa. Calculate its volume at 77 °C and 80.0 kPa.
(c) Explain why the value obtained in (a) would be too large if some of the liquid failed to vaporise.
Take \( R = 8.31\ \mathrm{J\,K^{-1}\,mol^{-1}} \).

(a) Convert everything to SI first, before touching the equation:

\( P = 101\ \mathrm{kPa} = 1.01 \times 10^{5}\ \mathrm{Pa} \)
\( V = 102\ \mathrm{cm^3} = 1.02 \times 10^{-4}\ \mathrm{m^3} \)
\( T = 90.0 + 273 = 363\ \mathrm{K} \)

\[ M = \frac{mRT}{PV} = \frac{(0.250)(8.31)(363)}{(1.01 \times 10^{5})(1.02 \times 10^{-4})} = \frac{754.1}{10.30} = 73.2\ \mathrm{g\,mol^{-1}} \]

So \( M \approx 73\ \mathrm{g\,mol^{-1}} \).

(b) A fixed mass changing conditions — use the combined gas law, with temperatures in kelvin but pressure and volume left in kPa and cm³ since they appear on both sides:

\[ V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{(120)(450)(350)}{(300)(80.0)} = \frac{18\,900\,000}{24\,000} = 788\ \mathrm{cm^3} \]

So 788 cm³.

(c) If some liquid remained unvaporised, the mass \( m \) used in the calculation would be the whole sample, but the volume \( V \) measured would be produced by less than that mass. Since \( M = mRT/PV \), \( m \) is too large for the \( V \) recorded, and the calculated molar mass is therefore too high.

Check it. In (b), predict the direction of each change before calculating. The temperature rises from 300 K to 350 K, which alone would expand the gas by a factor \( 350/300 = 1.17 \); the pressure falls from 120 kPa to 80 kPa, which alone would expand it by \( 120/80 = 1.5 \). Both push the same way, so the answer must be bigger than 450 cm³ by a factor of about \( 1.17 \times 1.5 = 1.75 \), and \( 450 \times 1.75 = 788 \). Getting a smaller volume means a ratio has been inverted. In (a), a plausibility check: 73 g mol⁻¹ is a sensible size for a small volatile organic molecule; an answer of 0.073 or 73 000 signals a unit conversion missed by a factor of \( 10^3 \).

Leaving pressure in kPa and volume in dm³ in \( PV = nRT \). That single mistake changes the answer by a factor of \( 10^6 \), and it is the most common error in the whole of Structure 1. The gas constant is quoted in J K⁻¹ mol⁻¹, and a joule is a pascal metre cubed — so pascals and cubic metres are compulsory. The second trap is subtler: using 90 °C instead of 363 K here would give a molar mass of about 18 g mol⁻¹, a number that looks perfectly reasonable, which is precisely why it is dangerous. Convert temperature first, every time, before you write anything else down.

📝Practise

Work through these on paper, then reveal the answer.

1. State the four assumptions of the ideal gas model, and identify which assumption fails when a real gas condenses to a liquid on cooling.

The assumptions: (i) the particles have negligible volume compared with the container; (ii) there are no intermolecular forces between the particles; (iii) all collisions are perfectly elastic, so no kinetic energy is lost; (iv) the particles are in constant random motion with average kinetic energy proportional to absolute temperature. The assumption that fails on condensation is (ii), no intermolecular forces. An ideal gas could never condense at all: condensation happens precisely because attractive forces exist between the particles, and as the temperature falls the particles no longer have enough kinetic energy to overcome them.

2. Calculate the volume occupied by 3.20 g of methane, CH₄, at STP (273 K, 100 kPa).

\( M(\mathrm{CH_4}) = 12.01 + 4(1.01) = 16.05\ \mathrm{g\,mol^{-1}} \), so \( n = \dfrac{3.20}{16.05} = 0.199\ \mathrm{mol} \). At STP the molar volume is 22.7 dm³ mol⁻¹ (data booklet), so \( V = 0.199 \times 22.7 = \mathbf{4.52\ dm^3} \). Alternatively, from \( PV = nRT \): \( V = \dfrac{(0.199)(8.31)(273)}{1.00 \times 10^{5}} = 4.52 \times 10^{-3}\ \mathrm{m^3} = 4.52\ \mathrm{dm^3} \) — the two routes must agree, which is a useful way of confirming you have the molar volume right.

3. Explain why a plot of volume against temperature in °C for a fixed mass of gas at constant pressure gives a straight line that does not pass through the origin, and what the intercept on the temperature axis represents.

Volume is directly proportional to absolute temperature, \( V \propto T \) with \( T \) in kelvin. The Celsius scale is the same scale shifted by 273.15 units, so \( V \propto (\theta + 273.15) \). A plot against \( \theta \) is therefore still a straight line — the gradient is unchanged — but it is displaced, cutting the volume axis at a positive value when \( \theta = 0 \). Extrapolating the line to \( V = 0 \) gives an intercept on the temperature axis at −273 °C, which is absolute zero: the temperature at which the particles would have minimum kinetic energy and, on this model, occupy no volume. The remarkable feature is that every gas gives the same intercept, which is strong evidence that absolute zero is a property of temperature itself rather than of any particular substance.

4. A gas cylinder of fixed volume contains gas at 15.0 °C and 2.50 × 10³ kPa. It is left in the sun and the temperature rises to 45.0 °C. Calculate the new pressure.

Volume is fixed, so the combined gas law reduces to \( \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \). Convert temperatures: \( T_1 = 15.0 + 273 = 288\ \mathrm{K} \), \( T_2 = 45.0 + 273 = 318\ \mathrm{K} \). Then \( P_2 = P_1 \times \dfrac{T_2}{T_1} = 2.50 \times 10^{3} \times \dfrac{318}{288} = \mathbf{2.76 \times 10^{3}\ kPa} \). Note how modest the increase is — about 10% — even though the Celsius temperature tripled. Using Celsius here would predict a threefold pressure rise, which is nonsense, and is the clearest demonstration of why kelvin is compulsory.

5. Under identical conditions of temperature and pressure, ammonia deviates from ideal behaviour considerably more than helium. Explain why.

Two of the ideal assumptions fail more badly for ammonia. First, intermolecular forces: \( \mathrm{NH_3} \) is a polar molecule capable of hydrogen bonding, which is a comparatively strong intermolecular attraction, whereas helium atoms experience only very weak London (dispersion) forces because they have just two electrons. Attractions between the ammonia molecules reduce the force with which they strike the container walls, so the observed pressure falls below the ideal prediction. Second, particle volume: an \( \mathrm{NH_3} \) molecule is substantially larger than a helium atom, so its own volume is a larger fraction of the container volume, and the assumption of negligible particle volume is a worse approximation. Helium is close to the best approximation to an ideal gas there is, which is why it liquefies only at about 4 K.

6. A student vaporises 0.180 g of a volatile liquid and measures a volume of 90.5 cm³ at 100.0 °C and 99.0 kPa. (a) Calculate the molar mass. (b) The liquid has the empirical formula CH₃O. Deduce its molecular formula. (c) Suggest one reason why the experimental molar mass might come out lower than the true value.

(a) Convert: \( P = 9.90 \times 10^{4}\ \mathrm{Pa} \), \( V = 9.05 \times 10^{-5}\ \mathrm{m^3} \), \( T = 373\ \mathrm{K} \). \( M = \dfrac{mRT}{PV} = \dfrac{(0.180)(8.31)(373)}{(9.90 \times 10^{4})(9.05 \times 10^{-5})} = \dfrac{557.9}{8.960} = \mathbf{62.3\ g\,mol^{-1}} \). (b) The empirical unit \( \mathrm{CH_3O} \) has mass \( 12.01 + 3(1.01) + 16.00 = 31.04 \). \( k = \dfrac{62.3}{31.04} = 2.01 \approx 2 \), so the molecular formula is C₂H₆O₂ — ethane-1,2-diol. (c) Since \( M = mRT/PV \), the result is too low if the recorded volume is too large or the recorded mass too small. The standard answer is that air was already present in the syringe, or leaked in, so the measured volume is greater than the volume of vapour alone. Equally acceptable: some of the volatile liquid evaporated after weighing but before injection, so less than 0.180 g actually produced the reading. Both are systematic errors, so repeating the experiment would not remove them — which is the point Paper 1B usually wants made.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

PhET — Gas Properties, which lets you hold one variable constant and vary the others, and shows the particle motion at the same time; the fastest way to make the P–V hyperbola feel inevitable.
RSC Learn Chemistry — the gas syringe determination of the molar mass of a volatile liquid, with the standard error analysis.
Your data booklet — find \( R \), the molar volume at STP, and the two gas equations now, and note that STP here is 100 kPa, not 1 atm.