Ideal Gases and pV = nRT

AS · 12 min

A gas is the simplest state of matter to describe mathematically: its particles are far apart and moving fast, and one equation, $pV = nRT$ , links its pressure, volume, temperature and amount. This note explains where gas pressure comes from, what makes a gas "ideal" and why real gases fall short, and then drills the ideal gas equation, including the unit conversions that cost students the most marks and its classic use in finding the relative molecular mass of a volatile liquid. Expect a calculation in Paper 2 and possibly a practical question in Paper 3.

The kinetic model of a gas

The particles in a gas (molecules, or atoms for the noble gases) are:

in rapid, random motion, travelling in straight lines until they collide;
far apart compared with their own size, so most of a gas is empty space;
colliding with each other and with the walls of their container.

Where pressure comes from

Key result

Gas pressure is caused by gas particles colliding with the walls of the container. Each collision exerts a tiny force on the wall as the particle's direction (and momentum) changes. Pressure is the total force from all these collisions per unit area of wall.

This explains the everyday gas laws:

More particles in the same volume: more collisions with the walls per second, so higher pressure.
Smaller volume for the same particles: the particles hit any given area of wall more often, so higher pressure.
Higher temperature: the particles move faster, so they hit the walls more often and with more force, so higher pressure (or, if the pressure is held constant, the gas expands).

Ideal gases

An ideal gas is a model: a gas that obeys $pV = nRT$ exactly under all conditions. For that to be true, two assumptions must hold.

Key result

An ideal gas has:

zero particle volume: the particles themselves take up no space;
no intermolecular forces of attraction between the particles.

(The model also assumes the particles are in random motion and that collisions are perfectly elastic: no kinetic energy is lost overall.)

Real gases

No real gas is ideal, but most gases are close to ideal at room temperature and atmospheric pressure, because the particles are far apart (so their volume is a tiny fraction of the total) and moving fast (so the weak attractions between them hardly matter).

Real gases deviate most from ideal behaviour at:

high pressure: the particles are pushed close together, so the volume of the particles themselves becomes a significant fraction of the total volume, and intermolecular forces become significant because the particles are close;
low temperature: the particles move more slowly, so intermolecular attractions have more effect during collisions and pull particles towards each other, reducing the force with which they strike the walls. Near the boiling point, the gas begins to condense.

The gases closest to ideal are those with the weakest intermolecular forces and smallest particles: helium and hydrogen. Gases with strong intermolecular forces, especially polar ones that hydrogen bond such as ammonia and steam, deviate most.

One way to see this is to plot $\dfrac{pV}{nRT}$ against pressure. For an ideal gas this is always exactly 1 (the solid horizontal line). For a typical real gas such as nitrogen at room temperature, it dips slightly below 1 at moderate pressure, where attractions dominate, and then rises above 1 at high pressure, where the particles' own volume dominates (dashed curve).

y = 1 y = 1 - 0.006x + 0.0002x^2

(Horizontal axis: pressure in MPa; vertical axis: $pV/nRT$ . A sketch of the shape, not data to read off.)

Tip

You only need to explain the deviations qualitatively. If asked "under what conditions does a gas behave most ideally?", the answer is high temperature and low pressure, with the reasons above.

The ideal gas equation

Key result

pV = nRT

symbol	quantity	SI unit
$p$	pressure	pascal, Pa ( $= \text{N m}^{-2}$ )
$V$	volume	$\text{m}^3$
$n$	amount of gas	mol
$R$	molar gas constant, $8.31\ \text{J K}^{-1}\ \text{mol}^{-1}$	$\text{J K}^{-1}\ \text{mol}^{-1}$
$T$	temperature	kelvin, K

The equation only works if every quantity is in SI units. Almost every lost mark in this topic is a unit error.

Key result

Unit conversions

$T/\text{K} = \theta/^\circ\text{C} + 273$
$1\ \text{kPa} = 1 \times 10^{3}\ \text{Pa}$ ; $1\ \text{MPa} = 1 \times 10^{6}\ \text{Pa}$ ; $1\ \text{atm} = 101\ \text{kPa}$ (101 325 Pa)
$1\ \text{dm}^3 = 1 \times 10^{-3}\ \text{m}^3$
$1\ \text{cm}^3 = 1 \times 10^{-6}\ \text{m}^3$

Linking to the molar volume

At $20\ ^\circ\text{C}$ ( $293\ \text{K}$ ) and $101\ \text{kPa}$ , one mole of an ideal gas occupies:

V = \frac{nRT}{p} = \frac{1 \times 8.31 \times 293}{101 \times 10^{3}} = 0.0241\ \text{m}^3 = 24.1\ \text{dm}^3

which is very close to the $24.0\ \text{dm}^3\ \text{mol}^{-1}$ for room conditions that you used in stoichiometry (at $298\ \text{K}$ the same calculation gives $24.5\ \text{dm}^3$ ; the data booklet value is a convenient rounded figure). The ideal gas equation is the general version: use it whenever the conditions are not room conditions or s.t.p.

Rearrangements you will need

Substitute $n = \dfrac{m}{M}$ to bring in mass and molar mass:

Key result

pV = \frac{m}{M}RT \qquad\Longrightarrow\qquad M = \frac{mRT}{pV}

If $m$ is in grams, $M$ comes out in $\text{g mol}^{-1}$ , numerically equal to $M_r$ .

Dividing by $V$ gives the density form, $\rho = \dfrac{m}{V}$ :

M = \frac{\rho RT}{p}

For a fixed amount of gas ( $n$ constant) changing from one set of conditions to another, $\dfrac{pV}{T} = nR$ is constant, so:

\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}

Here the units of $p$ and $V$ only need to match on both sides, but $T$ must still be in kelvin.

Method

Solving a pV = nRT problem

List the given values with their units.
Convert each to SI: Pa, $\text{m}^3$ , K. Write the converted values down.
Rearrange the equation for the unknown before substituting.
Substitute and calculate. Keep extra figures in intermediate steps.
Convert the answer to the units asked for (often $\text{dm}^3$ or $\text{cm}^3$ ), and give it to the same number of significant figures as the least precise data, usually 3.

Worked examples

Volume of a gas

Calculate the volume, in $\text{dm}^3$ , occupied by $0.500\ \text{mol}$ of an ideal gas at $25\ ^\circ\text{C}$ and $100\ \text{kPa}$ .

Solution

Convert: $T = 25 + 273 = 298\ \text{K}$ ; $p = 100 \times 10^{3} = 1.00 \times 10^{5}\ \text{Pa}$ .

V = \frac{nRT}{p} = \frac{0.500 \times 8.31 \times 298}{1.00 \times 10^{5}} = 0.01238\ \text{m}^3

Convert to $\text{dm}^3$ : $0.01238 \times 10^{3} = 12.4\ \text{dm}^3$ .

Amount from small volumes

How many moles of gas are in a $250\ \text{cm}^3$ flask at $150\ \text{kPa}$ and $27\ ^\circ\text{C}$ ?

Solution

$V = 250 \times 10^{-6} = 2.50 \times 10^{-4}\ \text{m}^3$ ; $p = 1.50 \times 10^{5}\ \text{Pa}$ ; $T = 300\ \text{K}$ .

n = \frac{pV}{RT} = \frac{1.50 \times 10^{5} \times 2.50 \times 10^{-4}}{8.31 \times 300} = \frac{37.5}{2493} = 0.0150\ \text{mol}

Relative molecular mass of a volatile liquid

A $0.200\ \text{g}$ sample of a volatile hydrocarbon liquid was injected into a gas syringe in an oven at $100\ ^\circ\text{C}$ . It vaporised completely, giving $72.0\ \text{cm}^3$ of vapour at $101\ \text{kPa}$ . Calculate $M_r$ and suggest a molecular formula for the hydrocarbon.

Solution

Convert: $T = 373\ \text{K}$ ; $p = 1.01 \times 10^{5}\ \text{Pa}$ ; $V = 72.0 \times 10^{-6}\ \text{m}^3$ .

M = \frac{mRT}{pV} = \frac{0.200 \times 8.31 \times 373}{1.01 \times 10^{5} \times 72.0 \times 10^{-6}} = \frac{619.9}{7.272} = 85.2\ \text{g mol}^{-1}

$M_r \approx 85$ . An alkane $\ce{C_nH_{2n+2}}$ with $14n + 2 = 86$ gives $n = 6$ : hexane, $\ce{C6H14}$ ( $M_r = 86.0$ ), within experimental error.

Gas volume from a reaction at high temperature

$4.20\ \text{g}$ of sodium hydrogencarbonate ( $M_r = 84.0$ ) is heated to $120\ ^\circ\text{C}$ and decomposes completely:

\ce{2NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)}

Calculate the total volume of gas produced, in $\text{dm}^3$ , at $120\ ^\circ\text{C}$ and $101\ \text{kPa}$ .

Solution

$n(\ce{NaHCO3}) = \dfrac{4.20}{84.0} = 0.0500\ \text{mol}$ .

From the equation, 2 mol $\ce{NaHCO3}$ gives 1 mol $\ce{H2O}$ and 1 mol $\ce{CO2}$ : 2 mol of gas. So total gas $= 0.0500\ \text{mol}$ . (At $120\ ^\circ\text{C}$ the water is a gas and counts.)

$T = 393\ \text{K}$ , $p = 1.01 \times 10^{5}\ \text{Pa}$ :

V = \frac{0.0500 \times 8.31 \times 393}{1.01 \times 10^{5}} = 1.617 \times 10^{-3}\ \text{m}^3 = 1.62\ \text{dm}^3

Exam-hard: identifying a gas from its density

A gaseous hydrocarbon contains $85.7\%$ carbon by mass. At $273\ \text{K}$ and $101\ \text{kPa}$ its density is $1.25\ \text{g dm}^{-3}$ . Determine its molecular formula.

Solution

Empirical formula. C: $\dfrac{85.7}{12.0} = 7.14$ ; H: $\dfrac{14.3}{1.0} = 14.3$ . Ratio $1 : 2$ , so the empirical formula is $\ce{CH2}$ (empirical mass 14.0).

Molar mass. $1.25\ \text{g dm}^{-3} = 1.25\ \text{kg m}^{-3} = 1250\ \text{g m}^{-3}$ .

M = \frac{\rho RT}{p} = \frac{1250 \times 8.31 \times 273}{1.01 \times 10^{5}} = 28.1\ \text{g mol}^{-1}

Molecular formula. $\dfrac{28.1}{14.0} = 2$ , so the molecular formula is $\ce{C2H4}$ (ethene).

Determining the Mr of a volatile liquid or a gas

Method 1: gas syringe in an oven (volatile liquid).

Fill a hypodermic syringe with the liquid and weigh it.
Inject a small amount (about $0.1$ to $0.2\ \text{g}$ ) through a self-sealing cap into a gas syringe kept in an oven at a known temperature well above the liquid's boiling point (for example $100\ ^\circ\text{C}$ for hexane).
Reweigh the hypodermic syringe: the difference is the mass injected.
When the plunger stops moving, record the volume of vapour, the oven temperature and the atmospheric pressure.
Calculate $M = \dfrac{mRT}{pV}$ .

Method 2: butane from a lighter refill (a gas).

Weigh the canister. Fill a measuring cylinder with water and invert it in a trough of water.
Release butane under the water into the cylinder until about $100$ to $200\ \text{cm}^3$ is collected; level the water inside and outside the cylinder so the gas is at atmospheric pressure.
Dry the canister carefully and reweigh it. Record room temperature and pressure.
Calculate $M$ ; for butane, expect about $58$ .

Variables and errors

source of error	effect on calculated $M_r$	improvement
liquid not fully vaporised (Method 1)	$V$ too small, $M_r$ too high	use a higher oven temperature
vapour condenses on cold parts of the syringe	$V$ too small	allow the syringe to reach oven temperature first
some butane dissolves in water (Method 2)	$V$ too small, $M_r$ too high	use water already saturated with the gas
canister wet when reweighed	mass loss too small, $M_r$ too low	dry thoroughly before reweighing
gas collected contains water vapour	$V$ includes water vapour	correct for the vapour pressure of water
real gas, not ideal (close to boiling point)	equation slightly inaccurate	work at higher temperature, lower pressure

The largest percentage uncertainty is usually in the small mass (for example $\pm 0.01\ \text{g}$ in $0.20\ \text{g}$ is $\pm 5\%$ ), so use a balance reading to $0.001\ \text{g}$ or a larger sample.

Watch out

Celsius in the equation. $T$ must be in kelvin. Using $25$ instead of $298$ gives an answer about twelve times too small.
kPa and dm³ together. If you put $p$ in kPa and $V$ in $\text{dm}^3$ , the factors of $10^3$ cancel and you get the right $n$ , but only by luck: as soon as $\text{cm}^3$ or mass is involved, mixing units fails. Always convert to Pa and $\text{m}^3$ .
$\text{cm}^3$ to $\text{m}^3$ . The factor is $10^{-6}$ , not $10^{-3}$ .
Density units. $1\ \text{g dm}^{-3} = 1\ \text{kg m}^{-3} = 1000\ \text{g m}^{-3}$ .
Forgetting that water is a gas above $100\ ^\circ\text{C}$ when counting moles of gas in a reaction.

Exam tip

"Explain the origin of gas pressure": particles collide with the walls of the container, exerting a force; pressure is force per unit area. Two marks: collisions, with the walls.
"State the assumptions of an ideal gas": the syllabus pair is zero particle volume and no intermolecular forces. Give these two first.
Show every conversion as a separate line. If the final answer is wrong, the conversions still earn method marks.
Read the question for the units of the answer: $\text{dm}^3$ , $\text{cm}^3$ or $\text{m}^3$ . Quote 3 significant figures unless told otherwise.
In Paper 3, an $M_r$ experiment question often asks why the result is higher or lower than expected: link each error to whether it makes $V$ or $m$ too big or small, then to $M = mRT/pV$ .

Summary

Gas pressure: particles colliding with the container walls; force per unit area.
Ideal gas: zero particle volume, no intermolecular forces. Real gases are nearly ideal at high temperature and low pressure; deviate at high pressure and low temperature, especially polar molecules.
$pV = nRT$ with $p$ in Pa, $V$ in $\text{m}^3$ , $T$ in K, $R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1}$ .
$M = \dfrac{mRT}{pV}$ and $M = \dfrac{\rho RT}{p}$ find relative molecular masses.
For fixed $n$ : $\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}$ .
Conversions: $+273$ ; $\text{kPa} \times 10^3$ ; $\text{dm}^3 \times 10^{-3}$ ; $\text{cm}^3 \times 10^{-6}$ .

Practice

Question

Explain, in terms of particles, why the pressure of a gas in a sealed rigid container increases when it is heated.
State the two assumptions about particles in an ideal gas, and explain why real gases deviate from ideal behaviour at high pressure.
Calculate the pressure, in kPa, exerted by $2.00\ \text{mol}$ of gas in a $40.0\ \text{dm}^3$ container at $300\ \text{K}$ .
A $5.00\ \text{dm}^3$ steel cylinder contains oxygen at $2.00\ \text{MPa}$ and $20\ ^\circ\text{C}$ . Calculate the mass of oxygen in the cylinder.
$0.640\ \text{g}$ of a gaseous oxide occupies $0.250\ \text{dm}^3$ at $300\ \text{K}$ and $100\ \text{kPa}$ . Calculate its $M_r$ and suggest its identity.
Which gas would you expect to behave less ideally at room temperature, ammonia or helium? Explain.
In a syringe experiment, $0.150\ \text{g}$ of a volatile liquid gave $62.5\ \text{cm}^3$ of vapour at $98\ ^\circ\text{C}$ and $102\ \text{kPa}$ . Calculate its $M_r$ and suggest the formula of an alkane that fits.
$500\ \text{cm}^3$ of gas at $20\ ^\circ\text{C}$ and $100\ \text{kPa}$ is heated to $100\ ^\circ\text{C}$ and compressed to $150\ \text{kPa}$ . Calculate its new volume.
Ammonium nitrate decomposes on strong heating: $\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$ . Calculate the total volume of gas, in $\text{dm}^3$ , formed from $1.00\ \text{g}$ of ammonium nitrate at $500\ \text{K}$ and $101\ \text{kPa}$ . In a sealed container, explain why the pressure falls considerably when the products are cooled to room temperature.
A compound contains $24.3\%$ C, $4.1\%$ H and $71.6\%$ Cl by mass. $0.484\ \text{g}$ of the compound, completely vaporised at $100\ ^\circ\text{C}$ and $101\ \text{kPa}$ , occupies $150\ \text{cm}^3$ . Determine the molecular formula and draw the structures of two possible isomers.

Answers

Heating increases the average kinetic energy of the particles, so they move faster. They collide with the walls more frequently and with greater force per collision. The total force on the walls per unit area increases, so the pressure rises (the volume cannot change).
Zero particle volume; no intermolecular forces of attraction. At high pressure the particles are close together, so their own volume is a significant fraction of the container's volume, and intermolecular attractions become significant because the particles are close; both assumptions fail.
$V = 40.0 \times 10^{-3} = 0.0400\ \text{m}^3$ . $p = \dfrac{nRT}{V} = \dfrac{2.00 \times 8.31 \times 300}{0.0400} = 1.25 \times 10^{5}\ \text{Pa} = 125\ \text{kPa}$ .
$p = 2.00 \times 10^{6}\ \text{Pa}$ , $V = 5.00 \times 10^{-3}\ \text{m}^3$ , $T = 293\ \text{K}$ . $n = \dfrac{2.00 \times 10^{6} \times 5.00 \times 10^{-3}}{8.31 \times 293} = 4.11\ \text{mol}$ . Mass $= 4.11 \times 32.0 = 131\ \text{g}$ .
$M = \dfrac{0.640 \times 8.31 \times 300}{1.00 \times 10^{5} \times 2.50 \times 10^{-4}} = \dfrac{1595.5}{25.0} = 63.8$ . $M_r \approx 64$ : sulfur dioxide, $\ce{SO2}$ ( $M_r = 64.1$ ).
Ammonia. Its molecules are polar and form hydrogen bonds, so there are significant intermolecular attractions; its molecules are also larger. Helium atoms are tiny and have only very weak id-id forces between them, so helium is close to ideal.
$T = 371\ \text{K}$ . $M = \dfrac{0.150 \times 8.31 \times 371}{1.02 \times 10^{5} \times 62.5 \times 10^{-6}} = \dfrac{462.4}{6.375} = 72.5$ . An alkane with $14n + 2 = 72$ gives $n = 5$ : pentane, $\ce{C5H12}$ ( $M_r = 72.0$ ).
$V_2 = V_1 \times \dfrac{p_1}{p_2} \times \dfrac{T_2}{T_1} = 500 \times \dfrac{100}{150} \times \dfrac{373}{293} = 424\ \text{cm}^3$ .
$M_r(\ce{NH4NO3}) = 80.0$ ; $n = \dfrac{1.00}{80.0} = 0.0125\ \text{mol}$ . Moles of gas $= 3 \times 0.0125 = 0.0375\ \text{mol}$ . $V = \dfrac{0.0375 \times 8.31 \times 500}{1.01 \times 10^{5}} = 1.54 \times 10^{-3}\ \text{m}^3 = 1.54\ \text{dm}^3$ . On cooling to room temperature the steam condenses to liquid water, so two-thirds of the gas molecules are removed from the gas phase; the temperature is also lower. Both reduce the number and energy of collisions with the walls, so the pressure falls considerably.
Empirical: C $\dfrac{24.3}{12.0} = 2.03$ ; H $\dfrac{4.1}{1.0} = 4.1$ ; Cl $\dfrac{71.6}{35.5} = 2.02$ . Ratio $1 : 2 : 1$ : $\ce{CH2Cl}$ , empirical mass $49.5$ . $M = \dfrac{0.484 \times 8.31 \times 373}{1.01 \times 10^{5} \times 150 \times 10^{-6}} = 99.0$ . $\dfrac{99.0}{49.5} = 2$ : molecular formula $\ce{C2H4Cl2}$ . Isomers: 1,1-dichloroethane, $\ce{CH3CHCl2}$ , and 1,2-dichloroethane, $\ce{CH2ClCH2Cl}$ .

The kinetic model of a gas

Where pressure comes from

Ideal gases

Real gases

The ideal gas equation

Linking to the molar volume

Rearrangements you will need

Worked examples

Practice

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