The gaseous state: ideal and real gases and pV = nRT

1. Overview

A gas is mostly empty space, with its particles moving fast and at random. This section explains how those particles create pressure, what makes a gas ideal, and why real gases deviate from ideal behaviour, most of all at high pressure and low temperature. It also covers the ideal gas equation, $pV = nRT$, which links the amount of a gas to its pressure, volume and temperature, and how to use it to find the relative molecular mass, $M_{\text{r}}$.

Key Definitions

• Pressure of a gas: the force per unit area that a gas exerts on the walls of its container, caused by its particles colliding with the walls; it is measured in pascals, where 1 Pa = 1 N m⁻².
• Ideal gas: a model gas whose particles have zero volume and no intermolecular forces of attraction between them, so it obeys pV = nRT exactly under all conditions.
• Ideal gas equation: pV = nRT, where p is the pressure in Pa, V is the volume in m³, n is the amount of gas in mol, R is the molar gas constant and T is the temperature in K.
• Real gas: an actual gas, whose particles do take up some volume and do attract one another, so it deviates from pV = nRT, most at high pressure and low temperature.
• Kinetic theory of gases: the model that explains the behaviour of a gas in terms of its particles moving constantly and randomly and colliding elastically with each other and with the walls of the container.
• Molar gas constant (R): the constant in the ideal gas equation, which has the same value for every gas: 8.31 J K⁻¹ mol⁻¹.
• Thermodynamic temperature (T): temperature on the kelvin scale, which starts at absolute zero, 0 K; add 273 to a temperature in °C to convert it to kelvin.
• Molar volume of a gas (Vm): the volume occupied by one mole of any gas that behaves ideally: 22.4 dm³ mol⁻¹ at s.t.p. (273 K and 101 kPa) and 24.0 dm³ mol⁻¹ at room conditions.

Content

How a gas exerts pressure

The particles of a gas (atoms or molecules) move in random directions, in straight lines between collisions, typically at several hundred metres per second at room temperature.

When a particle hits a wall, it bounces back, so its momentum changes. The wall must push on the particle to cause that change. By Newton's third law, the particle pushes on the wall with an equal and opposite force. One collision is a tiny push, but a huge number of particles hit the wall every second, and together they give a steady force. Pressure is that force divided by the area of wall:

$$p = \frac{F}{A}$$

Here $p$ is in pascals (Pa), $F$ in newtons (N) and $A$ in m², so 1 Pa = 1 N m⁻². Because the motion is random, the pressure is the same on every wall.

Pressure depends on how often particles hit each unit area of wall and how hard each one hits:

Change (everything else fixed)	Effect on the particles	Collisions with the walls	Pressure
Temperature raised (fixed volume)	move faster on average	more frequent and harder	increases
Volume reduced (fixed temperature)	same average speed, more particles per unit volume	more frequent per unit area	increases
More gas added (fixed volume and temperature)	more particles	more frequent	increases

Pressure comes from collisions with the walls, not from collisions between particles.

The kinetic model of an ideal gas

An ideal gas fits the model of the kinetic theory of gases exactly:

1. The gas is a very large number of particles in constant, random motion.
2. The particles have zero volume. Their size is negligible compared with the volume of the container.
3. There are no intermolecular forces of attraction between the particles, so between collisions they move in straight lines at constant speed.
4. Collisions between particles, and with the walls, are perfectly elastic (no kinetic energy is lost overall) and last a negligible time.

Points 2 and 3 are the two you must know. The particles still have mass. "Zero volume" describes each particle, not the gas: one mole of an ideal gas still fills 24.0 dm³ at room conditions.

The average kinetic energy of the particles' movement is proportional to the temperature in kelvin. That is why gas calculations always use $T$ in kelvin, never in °C.

Real gases: when and why they deviate

No real gas is ideal: real particles take up some space, and they attract one another through van der Waals' forces.

Ideal assumption	Fails badly when	Why
Zero particle volume	the pressure is very high	the particles are squeezed close together, so their own volume is a significant fraction of the container's volume and the free space is less than $V$
No intermolecular attraction	the temperature is low	the particles move more slowly, with less kinetic energy, so the attractions between them have a significant effect
No intermolecular attraction	the pressure is high	the particles are close together, and attractions are significant only over short distances

Attractions lower the pressure. A particle about to hit a wall is pulled back slightly by the particles behind it, so it hits less hard, and the pressure is lower than an ideal gas would give. At a low enough temperature the attractions hold the particles together and the gas condenses to a liquid, the most extreme failure of the model.

The two failures push in opposite directions. Attractions make $pV$ smaller than $nRT$. The particles' own volume makes $pV$ larger than $nRT$, because only the free space between the particles can be squeezed. At room conditions both effects are tiny: for nitrogen or oxygen, $pV$ and $nRT$ differ by well under 1%.

So a real gas behaves most like an ideal gas at high temperature and low pressure:

• Low pressure: the particles are far apart, so their own volume is negligible compared with the container, and the attractions between them are negligible.
• High temperature: the particles move so fast that the attractions have little effect.

Which real gases are closest to ideal?

Under the same conditions, a gas is closer to ideal when its intermolecular forces are weaker and its particles are smaller.

• Type of force. For molecules with similar numbers of electrons, instantaneous dipole-induced dipole (id-id) forces are weakest, permanent dipole-permanent dipole (pd-pd) forces stronger and hydrogen bonds strongest.
• Number of electrons. More electrons mean stronger id-id forces. Helium (2 electrons per atom, tiny atoms) has the weakest attractions of any gas, so it fits the ideal model most closely; xenon (54 electrons) deviates much more.
• Same electrons, different forces. Methane, CH₄, and ammonia, NH₃, both have 10 electrons, but only ammonia forms hydrogen bonds, so ammonia deviates much more.
• Only forces between particles count. The strength of the covalent bonds inside a molecule plays no part.

The ideal gas equation, pV = nRT

$$pV = nRT$$

Symbol	Quantity	Unit to use	How to convert
$p$	pressure	Pa	1 kPa = 1000 Pa
$V$	volume	m³	1 dm³ = $10^{-3}$ m³; 1 cm³ = $10^{-6}$ m³
$n$	amount of gas	mol	$n = m/M$ for a mass $m$ in g
$R$	molar gas constant	J K⁻¹ mol⁻¹	8.31, from the Data Booklet
$T$	thermodynamic temperature	K	add 273 to the temperature in °C

Why these units. The unit of $R$ contains joules, and 1 Pa × 1 m³ = 1 N m⁻² × 1 m³ = 1 N m = 1 J. So $pV$ is in joules, matching $nRT$ (mol × J K⁻¹ mol⁻¹ × K = J), only when $p$ is in Pa and $V$ in m³.

It holds at any temperature and pressure, not just room conditions, as long as the gas behaves ideally. For a mixture, $n$ is the total amount of gas.

Sense check. One mole of gas at room conditions fills about 0.024 m³. If an answer is out by a factor of 1000 or a million, a volume or pressure conversion is missing.

Molar volume shortcut. At s.t.p. (273 K and 101 kPa), one mole of gas occupies 22.4 dm³; at room conditions it occupies 24.0 dm³. Use $n = V/V_{\text{m}}$ only when a question names one of those conditions. If it gives a temperature and pressure instead, use $pV = nRT$: at 25 °C and 101 kPa, for example, one mole occupies 24.5 dm³, not 24.0. (Working out the s.t.p. value from $V = RT/p$ with 8.31, 273 K and 101 000 Pa gives 22.5 dm³; the Data Booklet's 22.4 uses the unrounded pressure, 101.325 kPa.)

How p, V and T are linked

For a fixed amount of gas, $pV/T = nR$ stays constant, so for a change from state 1 to state 2:

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

Here $p$ and $V$ just need the same units on both sides, but $T$ must be in kelvin.

Held constant	Relationship	Graph
$n$ and $T$	$p \propto 1/V$: double the pressure, halve the volume	$p$ against $V$ is a curve; $p$ against $1/V$ is a straight line through the origin
$n$ and $p$	$V \propto T$	$V$ against $T$ is a straight line through the origin, gradient $nR/p$ (less gas gives a shallower line)
$n$ and $V$	$p \propto T$	$p$ against $T$ is a straight line through the origin
$p$ and $T$	$V \propto n$	equal volumes of gases at the same $T$ and $p$ contain equal numbers of particles

The $V$ against $T$ and $p$ against $T$ lines pass through the origin only when $T$ is in kelvin. On a Celsius scale, the line of $V$ against temperature reaches zero volume at −273 °C, absolute zero. Real gases liquefy long before that, so the low-temperature part is an extrapolation.

Worked example 1: pressure of ethane in a sealed vessel

Question: A sealed steel vessel of volume 1.25 dm³ holds 1.35 g of ethane, C₂H₆, at 58 °C. Calculate the pressure inside the vessel, in kPa. Assume ethane behaves as an ideal gas. ($M$ of C₂H₆ = 30.0 g mol⁻¹)

Step 1: put every quantity in SI units. $$n = \frac{1.35}{30.0} = 0.0450\ \text{mol} \qquad V = 1.25 \times 10^{-3}\ \text{m}^3 \qquad T = 58 + 273 = 331\ \text{K}$$

Step 2: rearrange, then substitute. $$p = \frac{nRT}{V} = \frac{0.0450 \times 8.31 \times 331}{1.25 \times 10^{-3}} = 99\,022\ \text{Pa}$$

Step 3: convert to kPa. 99 022 Pa ÷ 1000 = 99.0 kPa.

Answer: $p = 99.0$ kPa (3 s.f.).

If the vessel is then heated to 140 °C (413 K), $n$ and $V$ stay fixed, so $p \propto T$: $p = 99.0 \times \frac{413}{331} = 124$ kPa. Using Celsius temperatures, $99.0 \times \frac{140}{58}$, would give 239 kPa, which is wrong.

Worked example 2: volume of gas made in a reaction

Question: Calcium hydride reacts with water to release hydrogen.

CaH₂(s) + 2H₂O(l) → Ca(OH)₂(s) + 2H₂(g)

Calculate the volume of hydrogen, in dm³, made when 2.10 g of calcium hydride reacts completely. The hydrogen is at 22 °C and 98.5 kPa. ($A_{\text{r}}$: Ca 40.1, H 1.0)

Step 1: moles of CaH₂. $M = 40.1 + 2(1.0) = 42.1$ g mol⁻¹. $$n(\text{CaH}_2) = \frac{2.10}{42.1} = 0.04988\ \text{mol}$$

Step 2: use the equation. 1 mol of CaH₂ gives 2 mol of H₂, so $n(\text{H}_2) = 2 \times 0.04988 = 0.09976$ mol.

Step 3: SI units. $T = 22 + 273 = 295$ K and $p = 98.5 \times 10^3 = 98\,500$ Pa.

Step 4: rearrange, substitute and convert. $$V = \frac{nRT}{p} = \frac{0.09976 \times 8.31 \times 295}{98\,500} = 2.483 \times 10^{-3}\ \text{m}^3 = 2.48\ \text{dm}^3$$

Answer: $V = 2.48$ dm³ (3 s.f.).

Finding Mr with the ideal gas equation

Substitute $n = m/M$ into $pV = nRT$ and rearrange:

$$pV = \frac{m}{M}RT \qquad\Rightarrow\qquad M = \frac{mRT}{pV}$$

With $m$ in g, $p$ in Pa, $V$ in m³ and $T$ in K, $M$ comes out in g mol⁻¹. $M_{\text{r}}$ is the same number with no units.

• For a gas: weigh an evacuated flask of known volume, fill it with the gas at a known temperature and pressure, and weigh it again. The increase in mass is $m$.
• For a volatile liquid: keep a gas syringe at a steady temperature above the liquid's boiling point. Inject a sample from a hypodermic syringe weighed before and after (the difference is $m$). The liquid evaporates completely, and the rise in the gas-syringe reading is the volume of vapour.
• From a graph: a plot of mass of gas against volume at fixed $T$ and $p$ has gradient equal to the density, $\rho$, and $M = \rho RT/p$. Convert the gradient to g m⁻³ first (1 g cm⁻³ = $10^6$ g m⁻³).

Worked example 3: Mr of a volatile liquid

Question: A gas syringe is kept at 100 °C. A hypodermic syringe containing liquid X weighs 13.207 g. Some X is injected into the gas syringe, and the hypodermic syringe then weighs 13.077 g. All the X evaporates, and the gas-syringe reading rises from 6.5 cm³ to 60.5 cm³. The pressure is 99.8 kPa. X is a ketone with formula CₙH₂ₙO. Assume the vapour behaves as an ideal gas. Calculate $M_{\text{r}}$ of X and deduce its formula. ($A_{\text{r}}$: C 12.0, H 1.0, O 16.0)

Step 1: put every quantity in SI units.

• $m = 13.207 - 13.077 = 0.130$ g
• $V = 60.5 - 6.5 = 54.0$ cm³ $= 54.0 \times 10^{-6}$ m³
• $T = 100 + 273 = 373$ K
• $p = 99.8 \times 10^3 = 99\,800$ Pa

Step 2: substitute. $$M = \frac{mRT}{pV} = \frac{0.130 \times 8.31 \times 373}{99\,800 \times 54.0 \times 10^{-6}} = \frac{402.95}{5.3892} = 74.77\ \text{g mol}^{-1}$$

Step 3: deduce the formula. For CₙH₂ₙO, $M_{\text{r}} = 14.0n + 16.0$, so $n = \frac{74.8 - 16.0}{14.0} = 4.2$. The nearest whole number is 4.

Answer: $M_{\text{r}} = 74.8$ (3 s.f.), so X is C₄H₈O, butanone ($M_{\text{r}}$ = 72.0).

The result is about 4% above the true $M_{\text{r}}$, and this method tends to read high. One reason is that the vapour is not ideal. At 100 °C, butanone is only about 20 K above its boiling point (80 °C), so the attractions between its polar molecules are still significant. In a gas syringe the pressure is fixed, so the attractions show up as a volume slightly smaller than an ideal gas would fill, and a smaller $V$ in $M = mRT/(pV)$ gives a larger $M$.

Key Equations

Pressure ($F$ = force in N; $A$ = area in m²; $p$ in Pa, 1 Pa = 1 N m⁻²): $$p = \frac{F}{A}$$

Ideal gas equation ($p$ in Pa; $V$ in m³; $n$ in mol; $R$ = 8.31 J K⁻¹ mol⁻¹; $T$ in K): $$pV = nRT$$

Molar mass of a gas or vapour (from $n = m/M$; $m$ in g; $M$ in g mol⁻¹, numerically equal to $M_{\text{r}}$; $\rho$ = density in g m⁻³): $$M = \frac{mRT}{pV} \qquad M = \frac{\rho RT}{p}$$

Fixed amount of gas (same units for $p$ and $V$ on both sides; $T$ in K): $$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$$

Molar volume ($V_{\text{m}}$ = 22.4 dm³ mol⁻¹ at s.t.p., 24.0 dm³ mol⁻¹ at room conditions): $$n = \frac{V}{V_{\text{m}}}$$

Conversions: $T$ (K) = temperature (°C) + 273; 1 cm³ = $10^{-6}$ m³; 1 dm³ = $10^{-3}$ m³; 1 kPa = $10^3$ Pa.

Common Mistakes to Avoid

• Mistake: putting cm³, dm³, kPa or °C straight into $pV = nRT$. Fix: convert every quantity first. In Worked example 3, 54.0 cm³ is $5.40 \times 10^{-5}$ m³; using 54.0 as if it were in m³ gives $M_{\text{r}} = 7.48 \times 10^{-5}$, a million times too small.
• Mistake: stopping at the moles of the substance you were given. Fix: use the reaction equation to get the moles of gas. In Worked example 2, 0.0499 mol of CaH₂ gives 0.0998 mol of H₂; skipping this halves the volume to 1.24 dm³.
• Mistake: rounding $n$ early. Fix: keep the calculator value until the end. In Worked example 3, $n = pV/(RT) = 1.739 \times 10^{-3}$ mol; rounding it to $1.7 \times 10^{-3}$ mol changes $M_{\text{r}}$ from 74.8 to 76.5.
• Mistake: saying ideal gas particles have no mass, or that "the gas" has no volume. Fix: the particles have mass. Write "the volume of the particles is negligible", not "the volume of the gas".
• Mistake: giving the assumptions when asked for the conditions, or getting the conditions backwards. Fix: the conditions for near-ideal behaviour are high temperature and low pressure. The assumptions are zero particle volume and no intermolecular forces of attraction.
• Mistake: explaining non-ideal behaviour with the strength of covalent bonds, or by comparing $M_{\text{r}}$ alone. Fix: compare intermolecular forces. N₂ has a very strong N≡N bond yet behaves almost ideally at room conditions, while, as gases at the same temperature and pressure, H₂O ($M_{\text{r}}$ 18.0) deviates far more than O₂ ($M_{\text{r}}$ 32.0) because of hydrogen bonding.
• Mistake: using 24.0 dm³ mol⁻¹ at any conditions. Fix: use 24.0 only for room conditions and 22.4 only for s.t.p.; otherwise use $pV = nRT$. At 100 °C and 101 kPa, one mole of gas occupies 30.7 dm³.
• Mistake: saying pressure comes from particles colliding with each other, or writing "more collisions" with no time. Fix: pressure comes from collisions with the walls. On heating, the particles hit the walls more often (more collisions per second) and harder.

Exam Tips

• State the ideal gas equation as $pV = nRT$ and define every symbol with its unit: $p$ in Pa, $V$ in m³, $n$ in mol, $R$ = 8.31 J K⁻¹ mol⁻¹, $T$ in K.
• Explain gas pressure as a chain: particles collide with the walls, each collision exerts a force, and pressure is the total force per unit area. For any change, say whether collisions with the walls become more frequent, harder, or both.
• State two assumptions of an ideal gas as "the particles have zero (negligible) volume" and "there are no intermolecular forces of attraction between the particles". Always write "particles", not "gas".
• Explain a deviation by naming the condition (high pressure or low temperature), the assumption that fails and the reason in terms of particles. To compare two gases, name the intermolecular forces in each.
• Calculate with each conversion on its own line, rearrange before substituting, and carry unrounded values. Give the final answer with its unit, to as many s.f. as the data justify (usually 3, the same as $R$ = 8.31); $M_{\text{r}}$ has no unit.