Behaviour of gases

Squeeze a balloon and it fights back, heat a sealed tin and it can burst, let a bicycle tyre lose air on a cold morning and the pressure gauge reads noticeably less than it did the day before. All three are the same story told three ways: a fixed amount of gas trades pressure, volume and temperature against each other according to one compact rule. Chemists, engine designers and the person filling your LPG cylinder all lean on the same equation. Three separate experimenters, working across roughly two centuries, each froze one variable and watched the other two dance. Robert Boyle squeezed gas at constant temperature and found pressure and volume trade off exactly. Jacques Charles heated gas at constant pressure and found volume grows in step with temperature. Amedeo Avogadro compared equal volumes of different gases and found they hold equal numbers of molecules, regardless of what gas it is. Stitch these three findings together and a single macroscopic law falls out: $PV = nRT$ , the equation of state of a perfect (ideal) gas. It says nothing yet about molecules bouncing around — that microscopic picture is kinetic theory, the next stop in this chapter — but it is the experimentally verified law every molecular model must reproduce. This article builds the equation of state from its three empirical pillars, shows how to read a P–V diagram and its isotherms, derives the work done when a gas is compressed or expanded isothermally, and works through the arithmetic a board exam or competitive paper actually asks: finding an unknown pressure, volume or temperature, and computing work done during compression. Keep sight of the practical anchor throughout — one mole of any ideal gas at standard temperature and pressure occupies about 22.4 litres, a number worth memorising cold.

Key takeaways

One equation governs a fixed amount of ideal gas: $PV = nRT$ , where $n$ is the number of moles and $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ is the universal gas constant, the same for every gas.
Boyle's law (constant $T$ ): $PV = \text{constant}$ — pressure and volume are inversely proportional.
Charles's law (constant $P$ ): $V/T = \text{constant}$ — volume grows in direct proportion to absolute temperature.
Avogadro's law: equal volumes of different ideal gases, at the same $P$ and $T$ , contain equal numbers of molecules — this is why the equation of state uses moles $n$ , not mass or gas identity.
Isotherms on a P–V diagram are rectangular hyperbolas ( $P \propto 1/V$ at fixed $T$ ); higher-temperature isotherms sit farther from the origin and different isotherms never cross.
Work done in an isothermal change: $W = nRT\ln(V_2/V_1)$ , positive when the gas expands and negative when it is compressed.
Temperature in this equation must always be the absolute (kelvin) temperature — the single most common source of lost marks in this topic.
Key idea: for a fixed amount of ideal gas, $PV = nRT$ ties pressure, volume and absolute temperature into one law that reduces to Boyle's and Charles's laws as special cases, and the work done in an isothermal change is $W = nRT\ln(V_2/V_1)$ .

The core insight

A fixed amount of ideal gas obeys $PV = nRT$ at every instant, with Boyle's law and Charles's law as its constant- $T$ and constant- $P$ special cases, and this single relation fixes both the shape of its P–V isotherms and the work $W = nRT\ln(V_2/V_1)$ done when the gas is compressed or expanded isothermally.

Where it fits in the syllabus

In the CBSE Kinetic Theory chapter, "behaviour of gases" covers the macroscopic, experimentally-derived side of the topic — Boyle's law, Charles's law and Avogadro's law combining into the equation of state $PV = nRT$ — before the chapter turns to the microscopic kinetic-theory model that explains why this equation holds. A Class 11 student is expected to state the equation of state, know $R$ 's value and units, sketch and read P–V and P–T diagrams for a fixed amount of gas, and compute the work done in an isothermal compression or expansion. This is squarely within the prescribed 2025-26 scope; van der Waals corrections for real gases and any transport-coefficient material are explicitly out of scope and are not touched here.

The physical picture

For a fixed amount of an ideal gas, pressure $P$ , volume $V$ and absolute temperature $T$ are locked together by the equation of state:

PV = nRT

where: $P$ is the pressure (Pa), $V$ is the volume (m³), $n$ is the number of moles, $T$ is the absolute temperature (K), and $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ is the universal gas constant — the same number for every gas, which is the whole point of writing the law in terms of moles rather than mass. The physical reading is that these four quantities cannot vary independently: fix any three and the fourth is determined. A real gas obeys this relation closely at low pressure and moderately high temperature, well away from the point where it would liquefy; that idealisation is what "perfect" or "ideal" gas means in this chapter. Two named special cases fall directly out of the equation of state when one variable is held fixed, alongside a third empirical law about the gas itself:

PV = \text{constant} \quad (\text{Boyle's law, fixed } n, T) \qquad\qquad \frac{V}{T} = \text{constant} \quad (\text{Charles's law, fixed } n, P)

where: in Boyle's law, squeezing the gas into a smaller volume raises its pressure in exact inverse proportion, so long as temperature does not change. In Charles's law, heating the gas at constant pressure makes it expand in direct proportion to the absolute temperature — and extrapolating that straight-line graph of $V$ against $T$ back to $V = 0$ is historically how physicists first located absolute zero near $-273\,^\circ\text{C}$ . Neither law is a separate fact about nature; each is simply $PV = nRT$ with one variable frozen.

Avogadro's law supplies the third pillar and explains why $n$ , rather than mass, is the correct variable: equal volumes of different ideal gases, at the same temperature and pressure, contain equal numbers of molecules. One mole of any gas — whether it is light like hydrogen or heavy like carbon dioxide — occupies the same volume as one mole of any other gas under identical $P$ and $T$ , about $22.4$ litres at standard temperature and pressure ( $0\,^\circ\text{C}$ , $1$ atm). This is why the equation of state does not care what gas you have put in the cylinder, only how many moles of it there are.

Figure

Three panels showing Boyle's law as a curve of pressure falling as volume rises at fixed temperature, Charles's law as a straight line of volume rising with absolute temperature at fixed pressure, and a family of isotherms on a pressure volume diagram with higher temperature curves lying farther from the origin

Work done in compressing or expanding a gas. When a gas changes volume quasi-statically (slowly enough to stay in equilibrium at every step) at constant temperature, the work done by the gas is the area under the P–V curve:

W = \int_{V_1}^{V_2} P\,dV = nRT\ln\!\left(\frac{V_2}{V_1}\right)

where: $V_1$ and $V_2$ are the initial and final volumes, and the substitution $P = nRT/V$ (valid at fixed $T$ from the equation of state) turns the integral of $P\,dV$ into a logarithm. If $V_2 > V_1$ (expansion), $W$ comes out positive — the gas does positive work on its surroundings. If $V_2 < V_1$ (compression), $\ln(V_2/V_1)$ is negative, so $W$ is negative, meaning work is done on the gas by whatever is squeezing it. On a P–V diagram an isotherm is a rectangular hyperbola, since $P \propto 1/V$ at fixed $T$ ; different isotherms never cross, and higher-temperature isotherms sit farther out from the origin, exactly as sketched above.

Derivation AWork done by an ideal gas during an isothermal change

The goal is to find, from first principles, how much work a fixed amount of ideal gas does on its surroundings when it expands or is compressed quasi-statically at constant temperature from volume $V_1$ to volume $V_2$ , using only the equation of state and the definition of work as the area under a P–V curve — no calculus beyond a single standard integral. This is the one formula in the chapter that a board paper is most likely to ask a student to actually derive rather than quote.

Step 1. Start from the definition of work done by a gas during a small volume change. When a gas pushes a piston out by an infinitesimal distance so that its volume changes by $dV$ at pressure $P$ , the small amount of work it does is $dW = P\,dV$ . Adding up (integrating) all these small contributions from the initial volume $V_1$ to the final volume $V_2$ gives the total work: $W = \displaystyle\int_{V_1}^{V_2} P\,dV$ .

Step 2. Express $P$ in terms of $V$ using the equation of state. Since the amount of gas $n$ and the temperature $T$ are both held fixed during an isothermal process, $PV = nRT$ can be rearranged to $P = \dfrac{nRT}{V}$ , with $nRT$ acting as a constant that can be pulled outside the integral.

Step 3. Substitute and integrate. $W = \displaystyle\int_{V_1}^{V_2} \frac{nRT}{V}\,dV = nRT \displaystyle\int_{V_1}^{V_2} \frac{dV}{V} = nRT\Big[\ln V\Big]_{V_1}^{V_2}$ . Evaluating the limits gives $W = nRT\big(\ln V_2 - \ln V_1\big)$ .

Step 4. Combine the logarithms and read the sign. Using $\ln V_2 - \ln V_1 = \ln(V_2/V_1)$ , the result is $W = nRT\ln\!\left(\dfrac{V_2}{V_1}\right)$ . If the gas expands, $V_2 > V_1$ , the ratio inside the logarithm exceeds 1, and $W$ is positive — work done by the gas on its surroundings. If the gas is compressed, $V_2 < V_1$ , the ratio is less than 1, $\ln(V_2/V_1)$ is negative, and $W$ comes out negative — meaning an external agent does positive work on the gas to squeeze it. Both signs match physical intuition directly, without any extra bookkeeping.

Worked examples

Board level - volume of a gas from the equation of state

A cylinder contains 2 mol of an ideal gas at a pressure of $2.0 \times 10^5\ \text{Pa}$ and temperature $300\ \text{K}$ . Find the volume occupied by the gas.

Step 1. Write down the equation of state and identify the known quantities: $n = 2\ \text{mol}$ , $P = 2.0\times10^5\ \text{Pa}$ , $T = 300\ \text{K}$ , $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ , and $V$ is the unknown.

Step 2. Rearrange $PV = nRT$ for $V$ and substitute: $V = \dfrac{nRT}{P} = \dfrac{2 \times 8.314 \times 300}{2.0 \times 10^5}$ .

Step 3. Evaluate the numerator first, then divide: $nRT = 2 \times 8.314 \times 300 = 4988.4\ \text{J}$ , so $V = \dfrac{4988.4}{2.0 \times 10^5} = 0.02494\ \text{m}^3$ .

Step 4. Convert to litres for a more readable answer, using $1\ \text{m}^3 = 1000\ \text{L}$ : $V = 0.02494 \times 1000 = 24.94\ \text{L}$ . Answer: the gas occupies about 24.9 litres under these conditions, sensibly close to the 22.4 L that one mole would occupy at STP, scaled up for 2 mol at a somewhat higher pressure and temperature.

Competitive level - work done during an isothermal compression, then check against the ideal-gas volume ratio

An ideal gas at temperature 300 K and volume $2.0\ \text{L}$ is compressed isothermally to $0.5\ \text{L}$ . If the amount of gas is 0.1 mol, find (a) the work done on the gas during this compression, and (b) verify the compression is consistent with the pressure ratio predicted by Boyle's law.

Step 1. Set up the work integral result with the given numbers: $n = 0.1\ \text{mol}$ , $T = 300\ \text{K}$ , $V_1 = 2.0\ \text{L}$ , $V_2 = 0.5\ \text{L}$ . Work done by the gas is $W = nRT\ln(V_2/V_1) = 0.1 \times 8.314 \times 300 \times \ln(0.5/2.0)$ .

Step 2. Evaluate the logarithm: $\ln(0.25) = -1.386$ . Multiply through: $W = 0.1 \times 8.314 \times 300 \times (-1.386) = -345.7\ \text{J}$ .

Step 3. Interpret the sign: since $W$ (work done by the gas) is negative, work is in fact done on the gas by the external agent compressing it, and its magnitude is about $345.7\ \text{J}$ . Answer (a): about 345.7 J of work is done on the gas.

Step 4. Cross-check with Boyle's law. At fixed $T$ and $n$ , $P_1V_1 = P_2V_2$ , so $P_2/P_1 = V_1/V_2 = 2.0/0.5 = 4$ — the pressure must have quadrupled as the volume dropped to a quarter. This is consistent with squeezing the gas hard doing a substantial amount of work on it: quartering the volume at constant temperature is a large compression, matching the sizeable $345.7\ \text{J}$ found above. Answer (b): the pressure rises by a factor of 4, consistent with the magnitude of work computed. Timing note: under 90 seconds for part (a) if $\ln(0.25) = -1.386$ is recalled or computed quickly on a calculator; the Boyle's-law cross-check in part (b) is a 15-second sanity test worth doing on every isothermal work problem, since a wildly inconsistent pressure ratio usually signals an arithmetic slip in $\ln(V_2/V_1)$ .

Common traps

Using temperature in degrees Celsius instead of kelvin in $PV = nRT$ — always convert to kelvin ( $T(\text{K}) = T(^\circ\text{C}) + 273$ ) before substituting; this is the single most common way to lose marks in this topic.
Forgetting that Boyle's and Charles's laws are special cases of the same equation of state at fixed $n$ , and trying to memorise them as unrelated rules instead of deriving them from $PV = nRT$ on demand.
Mixing up the units of $R$ — with $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ , pressure must be in pascals and volume in cubic metres for the equation to give consistent SI results; mixing atmospheres, litres and kelvin without converting is a frequent slip.
Dropping the sign in the work formula — writing $W = nRT\ln(V_1/V_2)$ instead of $nRT\ln(V_2/V_1)$ flips the sign of the answer and reverses whether work is done on or by the gas.
Assuming real gases obey $PV = nRT$ exactly under all conditions — it is only a good approximation at low pressure and away from the liquefaction point; real-gas corrections (the van der Waals equation) are beyond this chapter's scope and should not be quoted in a CBSE answer.

You should be able to…

State the ideal gas equation $PV = nRT$ and identify each symbol with its correct SI unit.
Derive Boyle's law and Charles's law as special cases of $PV = nRT$ at fixed temperature and fixed pressure respectively, and state Avogadro's law in words.
Sketch an isotherm on a P–V diagram, explain why it is a rectangular hyperbola, and explain why isotherms at different temperatures never cross.
Calculate an unknown among $P$ , $V$ , $T$ , $n$ given the other three, always converting temperature to kelvin first.
Derive and apply the isothermal work formula $W = nRT\ln(V_2/V_1)$ , correctly reading the sign to say whether work is done by or on the gas.

Ask the tutor: Want to walk through more P–V–T problems, practise reading isotherm diagrams, or see how this equation of state reappears once kinetic theory derives it from molecular motion — ask the AI tutor for extra worked examples.

Why does the equation of state use moles instead of mass or the number of molecules directly?

Because Avogadro's law says equal volumes of different gases, at the same pressure and temperature, hold equal numbers of molecules — regardless of whether those molecules are light like hydrogen or heavy like carbon dioxide. Using mass would force a different constant for every gas, since a gram of hydrogen contains far more molecules than a gram of carbon dioxide. Using moles instead means one universal constant $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ works for every ideal gas, which is exactly why the law is written as $PV = nRT$ and not, say, $PV = (\text{mass})\times(\text{something})\times T$ .

Is PV equals nRT true for a real gas like the air I breathe, or only for a fictional ideal gas?

It is an excellent approximation for real gases like air at ordinary room conditions — low pressure and moderate temperature, well away from the point where the gas would condense into a liquid. Air at atmospheric pressure and room temperature obeys $PV = nRT$ to within a small fraction of a per cent, which is why engineers use it confidently for everyday calculations. It starts breaking down at very high pressure or very low temperature, close to liquefaction, where molecules are packed close enough that their own size and their mutual attractions start to matter — corrections for that (the van der Waals equation) are beyond this chapter.

Why is the work done in an isothermal process nRT ln(V2/V1) and not just P times delta V?

Because pressure itself is changing throughout an isothermal compression or expansion — as the volume shrinks, the pressure rises (and vice versa), following $P = nRT/V$ . Using a single value of $P$ times the total change in volume, $P\,\Delta V$ , would only be correct if pressure stayed constant, which it does not here. The correct approach is to add up (integrate) the tiny bits of work $P\,dV$ done at each instant, with $P$ changing continuously, and that integral is exactly what produces the logarithm in $W = nRT\ln(V_2/V_1)$ .

My answer for work done came out negative — did I make a mistake?

Not necessarily — check what the sign is telling you before assuming an error. $W = nRT\ln(V_2/V_1)$ gives the work done by the gas. If the gas is being compressed ( $V_2 < V_1$ ), the ratio $V_2/V_1$ is less than one, its logarithm is negative, and so $W$ comes out negative. That is physically correct: it means the gas does negative work on its surroundings, or equivalently, the surroundings do positive work on the gas to squeeze it smaller. Only worry if the sign contradicts what actually happened physically — a negative $W$ during an expansion, or a positive $W$ during a compression, would signal a genuine sign error, most likely from writing the volume ratio upside down.