Ideal gas equation and absolute temperature

Squeeze a balloon and it pushes back; heat it near a stove and it swells; leave it in a snowy car boot overnight and it shrinks. Behind all three everyday observations sits one remarkably compact rule that low-density gases obey almost exactly, no matter what gas is inside the balloon — nitrogen, helium, ordinary air. That rule is the ideal gas equation, $PV = nRT$ , and it does something no single-substance formula for a solid or liquid can do: it ties pressure, volume, amount of substance, and temperature together in one equation that works, to good approximation, for every dilute gas. The equation also hides a deeper discovery. If you take a fixed volume of gas and plot its pressure against temperature in degrees Celsius, you get a straight line — and every gas, no matter how different chemically, gives a line that crosses zero pressure at almost the same temperature: $-273.15\,^\circ\text{C}$ . A gas cannot have negative pressure, so this crossing point is not just an algebraic curiosity; it marks a genuine lower limit on temperature itself. That limit is absolute zero, and building a temperature scale that starts there — the Kelvin scale — is what finally makes $PV = nRT$ true in its simplest form. This article builds the equation up from three older experimental laws you may already have met separately — Boyle's, Charles's, and Gay-Lussac's — shows how a constant-volume gas thermometer reveals absolute zero, and works through the arithmetic needed to move confidently between litres, atmospheres, Celsius, and the SI units the equation actually demands. By the end you should be able to look at any gas-law numerical and know exactly which version of the equation to reach for, and why.

Key takeaways

One equation, one scale. $PV = nRT$ links pressure $P$ , volume $V$ , moles $n$ , and absolute temperature $T$ in kelvin, using $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ .
Three old laws are one new law. At constant $T$ it is Boyle's law ( $PV = \text{const}$ ); at constant $P$ it is Charles's law ( $V/T = \text{const}$ ); at constant $V$ it is Gay-Lussac's law ( $P/T = \text{const}$ ).
A gas thermometer finds absolute zero. Pressure at constant volume varies linearly with Celsius temperature; every gas's line extrapolates to zero pressure at almost exactly $-273.15\,^\circ\text{C}$ .
Kelvin is defined from that intercept. $T(\text{K}) = T(^\circ\text{C}) + 273.15$ , so $0\ \text{K}$ is absolute zero and the Kelvin scale has no negative values for an ideal gas.
Only kelvin works in $PV = nRT$ . Because $P \propto T$ only when $T$ is measured from absolute zero, plugging in a Celsius value gives a wrong answer, not just an inconvenient one.
The combined gas law tracks one fixed sample through change. $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ lets you connect any two states of the same amount of gas without knowing $n$ or $R$ explicitly.
Key idea: for a fixed amount of gas at low pressure, $PV = nRT$ holds only when T is the Kelvin (absolute) temperature, because a gas's pressure at constant volume extrapolates linearly to zero at −273.15 °C — the universal origin from which the Kelvin scale is built.

The core insight

The ideal gas equation $PV = nRT$ unifies Boyle's, Charles's, and Gay-Lussac's laws into one equation of state, and it only holds in this simple proportional form when temperature is measured on the Kelvin scale, whose zero is fixed by the universal pressure-versus-temperature intercept of a constant-volume gas thermometer at $-273.15\,^\circ\text{C}$ .

Where it fits in the syllabus

This topic sits inside Thermal Properties of Matter, right after thermometry, and is the bridge between "how we measure temperature" and "how gases behave." CBSE expects you to state and use $PV = nRT$ , explain how a constant-volume gas thermometer's pressure-temperature graph points to absolute zero, convert correctly between Celsius and Kelvin, and solve numericals using the combined gas law for a fixed sample of gas. The why behind the equation — that it emerges from molecules colliding with the container walls — belongs to the separate Kinetic Theory unit later in the book; here you take the equation as an experimentally verified description of gas behaviour and use it as a working tool.

The physical picture

Three separate nineteenth-century experiments, done independently, each varied one gas property while holding another fixed. Robert Boyle compressed gas in a J-tube at constant temperature and found pressure and volume trade off exactly:

PV = \text{constant} \quad (\text{fixed } n, T)

where: $P$ is the gas pressure and $V$ its volume; squeezing the gas into half the space doubles the pressure, and expanding it to twice the space halves the pressure — the product stays fixed as long as the temperature and the amount of gas do not change. Jacques Charles heated gas at constant pressure and found volume grows in direct proportion to Kelvin temperature:

\frac{V}{T} = \text{constant} \quad (\text{fixed } n, P)

where: $V$ is volume and $T$ is absolute temperature; a gas heated from $300\ \text{K}$ to $600\ \text{K}$ at fixed pressure doubles in volume — but only if $T$ is measured in kelvin, since the proportionality is a statement about distance from absolute zero, not from the freezing point of water. Joseph Gay-Lussac sealed gas in a rigid container and found pressure itself grows in direct proportion to absolute temperature:

\frac{P}{T} = \text{constant} \quad (\text{fixed } n, V)

where: $P$ is pressure and $T$ is absolute temperature; this is exactly the relationship a constant-volume gas thermometer exploits, and it is the experiment that hands us absolute zero, described below. All three combine into a single equation of state that also brings in the amount of gas, measured in moles $n$ :

PV = nRT

where: $P$ is pressure in pascals (Pa), $V$ is volume in cubic metres (m $^3$ ), $n$ is the number of moles, $T$ is the absolute temperature in kelvin, and $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant — universal because the same value of $R$ works for any dilute gas, whatever its identity. Set $n$ constant and compare two states of the same sample and you get the practically useful combined gas law:

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

where: subscripts $1$ and $2$ label an initial and a final state of one fixed sample of gas; because $n$ and $R$ are the same in both states, they cancel, leaving a relation you can use without ever computing $n$ or plugging in $R$ directly. Now the deeper question: why kelvin, and not Celsius or Fahrenheit? Take a fixed volume of gas — say, a bulb connected to a pressure gauge — and measure its pressure at several known Celsius temperatures: ice point ( $0\,^\circ\text{C}$ ), steam point ( $100\,^\circ\text{C}$ ), and a few points in between or above. Plot pressure against Celsius temperature. Gay-Lussac's law says this plot is a straight line, and experiment confirms it, for any dilute gas. The remarkable part is what happens when you extend that straight line backward, below $0\,^\circ\text{C}$ , toward lower and lower pressures: it always crosses $P = 0$ at very nearly the same temperature, $-273.15\,^\circ\text{C}$ , regardless of whether the bulb contains nitrogen, hydrogen, or helium. A real gas cannot actually reach zero pressure this way — it liquefies first — but the extrapolated intercept is universal, and physics takes that universality seriously: it defines absolute zero, the coldest temperature conceivable, and builds a new scale, the Kelvin scale, with its zero exactly there.

T(\text{K}) = T(^\circ\text{C}) + 273.15

where: the size of one kelvin is defined to be identical to one degree Celsius, so the two scales differ only by a fixed shift, never a stretch — a temperature difference is the same number in both scales, but an absolute temperature value is not.

Figure

A graph of gas pressure at constant volume plotted against temperature in Celsius, showing three straight lines for three different gases all converging to zero pressure at the same point near minus 273 degrees Celsius, which becomes zero kelvin on the absolute scale

Derivation ABuilding the combined gas law from three separate experimental laws

The goal here is to show, using nothing more than algebra, that Boyle's law, Charles's law, and Gay-Lussac's law are not three unrelated facts to memorise but three special cases of a single underlying equation of state, and to arrive at the combined gas law that lets a fixed sample of gas be tracked from any one state $(P_1, V_1, T_1)$ to any other state $(P_2, V_2, T_2)$ without ever needing to know how many moles are present or look up the value of $R$ separately for the problem at hand.

Step 1. Start from the ideal gas equation as the given experimental fact for a fixed amount of gas: $PV = nRT$ , with $n$ and $R$ both constants for that sample. Rearrange it to isolate the combination that will not change as the gas is squeezed, heated, or cooled: $\dfrac{PV}{T} = nR$ .

Step 2. Because $n$ (the amount of gas, in moles) does not change — no gas is added or removed — and $R$ is a universal constant, the right-hand side $nR$ is the same number no matter what state the gas is in. This means $\dfrac{PV}{T}$ is a constant of the sample, true at the initial state and equally true at any later state.

Step 3. Write this constancy explicitly for two states, an initial state labelled $1$ and a final state labelled $2$ : $\dfrac{P_1V_1}{T_1} = nR$ and $\dfrac{P_2V_2}{T_2} = nR$ . Since both expressions equal the same $nR$ , they equal each other: $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ . This is the combined gas law, and notice that $n$ and $R$ have vanished from the final relation entirely — only the measurable quantities $P$ , $V$ , $T$ remain.

Step 4. Recover the three named laws as special cases by freezing one variable at a time. Hold temperature fixed, $T_1 = T_2$ : the combined law reduces to $P_1V_1 = P_2V_2$ , Boyle's law. Hold pressure fixed, $P_1 = P_2$ : it reduces to $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ , Charles's law. Hold volume fixed, $V_1 = V_2$ : it reduces to $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ , Gay-Lussac's law. Each of the three historical laws is therefore a slice through the single equation of state, and none of them needed to be discovered as an independent postulate — they are algebraic consequences of $PV = nRT$ once one variable is held constant.

Worked examples

Board level - heating gas at constant pressure

A sample of gas occupies $2.0\ \text{L}$ at a pressure of $1.0\times10^5\ \text{Pa}$ and temperature $300\ \text{K}$ . It is heated at constant pressure until its volume becomes $2.5\ \text{L}$ . Find the final temperature, in both kelvin and Celsius.

Step 1. Identify what stays fixed: pressure is constant, so $P_1 = P_2$ and the combined gas law reduces to Charles's law, $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ .

Step 2. Substitute the known values, keeping volume in the same unit on both sides (litres are fine here since they cancel): $\dfrac{2.0}{300} = \dfrac{2.5}{T_2}$ .

Step 3. Solve for $T_2$ by cross-multiplying: $T_2 = \dfrac{2.5 \times 300}{2.0} = \dfrac{750}{2.0} = 375\ \text{K}$ .

Step 4. Convert to Celsius for a physically intuitive answer: $T_2 = 375 - 273.15 = 101.85\,^\circ\text{C}$ . The gas must be heated to $375\ \text{K}$ , approximately $102\,^\circ\text{C}$ — just above the boiling point of water at sea level.

Competitive level - finding moles from a full set of state variables

Find the number of moles of an ideal gas that occupies $8.314\ \text{L}$ at a pressure of $2.0\times10^5\ \text{Pa}$ and temperature $500\ \text{K}$ , using $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ . Then estimate the mass of the sample if it is nitrogen gas ( $\text{N}_2$ , molar mass $28\ \text{g/mol}$ ).

Step 1. Convert every quantity to SI units before touching $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ , since that value of $R$ only gives correct numbers with pascals, cubic metres, and kelvin. Volume: $V = 8.314\ \text{L} = 8.314\times10^{-3}\ \text{m}^3$ . Pressure and temperature are already SI: $P = 2.0\times10^5\ \text{Pa}$ , $T = 500\ \text{K}$ .

Step 2. Rearrange $PV = nRT$ for the unknown: $n = \dfrac{PV}{RT}$ . Substitute: $n = \dfrac{(2.0\times10^5)(8.314\times10^{-3})}{(8.314)(500)} = \dfrac{1662.8}{4157}$ .

Step 3. Carry out the division carefully, since competitive exams often penalise rounding too early: $n = 0.400\ \text{mol}$ (the neat cancellation of the $8.314$ in numerator and denominator is a useful check — it should leave a clean-looking number, confirming no arithmetic slip).

Step 4. Convert moles to mass using the molar mass of nitrogen: $m = n \times M = 0.400\ \text{mol} \times 28\ \text{g/mol} = 11.2\ \text{g}$ . The sample contains $0.40\ \text{mol}$ of gas, roughly $11.2\ \text{g}$ if it is nitrogen. Timing note: this problem carries two unit-conversion traps — litres to cubic metres, and moles to grams via a separately given molar mass — so budget about $2$ minutes: roughly $45$ seconds to convert units cleanly, $45$ seconds for the $PV/RT$ arithmetic, and the rest to convert to mass and sanity-check that $0.40\ \text{mol}$ of gas occupying about $8$ litres at moderate pressure and $500\ \text{K}$ is a physically reasonable quantity.

Common traps

Using Celsius directly in $PV = nRT$ . The equation is only proportional in the simple form $P \propto T$ , $V \propto T$ when $T$ is absolute; a Celsius value plugged in raw gives a wrong, sometimes negative or nonsensical, result. Always convert first.
Mixing units with $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ . This value of $R$ demands SI throughout — pascals, cubic metres, kelvin. Leaving volume in litres or pressure in atmospheres while using this $R$ silently produces answers wrong by a large factor.
Treating absolute zero as experimentally reached. It is a theoretical extrapolated limit from the pressure-temperature graph; real gases liquefy or solidify long before $0\ \text{K}$ , so absolute zero is approached asymptotically, never attained, in any real experiment.
Forgetting that the combined gas law needs a fixed amount of gas. If gas leaks, is added, or is exchanged between containers, $n$ changes and $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ no longer applies — you must go back to the full $PV = nRT$ for each state separately.
Assuming the ideal gas law is exact for all real gases at all conditions. It is an excellent approximation at low pressure and moderate-to-high temperature, where molecules are far apart and interactions are negligible; real gases deviate noticeably near liquefaction or at high pressure.

You should be able to…

State the ideal gas equation $PV = nRT$ and identify each symbol with its correct SI unit.
Explain, using a constant-volume gas thermometer, how the pressure-temperature graph of a gas points to a universal absolute zero.
Convert confidently between Celsius and Kelvin, and explain why only Kelvin can be used directly in the gas equation.
Derive Boyle's, Charles's, and Gay-Lussac's laws as special cases of the combined gas law.
Use the combined gas law to relate two states of a fixed amount of gas, and solve for any one of $P$ , $V$ , $n$ , or $T$ given the other three.
Recognise when the ideal gas approximation is reasonable to use and when real-gas deviations should be flagged, even qualitatively.

Ask the tutor: if you are unsure why a gas's pressure-versus-temperature graph must be a straight line, or want more practice converting between Celsius, Kelvin, and mixed-unit gas-law numericals, ask the AI tutor to walk through it with you step by step.

Why is absolute zero exactly -273.15 degrees Celsius and not some rounder number?

It comes directly from experiment, not from a rule someone chose. When you plot the pressure of a fixed volume of gas against Celsius temperature and extend that straight line backward to zero pressure, careful measurement across many gases consistently lands the crossing point within a small fraction of a degree of $-273.15\,^\circ\text{C}$ . It is not exactly a round number because nature does not owe us one — the shift of $273.15$ was chosen precisely so that the resulting Kelvin scale starts at that measured intercept, with each kelvin the same size as one degree Celsius.

If real gases liquefy before reaching absolute zero, how do we know the intercept is right?

The extrapolation is the trick. You do not need to actually cool a gas anywhere near $-273\,^\circ\text{C}$ to find where its pressure-temperature line would hit zero; you only need enough data points at accessible temperatures to fix the straight line's slope and intercept accurately, then extend it on paper (or by calculation). Different dilute gases, measured this way, all give intercepts clustering around the same value, which is strong evidence that the intercept reflects something universal about temperature itself, not a quirk of any one gas.

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

Moles give a convenient bridge between the microscopic world of molecules and the macroscopic quantities you measure in a lab. One mole is defined as containing Avogadro's number of particles, so $n$ moles corresponds to a definite molecule count regardless of which gas you have, while the gas constant $R$ is defined to work with that same mole-based counting for any dilute gas. You could rewrite the equation using total molecule number $N$ instead of $n$ , replacing $R$ with Boltzmann's constant $k_B$ , but that version belongs to the kinetic theory treatment rather than this equation-of-state topic.

Does the ideal gas equation work for the air in a bicycle tyre or a pressure cooker?

Reasonably well, as a useful approximation, even though air is a mixture and tyre or cooker pressures are higher than the "low pressure" ideal gases prefer. For rough calculations — like estimating how much a tyre's pressure rises on a hot day, or how much a sealed cooker's internal pressure rises as it heats — $PV = nRT$ or the combined gas law gives answers close enough to be genuinely useful. For precise engineering work at high pressure, correction terms for real-gas behaviour become necessary, but that refinement is beyond this syllabus.