Ideal Gas Law — Physica

Equivalent forms

P = \frac{nRT}{V}

PV = Nk_BT

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

Four macroscopic quantities linked by a single constant — the bridge between the microscopic world of molecules and the macroscopic world of engines.

Symbol	Name	SI unit	Dimension	Range
$P$	Pressureoutput Absolute pressure of the gas	Pa	M·L⁻¹·T⁻²	1000 – 1000000
$V$	Volume Volume occupied by the gas	m³	L³	0.001 – 1
$n$	Amount of substance Number of moles of gas	mol	N	0.01 – 10
$T$	Temperature Absolute temperature of the gas	K	Θ	100 – 1000

Unit systems

Symbol	SI	CGS	Imperial
$P$	Pa	dyn/cm²	psi
$V$	m³	cm³	ft³
$n$	mol	mol	mol
$T$	K	K	°R

Where it holds

Valid for dilute gases far from condensation. Breaks down at high pressures or low temperatures where intermolecular forces and molecular volume matter (use van der Waals equation instead).

Dimensional analysis

$[P]\cdot [V] = [n]\cdot [R]\cdot [T] \to (M\cdot L^{-1}\cdot T^{-2})(L^{3}) = (mol)(J\cdot mol^{-1}\cdot K^{-1})(K) \to M\cdot L^{2}\cdot T^{-2} = M\cdot L^{2}\cdot T^{-2} \checkmark$

Discovery

Émile Clapeyron · 1834

Clapeyron unified Boyle's, Charles's, and Avogadro's laws into a single equation. The constant R was later determined precisely by experiment.

Try this

Why does a balloon pop when you leave it in a hot car?

A sealed balloon contains 0.05 mol of air at 20°C and 101.3 kPa. Find the new pressure if the temperature rises to 60°C.

Research status: stable

Real-world applications

Weather balloon altitude prediction — volume increases as atmospheric pressure drops
Scuba diving tank pressure calculations
Internal combustion engine cylinder analysis
Designing HVAC systems and pneumatic machinery

Common misconceptions

The ideal gas law is not a single discovery but a synthesis of Boyle's, Charles's, and Avogadro's laws
T must be absolute temperature (Kelvin), not Celsius — using $^{\circ}C$ gives wrong results
Real gases deviate significantly at high pressure or low temperature; the law is an idealization, not universal

Experimental verification

Verified by Regnault (1847) using precision gas thermometry. Modern confirmation via pVT measurements on dilute gases (He, Ar) at low pressure, where intermolecular forces vanish. Agreement to <0.1% for pressures below 1 atm.

Derivation

Combines three empirical laws: Boyle's law

(PV =

const at fixed T), Charles's law

(V/T =

const at fixed P), and Avogadro's law

(V/n =

const at fixed P,T).

Merging:

PV/nT =

const

\equiv R

.

The constant

R = 8.314\,\mathrm{J}/(mol\cdot K)

is determined experimentally.

Limiting cases

V \to \infty

⟶

P \to 0

Gas expands into infinite volume, pressure vanishes.

T \to 0

⟶

P \to 0

At absolute zero, ideal gas pressure drops to zero (real gases liquefy first).

n \to 0

⟶

PV \to 0

No gas means no pressure-volume product.

What if…

What if you double the temperature while keeping volume fixed?

Pressure doubles. At constant V and n, $P \propto T$ . Doubling T from 300 K to 600 K doubles P — this is why sealed containers can burst when heated.

What if you halve the volume at constant temperature?

Pressure doubles (Boyle's law). $P \propto 1/V$ at constant T and n. This is the principle behind syringes and hydraulic presses.

What if you use the ideal gas law for water vapor near

100^{\circ}C

?

It works reasonably well at low pressures (< 0.5 atm), but fails near the boiling point at 1 atm because intermolecular hydrogen bonds cause large deviations. Use the van der Waals equation instead.

1

Hot-car balloon pressure

Given ·

n:: 0.05
T i:: 293
T f:: 333
P i:: 101300

Find · P_f

Steps

At constant n and V, $PV = nRT$ gives $P_{1}/T_{1} = P_{2}/T_{2}$
Convert temperatures: $T_i = 20 + 273 = 293\,\mathrm{K}$ , $T_f = 60 + 273 = 333\,\mathrm{K}$
$P_f = 101300 \times (333 / 293) = 115100\,\mathrm{Pa}$

Result ·

P_f = P_i \times (T_f / T_i) = 101300 \times (333 / 293) = 115100\,\mathrm{Pa} \approx 115.1\,\mathrm{kPa}

2

Volume of 1 mol at STP

Given ·

n:: 1
T:: 273.15
P:: 101325

Find · V

Steps

$Use PV = nRT$ rearranged to $V = nRT/P$
$V = (1)(8.314)(273.15) / 101325$
$V = 2271.0 / 101325 = 0.02241\,\mathrm{m}^{3} \approx 22.41\,\mathrm{L}$

Result ·

V = nRT/P = 1 \times 8.314 \times 273.15 / 101325 = 0.02241\,\mathrm{m}^{3} = 22.41\,\mathrm{L}

First Law of Thermodynamics→Maxwell-Boltzmann Speed Distribution→Carnot Efficiency→