The Ideal Gas Equation

Learning Objectives (OCR Spec 2.1.3)

• State and apply the ideal gas equation pV = nRT
• Convert between SI units for pressure, volume and temperature
• Calculate moles, molar mass or volume of a gas from conditions
• Understand the limitations of the ideal gas model
• Apply the equation to gases produced in chemical reactions
• Explain deviations of real gases from ideal behaviour

A note for OCR students: The ideal gas equation is a distinctive feature of the OCR Chemistry A specification at this early stage — AQA uses molar gas volume at RTP (24 dm³ mol⁻¹) instead. At A-Level, OCR expects confident use of pV = nRT in SI units. This is a small but important difference between OCR and AQA syllabi: OCR students gain a flexible calculation tool that works at any temperature and pressure, while AQA students use a simpler lookup at one set of conditions.

Historical Background

The ideal gas equation is a combination of three gas laws discovered in the 17th–19th centuries:

• Boyle's law (1662): at constant T, pV = constant
• Charles's law (1787): at constant p, V/T = constant
• Avogadro's law (1811): at constant p and T, V/n = constant

Combining these gives pV = nRT, where R is the universal gas constant. This combined law was first stated by Benoît Clapeyron in 1834 and became the foundation of physical chemistry.

The Ideal Gas Equation

$pV = nRT$pV=nRT

Where:

Symbol	Quantity	SI Unit
p	pressure	Pa (pascals)
V	volume	m³ (cubic metres)
n	amount of substance	mol
R	gas constant	8.314 J K⁻¹ mol⁻¹
T	temperature	K (kelvin)

Critical: you must use SI units for the equation to give the correct answer. A mismatched unit is the single biggest source of errors in these problems.

The gas constant R has the value 8.314 J K⁻¹ mol⁻¹ (or equivalently 8.314 N m K⁻¹ mol⁻¹, since 1 J = 1 N m). It is universal — the same for any ideal gas. R is provided on the OCR data sheet; you do not need to memorise its value, but you must recognise it.

Unit Conversions

Pressure

Unit	Conversion to Pa
1 kPa	1000 Pa
1 MPa	1 000 000 Pa
1 atm (standard atmosphere)	101 325 Pa (often rounded to 101 000 Pa)
1 bar	100 000 Pa
1 mmHg (torr)	133.3 Pa

At sea level atmospheric pressure is approximately 101 kPa = 101 000 Pa. When a question says "atmospheric pressure" with no further qualification, use 101 000 Pa or 101 325 Pa.

Volume

Unit	Conversion to m³
1 dm³	10⁻³ m³ (÷ 1000)
1 cm³	10⁻⁶ m³ (÷ 1 000 000)
1 L	10⁻³ m³
1 mL	10⁻⁶ m³

Temperature

Temperature must be in kelvin, not celsius:

$T(\text{K}) = \theta(°\text{C}) + 273$T(K)=θ(°C)+273

• 0 °C = 273 K
• 20 °C = 293 K
• 25 °C = 298 K (standard)
• 100 °C = 373 K

OCR usually writes exact conversion as +273.15, but +273 is acceptable at A-Level.

Rearranging the Equation

The ideal gas equation can be rearranged for any variable:

$n = \frac{pV}{RT} \quad V = \frac{nRT}{p} \quad T = \frac{pV}{nR} \quad p = \frac{nRT}{V}$n=RTpV​V=pnRT​T=nRpV​p=VnRT​

And combined with n = m/M we can also find molar mass directly:

$M = \frac{mRT}{pV}$M=pVmRT​

This is derived by substituting n = m/M into pV = nRT:

$pV = \frac{m}{M}RT \Rightarrow M = \frac{mRT}{pV}$pV=Mm​RT⇒M=pVmRT​

Worked Example 1: Finding Moles of a Gas

What amount (in moles) of oxygen gas occupies 2.50 dm³ at 25.0 °C and 100 kPa?

Step 1: Convert to SI units

• V = 2.50 × 10⁻³ m³
• p = 100 × 10³ Pa = 1.00 × 10⁵ Pa
• T = 25.0 + 273 = 298 K

Step 2: Rearrange and substitute

$n = \frac{pV}{RT} = \frac{(1.00 \times 10^5)(2.50 \times 10^{-3})}{(8.314)(298)}$n=RTpV​=(8.314)(298)(1.00×105)(2.50×10−3)​

$n = \frac{250}{2477.6} = 0.1009 \text{ mol}$n=2477.6250​=0.1009 mol

Answer: n ≈ 0.101 mol

Worked Example 2: Finding Volume

What volume does 0.250 mol of nitrogen occupy at 101 kPa and 20.0 °C?

• p = 101 000 Pa
• n = 0.250 mol
• T = 293 K

$V = \frac{nRT}{p} = \frac{(0.250)(8.314)(293)}{101000}$V=pnRT​=101000(0.250)(8.314)(293)​

$V = \frac{608.9}{101000} = 6.03 \times 10^{-3} \text{ m}^3$V=101000608.9​=6.03×10−3 m3

Converting to dm³: V = 6.03 dm³

Worked Example 3: Finding Molar Mass of an Unknown Gas

A 0.500 g sample of an unknown gas occupies 250 cm³ at 101 000 Pa and 100 °C. Calculate the molar mass.

Step 1: SI units

• m = 0.500 g
• V = 250 × 10⁻⁶ = 2.50 × 10⁻⁴ m³
• p = 101 000 Pa
• T = 100 + 273 = 373 K

Step 2: Find moles

$n = \frac{pV}{RT} = \frac{(101000)(2.50 \times 10^{-4})}{(8.314)(373)}$n=RTpV​=(8.314)(373)(101000)(2.50×10−4)​

$n = \frac{25.25}{3101.1} = 8.143 \times 10^{-3} \text{ mol}$n=3101.125.25​=8.143×10−3 mol

Step 3: Find molar mass

$M = \frac{m}{n} = \frac{0.500}{8.143 \times 10^{-3}} = 61.4 \text{ g mol}^{-1}$M=nm​=8.143×10−30.500​=61.4 g mol−1

Worked Example 4: Finding Mr and Identifying the Gas

A sealed 1.00 dm³ flask at 298 K and 101 000 Pa contains 1.80 g of a gas. Identify the gas.

• V = 1.00 × 10⁻³ m³
• T = 298 K
• p = 101 000 Pa
• n = pV/RT = (101 000 × 1.00 × 10⁻³) / (8.314 × 298) = 101 / 2477.6 = 0.04077 mol
• M = m/n = 1.80 / 0.04077 = 44.1 g mol⁻¹

Mr(CO₂) = 44.0 → the gas is carbon dioxide.

Common gases with Mr values worth recognising: