The Ideal Gas Equation

In Lesson 5 we saw that Boyle's, Charles's and the Pressure Law can be rolled into a single relationship for a fixed mass of ideal gas: pV/T = constant. In this lesson we take the final step and generalise that constant across different amounts of gas, arriving at the most important equation in the kinetic theory of gases: the ideal gas equation

pV = nRT

Module 5.1.3 of the OCR A-Level Physics A specification (H556) requires you to know, derive qualitatively, and apply this equation with fluency. This lesson develops it carefully and works through a range of numerical examples.

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Generalising the Combined Gas Law

If you double the amount of gas in a container without changing p or T, you need twice the volume. This is intuitive: gas molecules occupy space, and more molecules need more space. Therefore the constant in the combined gas law pV/T = const must be proportional to the amount of gas.

We define the amount of gas as the number of moles n, and write

pV / T = n R

where R is a universal constant — the molar gas constant — that does not depend on which gas we are considering. Rearranging:

pV = nRT

This is the ideal gas equation.

Note the structure:

• p is in pascals (Pa).
• V is in cubic metres (m³).
• n is in moles.
• R is the molar gas constant.
• T is in kelvin (K).

The units of the left-hand side are Pa × m³ = N m⁻² × m³ = N m = J. So pV has units of energy (joules), and nRT must too. This gives us the units of R: J mol⁻¹ K⁻¹. The numerical value is

R = 8.31 J mol⁻¹ K⁻¹  (OCR data sheet value)

(More precisely 8.314462618..., but 8.31 is what OCR uses.)

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The Mole

The mole is the SI unit for the amount of substance. It is defined (since 2019) so that one mole contains exactly N_A = 6.022 × 10²³ elementary entities, where N_A is the Avogadro constant. For our purposes:

• 1 mole of any gas contains 6.022 × 10²³ molecules.
• 1 mole of hydrogen (H_2) has a mass of 2.0 g.
• 1 mole of nitrogen (N_2) has a mass of 28.0 g.
• 1 mole of oxygen (O_2) has a mass of 32.0 g.
• 1 mole of air (a mixture, averaged) has a mass of about 29 g.

To convert between the number of moles n, the mass m and the molar mass M, use

n = m / M

where M is in kg mol⁻¹ (or g mol⁻¹, as long as the units match). For instance, 14 g of nitrogen is n = 14/28 = 0.50 mol.

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The Ideal Gas Model

An ideal gas is a theoretical construct in which:

1. The gas obeys pV = nRT exactly, at all p and T.
2. Intermolecular forces are zero (except during collisions).
3. The molecules themselves have zero volume.
4. Collisions are perfectly elastic.

No real gas is ideal in this sense, but all real gases approach ideal behaviour when they are dilute (low density, high molar volume) and hot (well above their condensation temperature). Air at room temperature and atmospheric pressure is very close to ideal — the error in pV = nRT is well under 1%.

Real gases deviate most noticeably:

• At high pressure, when the molecules are packed closely enough for their own volume to matter.
• At low temperature, when intermolecular attractions pull the molecules together and the gas begins to liquefy.

The van der Waals equation and more sophisticated models correct for these effects, but they lie beyond A-Level. For this course, assume pV = nRT holds unless told otherwise.

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Worked Example 1: Basic Use of pV = nRT

A sealed cylinder of volume V = 0.020 m³ contains n = 0.80 mol of nitrogen gas at T = 293 K. Find the pressure.

p = nRT / V
  = (0.80)(8.31)(293) / (0.020)
  = (0.80)(8.31)(293) / 0.020
  = 1947.9 / 0.020
  ≈ 9.7 × 10⁴ Pa

So the pressure is about 97 kPa, comparable with atmospheric.

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Worked Example 2: Finding the Amount

A steel container of volume V = 0.50 m³ contains helium at p = 2.5 × 10⁵ Pa and T = 300 K. How many moles of helium are present?

n = pV / (RT)
  = (2.5 × 10⁵)(0.50) / ((8.31)(300))
  = 125 000 / 2493
  ≈ 50.1 mol

The mass of helium is n × M = 50.1 × 0.004 = 0.20 kg (since helium has molar mass 4.0 g mol⁻¹).

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Worked Example 3: Volume of an Ideal Gas at STP

What volume does 1 mol of an ideal gas occupy at standard temperature and pressure (STP), taken as T = 273 K and p = 1.01 × 10⁵ Pa?

V = nRT / p
  = (1)(8.31)(273) / (1.01 × 10⁵)
  = 2268 / 101 000
  ≈ 0.0225 m³
  = 22.5 L

This value — 22.4 or 22.5 litres per mole at STP — is worth remembering. It gives you a sense of how much space a mole of gas occupies. A 1-litre bottle of air at STP contains about 1/22.4 ≈ 0.045 moles, or about 2.7 × 10²² molecules.

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Worked Example 4: A Car Tyre

A car tyre with internal volume V = 0.020 m³ is pumped up to an absolute pressure of p = 3.0 × 10⁵ Pa at a temperature of T = 290 K. How many moles of air are inside? If the molar mass of air is 0.029 kg mol⁻¹, what is the mass of air in the tyre?

n = pV / (RT)
  = (3.0 × 10⁵)(0.020) / ((8.31)(290))
  = 6000 / 2410
  ≈ 2.49 mol

m = n M = (2.49)(0.029) ≈ 0.072 kg = 72 g

So a standard car tyre contains about 72 g of air — roughly the mass of a large kiwi fruit.

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Worked Example 5: A Hot-Air Balloon

A hot-air balloon contains 1500 m³ of air at 100 °C (373 K) while the outside air is at 20 °C (293 K). If the pressure is the same inside and outside (1.01 × 10⁵ Pa), what is the mass of air inside the balloon? What is the mass of air that would occupy the same volume if it were at the outside temperature?

Both calculations use n = pV / (RT) followed by m = nM with M = 0.029 kg mol⁻¹.

Inside: