Ideal Gas Law

Overview

The ideal gas law combines all gas variables into a single equation, relating pressure, volume, temperature, and amount of gas. It assumes gas molecules have no volume and no intermolecular forces.

The Ideal Gas Equation

$$PV = nRT$$

Where:

• $P$ = pressure
• $V$ = volume
• $n$ = moles of gas
• $R$ = gas constant
• $T$ = temperature (Kelvin)

Gas Constant (R) Values

R Value	Units
8.314	J/(mol·K) or Pa·L/(mol·K)
0.0821	L·atm/(mol·K)
62.4	L·mmHg/(mol·K)
8.314	kPa·L/(mol·K)

Choose R based on your pressure and volume units!

Rearranged Forms

Find	Formula
Pressure	$P = \frac{nRT}{V}$
Volume	$V = \frac{nRT}{P}$
Moles	$n = \frac{PV}{RT}$
Temperature	$T = \frac{PV}{nR}$

Examples

Example 1: Finding Volume

What volume does 2.0 mol of gas occupy at 1.0 atm and 300 K?

$$V = \frac{nRT}{P} = \frac{(2.0 \text{ mol})(0.0821 \text{ L·atm/mol·K})(300 \text{ K})}{1.0 \text{ atm}} = 49.3 \text{ L}$$

Example 2: Finding Moles

How many moles of gas are in a 10.0 L container at 2.0 atm and 400 K?

$$n = \frac{PV}{RT} = \frac{(2.0 \text{ atm})(10.0 \text{ L})}{(0.0821 \text{ L·atm/mol·K})(400 \text{ K})} = 0.61 \text{ mol}$$

Example 3: Finding Pressure

What is the pressure of 0.50 mol of gas in a 5.0 L container at 25°C?

$T = 25 + 273 = 298$ K

$$P = \frac{nRT}{V} = \frac{(0.50 \text{ mol})(0.0821)(298 \text{ K})}{5.0 \text{ L}} = 2.4 \text{ atm}$$

Molar Mass from Ideal Gas Law

Since $n = \frac{m}{M}$ (mass/molar mass):

$$PV = \frac{m}{M}RT$$

Rearranging:

$$M = \frac{mRT}{PV}$$

Example

A 1.00 L container holds 1.80 g of gas at 1.00 atm and 273 K. Find the molar mass.

$$M = \frac{mRT}{PV} = \frac{(1.80 \text{ g})(0.0821)(273 \text{ K})}{(1.00 \text{ atm})(1.00 \text{ L})} = 40.3 \text{ g/mol}$$

Density from Ideal Gas Law

$$d = \frac{m}{V} = \frac{PM}{RT}$$

Example

Find the density of CO₂ at STP.

$M(\text{CO}_2) = 44.01$ g/mol, $T = 273$ K, $P = 1.00$ atm

$$d = \frac{PM}{RT} = \frac{(1.00 \text{ atm})(44.01 \text{ g/mol})}{(0.0821)(273 \text{ K})} = 1.96 \text{ g/L}$$

Stoichiometry with Gases

At the same T and P, equal volumes contain equal moles (Avogadro's Law).

Volume Ratios = Mole Ratios

$$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g)$$

2 L : 1 L : 2 L

Example: Gas Stoichiometry

How many liters of O₂ at STP are needed to burn 5.0 L of CH₄?

$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$ $$V(\text{O}_2) = 5.0 \text{ L} \times \frac{2 \text{ L O}_2}{1 \text{ L CH}_4} = 10.0 \text{ L}$$

Example: Mass to Volume

What volume of H₂ at STP is produced from 13.0 g of Zn?

$$\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$$ $$\text{mol Zn} = \frac{13.0 \text{ g}}{65.38 \text{ g/mol}} = 0.199 \text{ mol}$$ $$\text{mol H}_2 = 0.199 \text{ mol (1:1 ratio)}$$ $$V(\text{H}_2) = 0.199 \text{ mol} \times 22.4 \text{ L/mol} = 4.46 \text{ L}$$

Real Gases vs Ideal Gases

Ideal Gas Assumptions

1. Gas molecules have no volume
2. No intermolecular forces
3. Collisions are perfectly elastic

When Ideal Gas Law Works Best

• Low pressure (molecules far apart)
• High temperature (KE >> intermolecular forces)

When It Fails

• High pressure
• Low temperature
• Polar molecules
• Large molecules

Van der Waals Equation (Real Gases)

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

Where:

• $a$ = correction for intermolecular attraction
• $b$ = correction for molecular volume

Quick Reference

At STP (0°C, 1 atm)

$$1 \text{ mol of any ideal gas} = 22.4 \text{ L}$$

Useful Relationships

$$PV = nRT$$ $$d = \frac{PM}{RT}$$ $$M = \frac{dRT}{P} = \frac{mRT}{PV}$$ $$n = \frac{PV}{RT} = \frac{m}{M}$$