Ideal Gas Law

Definition and Overview

Concept

Ideal Gas Law: equation of state for ideal gases. Relates pressure (P), volume (V), temperature (T), and moles (n). Assumes gas particles have negligible volume and no intermolecular forces.

Significance

Enables prediction of gas behavior under varying conditions. Foundation for thermodynamics and physical chemistry calculations.

Scope

Applicable to idealized gases. Approximates real gases at low pressure, high temperature.

Historical Background

Boyle’s Law (1662)

Pressure inversely proportional to volume at constant temperature: P ∝ 1/V.

Charles’s Law (1787)

Volume directly proportional to temperature at constant pressure: V ∝ T.

Avogadro’s Hypothesis (1811)

Equal volumes of gases contain equal moles at same T, P: V ∝ n.

Ideal Gas Law Formation

Combines Boyle’s, Charles’s, Avogadro’s laws into PV = nRT.

Mathematical Formulation

Basic Equation

PV = nRT

Parameters

P: pressure (Pa, atm), V: volume (m³, L), n: moles (mol), R: gas constant, T: temperature (K).

Alternate Forms

Expressed via density, molar volume, or using different units with appropriate R values.

Variables and Constants

Pressure (P)

Force per unit area exerted by gas particles on container walls. Units: Pascal (Pa), atmosphere (atm), torr.

Volume (V)

Space occupied by gas. Units: cubic meters (m³), liters (L).

Temperature (T)

Measure of average kinetic energy of gas particles. Must be absolute (Kelvin).

Amount of Substance (n)

Number of moles representing quantity of gas particles. 1 mol = 6.022×10²³ particles.

Universal Gas Constant (R)

Proportionality constant. Value depends on units used.

Derivation of the Ideal Gas Law

From Empirical Gas Laws

Combines Boyle’s (P ∝ 1/V), Charles’s (V ∝ T), Avogadro’s (V ∝ n) laws.

From Kinetic Molecular Theory

Gas particles: point masses, elastic collisions, no intermolecular forces. Derives PV = nRT via particle momentum transfer.

Mathematical Steps

P ∝ 1/V (Boyle)V ∝ T (Charles)V ∝ n (Avogadro)Combining:V ∝ nT/PIntroducing R:PV = nRT 

Applications

Calculating Gas Properties

Determine unknown P, V, T, or n given other variables.

Stoichiometry

Relates gas volume to moles for chemical reactions at standard conditions.

Engineering

Design of engines, compressors, HVAC systems.

Atmospheric Science

Modeling air behavior, weather prediction.

Laboratory Use

Gas collection, calibration of instruments.

Limitations and Assumptions

Idealization

Assumes zero molecular volume, no intermolecular forces, elastic collisions.

Real Gas Deviations

At high pressure, low temperature, gases deviate due to molecular size and attractions.

Phase Change Exclusion

Not valid near condensation or critical points.

Non-ideal Gas Models

Van der Waals, Redlich-Kwong equations correct for interactions.

Related Gas Laws

Boyle’s Law

P₁V₁ = P₂V₂ at constant T.

Charles’s Law

V₁/T₁ = V₂/T₂ at constant P.

Gay-Lussac’s Law

P₁/T₁ = P₂/T₂ at constant V.

Avogadro’s Law

V₁/n₁ = V₂/n₂ at constant T, P.

Combined Gas Law

(P₁V₁)/T₁ = (P₂V₂)/T₂ for fixed n.

Thermodynamic Implications

Equation of State

Defines state variables for ideal gases.

Internal Energy

Dependent only on temperature, not volume or pressure.

Heat Capacities

Constant volume (C_V) and constant pressure (C_P) related by gas constant R.

Entropy and Enthalpy

Calculated using ideal gas assumptions for thermodynamic cycles.

Experimental Verification

Historical Experiments

Boyle, Charles, Gay-Lussac’s measurements of P, V, T relationships.

Modern Techniques

Mass spectrometry, manometry, gas chromatography validate ideal gas behavior.

Data Consistency

Agreement with law at low pressures, high temperatures.

Deviations Documented

Critical for developing real gas models.

Sample Problems

Problem 1: Calculate Volume

Given 1 mol O₂ at 1 atm, 273 K, find volume.

P = 1 atmn = 1 molT = 273 KR = 0.08206 atm·L/mol·KV = nRT/P = (1)(0.08206)(273)/1 = 22.414 L 

Problem 2: Calculate Pressure

Gas in 10 L container, 2 mol, 300 K.

V = 10 Ln = 2 molT = 300 KR = 0.08206 atm·L/mol·KP = nRT/V = (2)(0.08206)(300)/10 = 4.924 atm 

Problem 3: Temperature Change

Initial: 1 atm, 10 L, 300 K. Volume constant, pressure doubles. Find new temperature.

P₁ = 1 atm, T₁ = 300 KP₂ = 2 atm, V constantT₂ = P₂T₁/P₁ = (2)(300)/1 = 600 K 

Tables and Constants

Gas Constant Values

Standard Temperature and Pressure (STP)

References

• Atkins, P., Physical Chemistry, 10th ed., Oxford University Press, 2014, pp. 120-145.
• Smith, J.M., Van Ness, H.C., Abbott, M.M., Introduction to Chemical Engineering Thermodynamics, 7th ed., McGraw-Hill, 2005, pp. 90-110.
• Laidler, K.J., Meiser, J.H., Physical Chemistry, 3rd ed., Benjamin/Cummings, 1999, pp. 210-235.
• McQuarrie, D.A., Statistical Mechanics, University Science Books, 2000, pp. 50-75.
• Reid, R.C., Prausnitz, J.M., Poling, B.E., The Properties of Gases and Liquids, 4th ed., McGraw-Hill, 1987, pp. 10-45.