Thermodynamics
Ideal gases, thermodynamic transformations, 1st and 2nd law, heat engines, Carnot and Otto cycles, entropy. Every topic with step-by-step theory, interactive diagrams and solved exercises.
Complete Theory
4Zeroth Law of Thermodynamics: if two bodies A and B are each in thermal equilibrium with a third body C, then A and B are in thermal equilibrium with each other. This seemingly obvious principle is the logical foundation of temperature measurement: it allows us to use a thermometer (C) as an objective reference. Without it, we could not assert that two objects have the same temperature without direct contact.
Thermometric scales:
Real-world examples: mercury or alcohol thermometers exploit liquid expansion in a capillary. Electronic thermistors use resistance variation with . Gas thermometers at constant volume measure (Gay-Lussac's law) and are the most precise.
Connection: the Zeroth Law defines the concept of temperature, the fundamental variable for the ideal gas equation of state (Theory §2) and for all subsequent developments in thermodynamics.
Thermometric scales:
- Kelvin (K) — absolute scale: . Absolute zero (0 K = -273.15°C) is the theoretical temperature where all thermal motion ceases. Every other scale is linear with respect to this one.
- Celsius (°C) — = ice melting at 1 atm, = water boiling at 1 atm. A difference of 1°C = 1 K.
- Fahrenheit (°F) — , used in the USA. = ice melting, = water boiling.
Real-world examples: mercury or alcohol thermometers exploit liquid expansion in a capillary. Electronic thermistors use resistance variation with . Gas thermometers at constant volume measure (Gay-Lussac's law) and are the most precise.
Connection: the Zeroth Law defines the concept of temperature, the fundamental variable for the ideal gas equation of state (Theory §2) and for all subsequent developments in thermodynamics.
Empirical gas laws (17th-19th century):
Microscopic form: , with the number of molecules and J/K Boltzmann's constant. The two constants are related by , where mol⁻¹ is Avogadro's number.
Model limits: the ideal gas equation holds for rarefied gases at room temperature. At high pressures or temperatures near liquefaction, more accurate models such as Van der Waals (Theory §4) are required.
- Boyle (1662): at constant . If you compress a gas to half its volume, the pressure doubles. Explanation: molecules hit the walls with the same kinetic energy, but in a smaller volume collisions are more frequent.
- Charles (1787): at constant . Heating a gas at constant pressure causes the volume to increase proportionally to . Example: a hot-air balloon — hot air expands, becomes less dense than cold air, and rises.
- Gay-Lussac (1802): at constant . In a rigid container (pressure cooker), increasing temperature makes pressure rise linearly.
Microscopic form: , with the number of molecules and J/K Boltzmann's constant. The two constants are related by , where mol⁻¹ is Avogadro's number.
Model limits: the ideal gas equation holds for rarefied gases at room temperature. At high pressures or temperatures near liquefaction, more accurate models such as Van der Waals (Theory §4) are required.
Microscopic model: kinetic theory explains macroscopic properties (P, T) in terms of molecular motion. Assumptions: point-like molecules in random motion, elastic collisions with walls, no mutual interactions.
Pressure arises from the momentum change of molecules hitting the walls. Each collision transfers to the wall; integrating over all molecules yields .
Temperature is proportional to mean translational kinetic energy: . Higher T means faster molecules. At 300 K, N₂ molecules have m/s, faster than the speed of sound (343 m/s)!
Equipartition of energy: each quadratic degree of freedom contributes to internal energy.
Pressure arises from the momentum change of molecules hitting the walls. Each collision transfers to the wall; integrating over all molecules yields .
Temperature is proportional to mean translational kinetic energy: . Higher T means faster molecules. At 300 K, N₂ molecules have m/s, faster than the speed of sound (343 m/s)!
Equipartition of energy: each quadratic degree of freedom contributes to internal energy.
- Monatomic gas (He, Ar): 3 translational degrees of freedom → , .
- Diatomic gas (N₂, O₂, H₂): 3 translational + 2 rotational (at room T) → , . Vibrational degrees are excited only at high temperatures.
- What and are: the molar specific heats, i.e. the heat to raise 1 mole by 1 K. is at constant volume (the gas does not expand, does no work: all the heat goes into internal energy, ); is at constant pressure (the gas expands and does work , so more heat is needed).
- Mayer's relation: . It follows from the first law at constant : , hence . is always larger than (by exactly ) because at constant pressure you also pay the expansion work.
Why do we need a more accurate model? The ideal gas equation fails when:
Connection: below , Van der Waals isotherms describe the phase transition, connecting to the study of phase changes (Theory Transformations §4 — latent heat).
- Pressure is high — molecules are close enough to feel intermolecular forces.
- Temperature is low — kinetic energy is insufficient to overcome mutual attractions.
- Volume is small compared to the total molecular volume.
- — internal pressure term: attractive forces between molecules reduce the measured pressure compared to the ideal gas. is specific to each gas (large for polar molecules).
- — covolume: the actually available volume is because molecules have a finite volume. is about 4× the Avogadro molecular volume.
- — critical temperature, above which the gas cannot be liquefied by compression alone.
- — critical pressure.
- — critical molar volume.
Connection: below , Van der Waals isotherms describe the phase transition, connecting to the study of phase changes (Theory Transformations §4 — latent heat).
Worked Examples
2Example 1Isobaric process — A gas expanding at constant pressure
Given
mol, atm
K → K
Find
,
Work done by the gas
Step-by-step solution
1Identify the process type: this is an isobar, so = const = 1.5 atm. Use converting pressure to pascals: Pa. The initial volume is m³ = 32.85 L. Why convert to pascals? Because is in J/(mol·K) and 1 J = 1 Pa·m³, so volume and pressure must be in SI units.
2The second state: at constant pressure, Charles' law tells us , so L. Temperature increases by 50% (300 to 450 K), so the volume increases by exactly 50%. This makes physical sense: molecules have more kinetic energy and occupy more space if pressure does not change.
3Work calculation: in an isobar, work is . Alternatively, J. Both formulas are equivalent because from the equation of state. Work is positive because the gas expands and pushes against the surroundings. Verification: J ✓.
✓ Final result: L, L, kJ
Example 2RMS speed of nitrogen at 300 K — How fast are molecules?
Given
kg/mol
K
Find
Mean KE per molecule
Step-by-step solution
1RMS speed calculation: from kinetic theory, . Plugging values: m/s. Why do we use molar mass in kg/mol rather than molecular mass? Because is the constant per mole. Using and the mass of a single molecule would give the same result: . N₂ molecules at room temperature travel at 517 m/s = 1860 km/h — faster than an airliner!
2Mean kinetic energy: by the equipartition principle, each translational degree of freedom contributes . Nitrogen is diatomic but at 300 K only the 3 translational degrees contribute (rotational ones give rotational energy, not translational). So J per molecule. For a whole mole: J/mol.
3Physical comparison: 517 m/s is greater than the speed of sound in air (343 m/s at 20°C). The speed of sound is m/s. The ratio , confirmed by our numbers: 517/353 ≈ 1.46.
✓ Final result: m/s; J
Exercises with Solutions
2Exercise 1Compound process (isobaric + isochoric)Hard
Problem to solve
A monatomic gas ( mol, ) undergoes: A( K, atm) →isobaric→ B( K) →isochoric→ C( atm). Find total work and total internal energy change .
Given data
n=3 molC_v=3R/2P_A = P_B = 2 atm (isobar)V_B = V_C (isochore)
Step-by-step solution
1State A: m³.
2Leg AB (isobar): m³. Work: J.
3State C: from the isochore , so K.
4Leg BC (isochore): because .
5Internal energy changes: J. J.
6Verification: J. J ≈ 41.2 kJ. Note: depends only on , as for any ideal gas.
✓ Final answer: J; kJ
Exercise 2Dalton's law — Gas mixturesHard
Problem to solve
A container of volume L at K holds mol of O₂ and mol of N₂. Find total pressure, partial pressures, and mole fractions.
Given data
V=0.010 m³T=300 Kn_O₂=0.5 moln_N₂=1.2 mol
Step-by-step solution
1Total pressure: mol. Pa = 4.19 atm. Conversion: 1 atm = 101325 Pa, so atm.
2Partial pressures (Dalton's law): where . , . atm, atm.
3Verification: atm ≈ 4.19 atm (rounding). In Pa: Pa = 1.231 atm; Pa = 2.954 atm. Sum: 124710 + 299304 = 424014 Pa = 4.186 atm ✓. Dalton's law holds because ideal gases do not interact with each other.
✓ Final answer: atm; , atm
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Integrative Problems
Problems combining all chapters — exam levelProblem 1The Thermoelectric Power Plant: from Gas to Molecules to EntropyEXTREME
A thermoelectric power plant uses of a diatomic gas (, ) running through the following cycle on a PV diagram:
State A: , . A→B: adiabatic compression to (compression ratio ). B→C: isochoric, heating to (combustion). C→D: adiabatic expansion to (return to original volume). D→A: isochoric, cooling (Otto cycle).
State A: , . A→B: adiabatic compression to (compression ratio ). B→C: isochoric, heating to (combustion). C→D: adiabatic expansion to (return to original volume). D→A: isochoric, cooling (Otto cycle).
📌 Problem data
(a)Ideal Gas — Thermodynamic States(b)First Law — Work and Heat for Each Process(c)Cycles — Otto vs Carnot Efficiency(d)Kinetic Theory — Molecules in Motion(e)Entropy — Second Law and Global Balance