2aLaws of Thermodynamics
Master the four laws of thermodynamics: internal energy, the First Law (ΔU = Q − W), entropy, and the Second and Third Laws.
The laws of thermodynamics are among the most fundamental in all of science. They govern every energy conversion — from car engines to power plants to biological metabolism — and set absolute limits on what is physically possible.
Why can't a perfect engine convert all heat into work — and why does disorder in the universe always increase?
Lesson Overview
The laws of thermodynamics govern all energy transformations. The Zeroth Law defines thermal equilibrium. The First Law (conservation of energy) states ΔU = Q − W: the change in internal energy equals heat added minus work done by the system. The Second Law states that heat flows spontaneously from hot to cold and that entropy of an isolated system never decreases. The Third Law states that the entropy of a perfect crystal approaches zero as temperature approaches absolute zero. Together these laws set fundamental limits on engines, refrigerators, and all natural processes.
Key Equations
Worked Examples
A gas absorbs Q = 500 J of heat and does W = 200 J of work on its surroundings. Find the change in internal energy ΔU.
A gas is compressed by an external force doing W = 400 J of work on the gas. The gas releases Q = 150 J of heat to the surroundings. Find ΔU.
1 mol of an ideal monatomic gas is at T = 300 K. Find its internal energy U. (R = 8.314 J/mol·K)
500 J of heat flows from a hot reservoir at T = 500 K to a cold reservoir at T = 250 K. Calculate the entropy change of each reservoir and the total entropy change of the universe.
A gas expands at constant pressure P = 2 × 10⁵ Pa from V₁ = 0.01 m³ to V₂ = 0.03 m³. It absorbs Q = 8000 J. Find (a) the work done by the gas, (b) ΔU.
Guided Problems
A system absorbs Q = 1200 J of heat and its internal energy increases by ΔU = 800 J. How much work did the system do?
Hint: Use ΔU = Q − W, rearranged as W = Q − ΔU.
A gas is compressed isothermally (constant temperature). Its internal energy does not change (ΔU = 0). If W = −300 J (work done on gas), what is Q?
Hint: If ΔU = 0, then Q = W. Determine the sign: if work is done ON the gas (W negative for gas), heat must leave the system.
200 J of heat flows reversibly into a system at T = 400 K. Calculate the entropy change of the system.
Hint: Use ΔS = Q_rev/T. Make sure T is in Kelvin.
Classify each process as consistent or inconsistent with the Second Law: (a) Heat spontaneously flows from a cold object to a hot object. (b) A gas expands into a vacuum and its entropy increases. (c) All heat from a reservoir is converted to work with no other effect.
Hint: The Second Law: heat flows hot → cold spontaneously; entropy of isolated system never decreases; no process is 100% efficient at converting heat to work.
2 mol of ideal monatomic gas is heated from T₁ = 200 K to T₂ = 500 K at constant volume. Find ΔU and Q. (No work is done at constant volume.)
Hint: ΔU = (3/2)nRΔT. At constant volume W = 0, so Q = ΔU.
Key Vocabulary
First Law of Thermodynamics
Energy conservation applied to thermodynamic systems: ΔU = Q − W. The change in internal energy equals heat added to the system minus work done by the system.
Example: A gas absorbing 500 J and doing 200 J of work gains ΔU = 300 J of internal energy.
Internal Energy (U)
The total microscopic kinetic and potential energy of all particles in a system. For an ideal monatomic gas: U = (3/2)nRT. It is a state function — depends only on the current state, not the path.
Example: 1 mol of ideal gas at 300 K has U = (3/2)(1)(8.314)(300) ≈ 3741 J.
Second Law of Thermodynamics
Heat flows spontaneously from hot to cold, never the reverse. Equivalently: the total entropy of an isolated system never decreases (ΔS_universe ≥ 0). No heat engine can be 100% efficient.
Example: A hot coffee cup cools in a room — heat flows from coffee to air. The reverse never happens spontaneously.
Entropy (S)
A measure of the disorder or number of microstates of a system. For a reversible process: ΔS = Q_rev/T. Entropy increases in irreversible processes.
Example: When ice melts at 273 K, absorbing 334 J/g, the entropy increases: ΔS = Q/T = 334/273 ≈ 1.22 J/(g·K).
Third Law of Thermodynamics
The entropy of a perfect crystal at absolute zero (0 K) is exactly zero. It is impossible to reach absolute zero in a finite number of steps.
Example: As diamond is cooled toward 0 K, its entropy approaches zero because there is only one possible microstate — perfect order.
State Function
A thermodynamic property whose value depends only on the current state of the system, not on the path taken to reach that state. Internal energy U and entropy S are state functions; heat Q and work W are not.
Example: Whether a gas is compressed quickly or slowly to the same final state, ΔU is the same — but Q and W individually depend on the path.
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
A gas absorbs Q = 600 J and does W = 250 J of work. What is ΔU?
Which statement correctly expresses the Second Law of Thermodynamics?
200 J of heat flows reversibly into a system at T = 500 K. What is the entropy change?
For an ideal gas at constant volume, which is true?
Which law states that the entropy of a perfect crystal at 0 K equals zero?
Independent Practice
A gas absorbs Q = 2000 J and its internal energy increases by ΔU = 1400 J. How much work did the gas do on its surroundings?
A system does W = 500 J of work and releases Q = 300 J of heat. Find ΔU. Did the internal energy increase or decrease?
1 mol of ideal monatomic gas is heated from 300 K to 600 K at constant volume. Find ΔU and Q. (R = 8.314 J/mol·K)
Heat Q = 1000 J flows from a reservoir at 800 K to one at 400 K. Calculate ΔS for each reservoir and ΔS_universe. Does this satisfy the Second Law?
★ A Carnot engine operates between T_H = 600 K and T_C = 300 K. It absorbs Q_H = 1000 J per cycle. (a) Find the efficiency e = 1 − T_C/T_H. (b) Find the work output W. (c) Find the heat rejected Q_C. (d) Find ΔS_universe per cycle.
ChallengeCommon Mistakes
Using the sign convention W = work done ON the system (chemistry convention) instead of W = work done BY the system
In physics, the First Law is ΔU = Q − W where W is work done BY the system. In chemistry, ΔU = Q + W where W is work done ON the system. Always clarify which convention is used.
Thinking the Second Law says entropy always increases everywhere
Entropy can decrease locally (e.g., in a refrigerator or living organism) as long as the total entropy of the universe (system + surroundings) does not decrease.
Treating heat Q and work W as state functions
Q and W are path functions — they depend on how a process occurs. Only ΔU (and ΔS) are state functions that depend only on initial and final states.
Forgetting that ΔU = 0 for an isothermal process of an ideal gas
For an ideal gas, U depends only on T. If T is constant (isothermal), ΔU = 0, so Q = W — all heat absorbed is converted to work.
Math Tips
First Law: ΔU = Q − W. Q > 0 means heat flows INTO the system. W > 0 means the system does work ON surroundings (gas expands).
Work by gas: W = PΔV. Positive when gas expands (ΔV > 0); negative when compressed (ΔV < 0).
Entropy change: ΔS = Q_rev/T (T in Kelvin). For the universe: ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0.
Ideal monatomic gas: U = (3/2)nRT. Diatomic gas: U = (5/2)nRT. These give ΔU = (3/2)nRΔT or (5/2)nRΔT.