Class 11 Physics · NCERT chapter 12 · 7 topics · free to read
Thermal Equilibrium & Zeroth Law
Thermal equilibrium means no net heat flow between two bodies. The zeroth law establishes temperature as a measurable quantity: if A and B are each in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. NEET asks conceptual questions on the zeroth law and thermodynamic variables.
PV = nRT (equation of state for ideal gas)
Thermodynamic System & Surroundings
A thermodynamic system is whatever part of the universe we choose to study. Everything else is the surroundings. The boundary separating them can be real (piston wall) or imaginary.
In thermodynamics, the first step is always: define your system. The system is the part of the universe you're interested in — a gas in a cylinder, water in a beaker, or air in a balloon. Everything outside the system boundary is the surroundings.
Open system: Exchanges both energy AND matter with surroundings (e.g., boiling water in an open pot — steam escapes)
Closed system: Exchanges energy but NOT matter (e.g., gas in a sealed piston — heat can enter, gas cannot leave)
Isolated system: Exchanges NEITHER energy NOR matter (e.g., ideal thermos flask)
NEET thermodynamics almost always deals with closed systems (gas in a cylinder with a movable piston)
Thermal Equilibrium
Two bodies in thermal contact are in thermal equilibrium when they reach the same temperature and no net heat flows between them. This simple idea is the foundation of all temperature measurement.
Put a cold spoon into a hot cup of chai. Heat flows from the chai to the spoon. The chai cools slightly, the spoon warms up. Eventually, both reach the same temperature — net heat flow = zero. This is thermal equilibrium.
Thermal equilibrium requires thermal contact — a conducting path for heat flow
At thermal equilibrium: temperature of system = temperature of surroundings
Heat always flows from higher temperature to lower temperature — never the reverse (this foreshadows the second law!)
An adiabatic wall prevents heat flow; a diathermic wall allows it
Quick recall
Two bodies A and B are separated by an adiabatic wall. Are they in thermal equilibrium?
Not necessarily. An adiabatic wall prevents heat exchange, so A and B could be at different temperatures. Thermal equilibrium requires that they CAN exchange heat and the net flow is zero. If the wall is changed to diathermic and still no heat flows, THEN they are in thermal equilibrium.
Zeroth Law of Thermodynamics
If body A is in thermal equilibrium with body C, and body B is in thermal equilibrium with body C, then A and B are in thermal equilibrium with each other. This law justifies the concept of temperature and makes measurement possible.
The Zeroth Law states: if system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other.
T_A = T_C and T_B = T_C T_A = T_B
The zeroth law was formulated AFTER the first and second laws but is logically more fundamental — hence 'zeroth'
It introduces temperature as a state variable — the property that is the same for all bodies in thermal equilibrium
Without the zeroth law, temperature would be undefined and thermometers would be meaningless
Named the 'zeroth' law by Fowler (1931) because it underpins the first and second laws
Thermodynamic Variables & Equation of State
A thermodynamic system in equilibrium is fully described by a few macroscopic variables (P, V, T). The equation of state connects them — for an ideal gas, it's PV = nRT.
A gas in equilibrium is fully described by just three macroscopic variables: pressure P , volume V , and temperature T . But these three are not independent — an equation connects them. This is the equation of state.
PV = nRT (ideal gas equation of state)
Here n = number of moles, R = 8.314 J mol ^ -1 K ^ -1 (universal gas constant). Given any two of P , V , T , the third is fixed. So an ideal gas has only two independent state variables.
Extensive variables depend on system size: volume V , internal energy U , entropy S , mass m
Intensive variables do NOT depend on system size: pressure P , temperature T , density
Quick test: split the system in half. If the variable halves → extensive. If it stays the same → intensive
The equation of state only applies in equilibrium. A gas in turbulent flow doesn't have a single well-defined P or T
First Law of Thermodynamics
The first law is energy conservation applied to thermodynamic systems: the heat supplied to a system equals the increase in internal energy plus the work done BY the system ( U = Q - W ). Internal energy is a state function; heat and work are path functions. NEET heavily tests sign conventions and applying the first law to different processes.
U = Q - W
Internal Energy
Internal energy U is the total kinetic + potential energy of all molecules in the system. It depends only on the state (T, P, V) — not on how the system got there. For an ideal gas, U depends only on temperature.
Every gas has molecules zooming around, colliding, rotating, vibrating. The sum total of all this microscopic kinetic energy plus intermolecular potential energy is the internal energy U . You can't see it or measure it directly — but you can measure changes in it.
U is a state function — depends only on the current state ( T , P , V ), not on the path taken
For an ideal gas: U depends ONLY on temperature. U = nC_vT (for n moles). If T doesn't change, U doesn't change
U in a cyclic process = 0 (system returns to the same state, so U returns to its original value)
Internal energy includes translational, rotational, and vibrational KE of molecules + intermolecular PE
Heat & Work
Heat (Q) and work (W) are both modes of energy transfer — but neither is 'stored' in the system. They are path functions: their values depend on HOW the process is carried out, not just the initial and final states.
Heat Q is energy transferred due to a temperature difference. Work W is energy transferred by a force acting through a displacement (like a piston moving). Both are transient — they exist only during the process. A system doesn't 'contain' heat or work; it contains internal energy.
Depends only on current state (P, V, T)
Depends on the path/process taken
Has a definite value at any equilibrium state
Only defined during a process, not at a state
U between two states is unique
Q and W between the same two states can differ
Example: Internal energy, entropy
Example: Heat, work
dU is an exact differential
Q and W are inexact differentials
Heat and work are measured in joules (SI) or calories ( 1 cal = 4.186 J)
Q and W are path functions — they don't have exact differentials ( Q , W , not dQ , dW )
A system can go from state A to state B via infinitely many paths — each path gives different Q and W , but the SAME U
This is precisely why the first law is powerful: U = Q - W links a state function to two path functions
The First Law
The first law says: heat supplied to a system goes partly into increasing internal energy and partly into doing work. ΔU = Q - W. The sign convention is critical and is a NEET favorite.
The First Law of Thermodynamics: Energy can neither be created nor destroyed — it can only be transformed. For a thermodynamic system:
U = Q - W
Where U = change in internal energy, Q = heat supplied TO the system, W = work done BY the system.
Sign convention (NCERT): Q > 0 → heat absorbed BY system; Q < 0 → heat released by system
Sign convention (NCERT): W > 0 → work done BY system (expansion); W < 0 → work done ON system (compression)
Alternative form: Q = U + W (heat supplied = increase in internal energy + work done by system)
For a cyclic process: U = 0 → Q = W (net heat absorbed = net work done)
Quick recall
A gas absorbs 200 J of heat and does 150 J of work. What is U ?
U = Q - W = 200 - 150 = 50 J . The internal energy increases by 50 J — the remaining 150 J of absorbed heat went into doing work on the surroundings.
Applying the First Law
By setting Q = 0, W = 0, or ΔU = 0, the first law immediately gives powerful results for adiabatic, isochoric, and isothermal processes.
The first law U = Q - W becomes extremely simple when one of the three quantities is zero:
Adiabatic ( Q = 0 ): U = -W . Work done BY the gas reduces its internal energy (gas cools during expansion)
Isochoric ( V = 0 W = 0 ): U = Q . All heat goes into increasing internal energy
Isothermal ideal gas ( T = 0 U = 0 ): Q = W . All heat absorbed is converted to work
Cyclic process (returns to initial state → U = 0 ): Q_ net = W_ net
Quick recall
In a cyclic process, a gas does 500 J of net work. How much net heat was absorbed?
In a cyclic process, U = 0 so Q_ net = W_ net = 500 J . The net heat absorbed equals the net work done — energy is perfectly converted from heat to work over each complete cycle.
Thermodynamic Processes
Four key processes — isothermal ( T = 0 ), adiabatic ( Q = 0 ), isochoric ( V = 0 ), and isobaric ( P = 0 ) — each have distinct work formulas and PV curve shapes. NEET frequently tests work done, heat exchanged, and internal energy change in each process. Knowing the differences between isothermal and adiabatic is especially high-yield.
A quasi-static process proceeds so slowly that the system remains in thermodynamic equilibrium at every instant. Only quasi-static processes can be plotted on a PV diagram. All NEET thermodynamic calculations assume quasi-static processes.
Imagine compressing a gas by pushing a piston extremely slowly — so slowly that at every instant, the gas has a well-defined uniform pressure and temperature throughout. This is a quasi-static process. In reality, no process is perfectly quasi-static, but slow processes approximate it well.
Quasi-static = system passes through a continuous sequence of equilibrium states
Only quasi-static processes are reversible (can be reversed by an infinitesimal change)
All NEET process formulas ( W = nRT (V_2/V_1) , etc.) assume quasi-static conditions
A non-quasi-static (sudden) process cannot be represented by a continuous curve on a PV diagram
Isothermal Process
In an isothermal process, the system exchanges heat with a reservoir to keep T constant. For an ideal gas, PV = constant (Boyle's law), ΔU = 0, and all heat goes into work.
Isothermal = same temperature throughout. The system is in thermal contact with a large heat reservoir. As the gas expands, it absorbs heat from the reservoir to keep T constant. As it compresses, it releases heat.
PV = constant (Boyle's law at constant T)
W_ isothermal = nRT V_2 V_1 = nRT P_1 P_2
T = 0 U = 0 (for ideal gas) Q = W (all absorbed heat converts to work)
PV diagram: rectangular hyperbola ( P = nRT/V , with T fixed)
Expansion ( V_2 > V_1 ): W > 0 , gas does positive work, absorbs heat Q > 0
Compression ( V_2 < V_1 ): W < 0 , work done ON gas, releases heat Q < 0
Requires very slow process (so heat exchange can keep up) or perfect thermal contact with reservoir
Adiabatic Process
In an adiabatic process, the system is thermally insulated — no heat enters or leaves. All work comes from (or goes into) internal energy. The gas cools during adiabatic expansion and heats during compression. PV^γ = constant.
Adiabatic = no heat transfer ( Q = 0 ). The system is either perfectly insulated or the process is so fast that heat doesn't have time to flow. From the first law: U = -W .
= C_P/C_V — the adiabatic index (also called heat capacity ratio). Monoatomic: = 5/3 ; Diatomic: = 7/5
Adiabatic expansion: gas does work using its own internal energy → temperature drops
Adiabatic compression: work done on gas increases internal energy → temperature rises
PV diagram: adiabatic curve is steeper than isothermal (because > 1 , so PV^ falls faster than PV )
Free expansion into vacuum: W = 0 and Q = 0 → U = 0 → T = 0 (for ideal gas)
PV = constant
PV^ = constant
System in contact with heat reservoir
System thermally insulated
Temperature unchanged
Temperature changes (drops on expansion, rises on compression)
U = 0 ; Q = W
U = -W ; Q = 0
Slow process (heat exchange needed)
Fast process (no time for heat exchange)
PV curve: gentle hyperbola
PV curve: steeper than isothermal
Isochoric & Isobaric Processes
Isochoric (ΔV = 0): no work done, all heat goes to internal energy. Isobaric (ΔP = 0): work = PΔV, and heat goes to both internal energy and work. These are the simplest processes to calculate.
Isochoric (constant volume): Think of heating a gas in a rigid, sealed container. The gas can't expand, so it does no work. All heat directly increases internal energy.
Isochoric: W = 0 U = Q = nC_v T
Isobaric (constant pressure): Think of heating a gas in a cylinder with a freely movable piston (open to atmosphere). The gas expands at constant atmospheric pressure.
Isobaric: W = P V = P(V_2 - V_1) = nR T
Q_ isobaric = nC_P T | U = nC_v T
Isochoric: PV diagram is a vertical line (V constant, P changes)
Isobaric: PV diagram is a horizontal line (P constant, V changes)
In isobaric process: Q = nC_P T and U = nC_v T , so W = Q - U = nR T (using C_P - C_v = R )
Isochoric process is why C_v is called 'heat capacity at constant volume' — all heat goes to raising temperature at constant V
Quick recall
A gas at constant pressure expands from 2 L to 5 L at P = 10^5 Pa. What is W ?
W = P V = 10^5 (5 - 2) 10^ -3 = 300 J . (Convert liters to m ^3 : 1 L = 10^ -3 m ^3 )
Work Done Summary
A single comparison table of work, heat, and ΔU for all four processes. This is the most high-yield summary for NEET — memorize this table.
Isothermal ( T = const)
W = nRT (V_2/V_1) ; U = 0 ; Q = W
Adiabatic ( Q = 0 )
W = nR(T_1 - T_2)/( - 1) ; Q = 0 ; U = -W
Isochoric ( V = const)
W = 0 ; U = Q = nC_v T
Isobaric ( P = const)
W = P V = nR T ; Q = nC_P T ; U = nC_v T
For the same expansion ( V_1 V_2 ): W_ isobaric > W_ isothermal > W_ adiabatic
For the same compression ( V_2 V_1 ): |W_ adiabatic | > |W_ isothermal | > |W_ isobaric |
The slope on a PV diagram: adiabatic is steepest, then isothermal, then isobaric (horizontal)
In ALL processes for ideal gas: U = nC_v T — this formula is universal, not just for isochoric
PV Diagrams
PV diagrams are graphical representations of thermodynamic processes. Work done = area under the PV curve. In cyclic processes, net work = enclosed area. NEET heavily tests PV diagram interpretation — reading process type from curve shape, calculating work from area, and determining direction (clockwise = heat engine).
W = _ V_1 ^ V_2 P dV = Area under PV curve
Reading PV Diagrams
Each point on a PV diagram represents an equilibrium state. Each curve represents a process. The shape of the curve tells you what type of process occurred. Learning to 'read' PV diagrams is essential for NEET.
A PV diagram plots pressure ( P ) on the y-axis and volume ( V ) on the x-axis. Each point represents a unique equilibrium state of the gas. A curve connecting two points represents a thermodynamic process — the path the system took.
Horizontal line ( P constant) → isobaric process
Vertical line ( V constant) → isochoric process
Gentle hyperbola ( PV = const) → isothermal process (also called an isotherm)
Steeper curve ( PV^ = const) → adiabatic process (also called an adiabat)
Movement rightward (increasing V ) → expansion (positive work)
Movement leftward (decreasing V ) → compression (negative work)
Work = Area Under the PV Curve
Work done by a gas in any process equals the area between the PV curve and the V-axis. This graphical interpretation makes calculating work from PV diagrams straightforward — even when the process is complex.
Since W = P dV , on a PV diagram, work equals the area between the curve and the volume axis. This is the single most useful fact for solving NEET PV diagram problems.
W = _ V_1 ^ V_2 P dV = Area under the PV curve (between V_1 and V_2)
Expansion ( V_1 V_2 , left to right): area is positive → W > 0 (work done BY gas)
Compression ( V_2 V_1 , right to left): area is negative → W < 0 (work done ON gas)
Isochoric process: vertical line → zero area → W = 0 (consistent with no volume change)
The SAME two endpoints can be connected by different paths → different areas → different work (work is path-dependent!)
Quick recall
A gas expands along a straight line on a PV diagram from (2 L, 4 atm) to (6 L, 1 atm). How do you find the work done?
The area under the straight line is a trapezoid. Area = 1 2 (P_1 + P_2)(V_2 - V_1) = 1 2 (4 + 1)(6 - 2) = 10 L atm. Convert: 1 L atm = 101.3 J, so W 1013 J .
Cyclic Process on PV Diagram
When a system returns to its initial state (cyclic process), ΔU = 0 and the net work done equals the area enclosed by the loop on the PV diagram. Clockwise loop = positive net work (heat engine); counterclockwise = negative (refrigerator).
In a cyclic process, the system starts and ends at the same state. On a PV diagram, this appears as a closed loop. Since the system returns to its original state: U = 0 → Q_ net = W_ net .
W_ net = P dV = Area enclosed by the loop on PV diagram
Clockwise loop: W_ net > 0 → gas does net positive work → this is a heat engine
Counterclockwise loop: W_ net < 0 → net work done ON gas → this is a refrigerator/heat pump
U = 0 in any cyclic process (state function returns to original value)
Q_ net = W_ net — net heat absorbed = net work done per cycle
Quick recall
A cyclic process on a PV diagram forms a rectangle with corners at (1 L, 2 atm), (3 L, 2 atm), (3 L, 5 atm), (1 L, 5 atm), traversed clockwise. Find the net work.
Area of rectangle = V P = (3-1) (5-2) = 6 L atm. Since clockwise, W = +6 L atm = 6 101.3 607.8 J .
Heat Engines
A heat engine converts heat into work by operating in a cycle between a hot source ( T_1 ) and a cold sink ( T_2 ). It absorbs Q_1 , does work W = Q_1 - Q_2 , and rejects Q_2 . Efficiency = W/Q_1 = 1 - Q_2/Q_1 . No real engine achieves 100% efficiency — this connects to the second law. NEET tests efficiency calculations and conceptual understanding of why heat engines can't be perfect.
= W Q_1 = 1 - Q_2 Q_1
How a Heat Engine Works
A heat engine takes in heat from a hot reservoir, converts part of it into useful work, and dumps the rest into a cold reservoir. The key insight: you CANNOT convert all heat into work in a cyclic process.
A heat engine is a device that operates in a cyclic process and converts heat energy into mechanical work. It needs three components:
Hot reservoir (source) at temperature T_1 : supplies heat Q_1 to the engine
Working substance (usually a gas): undergoes a cyclic process, doing net work W
Cold reservoir (sink) at temperature T_2 : receives rejected heat Q_2 from the engine
Energy balance: Q_1 = W + Q_2 → the heat absorbed equals work done plus heat rejected
Efficiency
Efficiency η = W/Q₁ = (Q₁ - Q₂)/Q₁ = 1 - Q₂/Q₁. It tells you what fraction of absorbed heat is converted to work. η < 1 always (second law). Typical values: car engine ~25%, power plant ~40%.
The efficiency of a heat engine measures what fraction of the input heat becomes useful work:
= W Q_1 = Q_1 - Q_2 Q_1 = 1 - Q_2 Q_1
is always less than 1 (less than 100%) for any real or ideal engine operating in a cycle
= 1 would require Q_2 = 0 — impossible by the second law
Higher efficiency means less waste heat ( Q_2 smaller relative to Q_1 )
Efficiency depends on the temperatures of the reservoirs — the greater the difference ( T_1 - T_2 ), the higher the possible efficiency
Quick recall
An engine absorbs 800 J from the source and rejects 600 J to the sink. Find W and η.
A refrigerator transfers heat from a cold body to a hot body by doing external work on the working substance. Its performance is measured by the coefficient of performance (COP), not efficiency.
A refrigerator is a heat engine in reverse. It absorbs heat Q_2 from a cold body (inside the fridge), requires external work W (compressor), and dumps heat Q_1 = Q_2 + W to the hot body (the room behind your fridge).
COP = Q_2 W = Q_2 Q_1 - Q_2
COP (Coefficient of Performance) replaces efficiency for refrigerators — it CAN be greater than 1
For a refrigerator: COP = Q_2/W (how much heat is removed per unit work input)
For a heat pump: COP = Q_1/W (how much heat is delivered per unit work input)
Carnot refrigerator: COP = T_2/(T_1 - T_2) — maximum possible COP
Carnot Engine
The Carnot engine is an idealized heat engine operating between two temperatures, using two isothermal and two adiabatic steps. It achieves the maximum possible efficiency = 1 - T_2/T_1 between any two given temperatures. Carnot's theorem states no engine can be more efficient than a Carnot engine operating between the same reservoirs. NEET loves numerical problems on Carnot efficiency.
_ Carnot = 1 - T_2 T_1
The Carnot Cycle
The Carnot cycle consists of four reversible steps: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. Each step is carefully designed to maximize efficiency by ensuring all processes are reversible.
The Carnot cycle is a theoretical cycle that achieves the highest possible efficiency. It consists of four steps, all perfectly reversible:
Step 1 — Isothermal expansion (A→B): Gas absorbs heat Q_1 from hot reservoir at T_1 . Gas expands slowly, doing work. Temperature stays at T_1
Step 2 — Adiabatic expansion (B→C): Gas is thermally insulated. Continues expanding, doing work. Temperature drops from T_1 to T_2 . No heat exchange
Step 3 — Isothermal compression (C→D): Gas in contact with cold reservoir at T_2 . Compressed slowly, releasing heat Q_2 to cold reservoir. Temperature stays at T_2
Step 4 — Adiabatic compression (D→A): Gas insulated again. Compressed further. Temperature rises from T_2 back to T_1 . System returns to initial state
Carnot Efficiency
The Carnot efficiency η = 1 - T₂/T₁ depends ONLY on the temperatures of the two reservoirs. Temperatures must be in Kelvin. This is the upper limit of efficiency for ANY heat engine between these two temperatures.
_ Carnot = 1 - T_2 T_1 = T_1 - T_2 T_1
This elegant formula says efficiency depends ONLY on the reservoir temperatures — not on the working substance, not on the engine design. Use Kelvin (absolute temperature) — NEVER Celsius.
T_1 and T_2 must be in Kelvin — using Celsius gives a completely wrong answer
= 1 (100%) only if T_2 = 0 K — impossible (can't reach absolute zero, third law)
= 0 if T_1 = T_2 — no temperature difference means no work can be extracted
Higher T_1 or lower T_2 → higher efficiency. This is why power plants use superheated steam ( T_1 high) and cold water condensers ( T_2 low)
Quick recall
A Carnot engine works between 500 K and 300 K. Find its efficiency. If it absorbs 1000 J, find work done and heat rejected.
Carnot's theorem states: (1) No engine operating between two temperatures can be more efficient than a Carnot engine. (2) All reversible engines between the same two temperatures have the same efficiency. This sets an absolute upper bound on heat engine performance.
Carnot's Theorem makes two powerful claims:
Statement 1: No heat engine operating between temperatures T_1 and T_2 can have efficiency greater than _ Carnot = 1 - T_2/T_1
Statement 2: ALL reversible engines operating between the same two temperatures have the SAME efficiency — regardless of working substance or engine design
An irreversible engine is always LESS efficient than the Carnot engine between the same reservoirs
This means: _ irreversible < _ reversible = _ Carnot
Quick recall
Two engines operate between the same two reservoirs. Engine A is reversible, Engine B is irreversible. Which is more efficient?
Engine A (the reversible one). By Carnot's theorem, _B < _A = _ Carnot . All reversible engines between the same temperatures share the same (maximum) efficiency.
Second Law & Entropy
The second law restricts the direction of natural processes — heat flows from hot to cold, not the reverse. Two equivalent statements: Kelvin-Planck (no engine can convert all heat to work) and Clausius (heat cannot flow from cold to hot without external work). Entropy is a measure of disorder that always increases in irreversible processes. NEET tests both statements and their equivalence.
S = Q_ rev T 0 (for isolated system)
Kelvin-Planck Statement
It is impossible to construct a heat engine that, operating in a cycle, converts ALL heat absorbed from a single reservoir entirely into work without rejecting any heat. In short: Q₂ = 0 is impossible in a cyclic engine.
Kelvin-Planck Statement: It is impossible to design a heat engine that works in a complete cycle and produces no effect other than the conversion of heat absorbed from a single reservoir entirely into work.
Applies specifically to cyclic processes. A single isothermal expansion CAN convert all Q to W — but that's not a complete cycle
This means Q_2 > 0 for every heat engine — some heat MUST be wasted
A device that violates this is called a Perpetual Motion Machine of the Second Kind (PMM-2) — it does not violate energy conservation but violates the second law
Clausius Statement
It is impossible for heat to spontaneously flow from a colder body to a hotter body without any external work being done. A refrigerator needs work input — heat doesn't climb the temperature gradient by itself.
Clausius Statement: It is impossible to construct a device that, operating in a cycle, transfers heat from a colder body to a hotter body without any work being done on the system.
Heat naturally flows from hot to cold — never the reverse, unless external work is supplied
A refrigerator does NOT violate the second law — it uses work input ( W ) to transfer heat from cold to hot
Clausius and Kelvin-Planck statements are equivalent — violating one automatically violates the other
Spontaneous (natural) processes are always from hot to cold — this defines the 'arrow of time' in thermodynamics
Quick recall
Are the Kelvin-Planck and Clausius statements equivalent? How?
Yes, completely equivalent. If you could violate Kelvin-Planck (100% efficient engine), you could couple it with a refrigerator to transfer heat from cold to hot without net work — violating Clausius. And vice versa. Violating one implies violating the other.
Reversible & Irreversible Processes
A reversible process can be exactly reversed, leaving no change in the universe. All natural processes are irreversible — they involve friction, finite temperature differences, or other dissipative effects. Irreversibility is the reason why entropy always increases.
A reversible process is an idealized process that can be reversed by an infinitesimal change, restoring both system and surroundings to their original states. An irreversible process cannot be perfectly reversed — the universe is permanently changed.
Infinitely slow (quasi-static)
Proceeds at finite speed
System always in equilibrium
System passes through non-equilibrium states
No friction, no turbulence
Friction, viscosity, turbulence present
Temperature difference = infinitesimal
Finite temperature difference
Can be exactly retraced
Cannot be exactly retraced
Theoretical idealization
ALL real processes
All natural processes are irreversible — heat conduction, free expansion, mixing, friction
Sources of irreversibility: friction, finite temperature gradient, turbulence, unresisted expansion, mixing of substances
Reversible processes give maximum work in expansion and require minimum work in compression
Carnot engine is the most efficient because its cycle is fully reversible
Entropy
Entropy (S) is a state function that measures the degree of disorder or randomness of a system. In any spontaneous process, the total entropy of the universe increases. Entropy provides the deepest explanation for why the second law works.
Entropy S is a thermodynamic state function that measures the disorder or randomness of a system. For a reversible process at temperature T :
S = Q_ rev T
Units: J/K (joules per kelvin). Higher entropy = more disorder = more microscopic arrangements.
Entropy is a state function — S depends only on initial and final states, not the path
Second law in entropy form: For an isolated system, S 0 . Equality holds only for reversible processes
Natural (irreversible) processes ALWAYS increase the total entropy of the universe: S_ universe > 0
Entropy increase = more disorder. Gas expanding into vacuum: molecules spread out → more disorder → S > 0
At absolute zero ( 0 K), entropy of a perfect crystal = 0 (third law of thermodynamics)
Quick recall
1000 J of heat flows from a body at 500 K to a body at 250 K. What is the entropy change of the universe?
S_ hot = -Q/T_1 = -1000/500 = -2 J/K. S_ cold = +Q/T_2 = +1000/250 = +4 J/K. S_ universe = -2 + 4 = +2 J/K . Positive, as expected for an irreversible process.