JEE Main chapters · Class 11 · Physics

Thermodynamics: JEE Main Physics formulas and notes

Thermodynamics is one of the most scoring chapters: JEE Mains usually asks 2 to 3 questions, almost all from the first law, the four processes, the P–V graph and the efficiency of the Carnot engine. If you know the five process formulas and the signs, you can do almost every question in a minute. About 10 hours is enough.

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Quick revision

• Q = ΔU + W, ΔU = nC(v)ΔT, W = area under P–V
• U = (f)/(2)nRT, C(v) = (fR)/(2), C(p) = C(v) + R, γ = 1 + (2)/(f)
• Isothermal: W = nRT ln(V₂)/(V₁), ΔU = 0, Q = W
• Adiabatic: PV^γ = constant, TV^γ − 1 = constant, W = (nR(T₁ − T₂))/(γ − 1)
• Isobaric W = PΔV; isochoric W = 0
• Monatomic isobaric: Q : ΔU : W = 5 : 3 : 2; diatomic: 7 : 5 : 2
• Polytropic PV^x: C = C(v) + (R)/(1 − x)
• η = 1 − (Q₂)/(Q₁), Carnot η = 1 − (T₂)/(T₁)
• Refrigerator β = (T₂)/(T₁ − T₂), η = (1)/(1 + β)
• ΔS = (Q)/(T), Otto η = 1 − r^1 − γ

Formulas with solved examples

Basics

Zeroth law, state variables and the ideal gas

• zeroth law: two bodies in thermal equilibrium with a third are in equilibrium with each other (this defines temperature)
• state functions: P, V, T, U (depend only on the state)
• path functions: Q and W
• PV = nRT, R = 8.31 J/mol K
• an adiabatic wall allows no heat, a diathermic wall does

How to apply. A state function has a change that depends only on the end points; heat and work depend on the path. Use PV = nRT with T in kelvin to connect the states.

Example. 1 mol at 300 K and 1 atm: V = 8.31 × 300 ÷ (1.013 × 10⁵) = 0.0246 m³ = 24.6 L.

Degrees of freedom and equipartition

• each quadratic term in the energy has ½kT per molecule
• f = 3 for a monatomic gas, 5 for a diatomic (3 translation + 2 rotation), 6 for a non-linear polyatomic
• vibration adds 2 at high temperature
• U = (f)/(2)nRT
• U depends on T only for an ideal gas

How to apply. Count the independent ways the molecule can store energy. Translation and rotation count at room temperature; vibration is frozen unless a question says that it is active.

Example. Argon (f = 3) at 300 K: U = 1.5 × 8.31 × 300 = 3740 J per mole.

Work and the first law

Work done by a gas

• dW = P dV
• W = ∫P dV = area under the P–V curve
• expansion: W > 0; compression: W < 0
• isobaric: W = PΔV
• a cycle traversed clockwise on the P–V diagram does positive work equal to the area enclosed (anticlockwise: negative)
• in an indicator diagram of a cycle W is the area of the loop

How to apply. Find the area under the curve from the initial to the final volume, and take its sign from the direction. For a cycle, use the area of the loop; for a triangle ½ × base × height.

Example. A gas expands at 2 × 10⁵ Pa from 1 L to 3 L: W = 2 × 10⁵ × 2 × 10⁻³ = 400 J.

First law of thermodynamics

• Q = ΔU + W
• Q is positive when heat enters the gas, W positive when the gas does work
• ΔU = nC(v)ΔT for any process of an ideal gas
• for a cycle ΔU = 0, so Q(net) = W(net)
• it is the law of conservation of energy including heat

How to apply. Write ΔU from the change of temperature, W from the area, and then find Q. Keep the signs: heat out of the gas is negative, work done on the gas is negative.

Example. 100 J of heat is given to a gas and it does 40 J of work: ΔU = 100 − 40 = 60 J.

Mechanical equivalent of heat

• W = JQ, J = 4.18 J/cal
• mgh = ms ΔT when a falling body is brought to rest and all the energy becomes heat
• the specific heat of water is 4200 J/kg K

How to apply. Convert between joules and calories first, and use a fraction if only part of the energy becomes heat.

Example. A kilogram of water falling through 420 m: mgh = 1 × 10 × 420 = 4200 J, enough to warm it by 1 K.

Processes

Isothermal process

• T constant, PV = constant
• ΔU = 0, so Q = W
• W = nRT ln(V₂)/(V₁) = nRT ln(P₁)/(P₂) = 2.303 nRT log₁₀(V₂)/(V₁)
• slope of the P–V curve dP/dV = −(P)/(V)
• isothermal bulk modulus B = P
• happens slowly, in contact with a reservoir

How to apply. All the heat taken in is turned into work, because the internal energy of an ideal gas does not change when T does not change. Take care to use the natural logarithm unless the log₁₀ form is written.

Example. 1 mol at 300 K doubles its volume: W = 8.31 × 300 × ln 2 = 1730 J, and the gas takes in 1730 J of heat.

Adiabatic process

• Q = 0
• PV^γ = constant
• TV^γ − 1 = constant
• T^γ P^1 − γ = constant
• W = (P₁V₁ − P₂V₂)/(γ − 1) = (nR(T₁ − T₂))/(γ − 1)
• slope = −γ(P)/(V), steeper than the isothermal curve
• adiabatic bulk modulus B = γP
• happens quickly or in an insulated vessel
• ΔU = −W

How to apply. With no heat exchange, the work comes from the internal energy: the gas cools when it expands and heats when it is compressed. Use the three power laws to relate P, V and T; do not use the isothermal law here.

Example. A monatomic gas (γ = 5/3) compressed to 1/8 of its volume: T₂ = T₁ × 8^2/3 = 4T₁ and P₂ = P₁ × 8^5/3 = 32P₁.

Isochoric and isobaric processes

• isochoric (V constant): W = 0, Q = ΔU = nC(v)ΔT
• isobaric (P constant): W = PΔV = nRΔT, Q = nC(p)ΔT, ΔU = nC(v)ΔT
• for a monatomic gas in an isobaric process: Q : ΔU : W = 5 : 3 : 2
• for a diatomic gas: 7 : 5 : 2
• the fraction of heat turned into work = (R)/(C(p)) = (γ − 1)/(γ)

How to apply. Choose the heat capacity to match the process: C(v) at constant volume, C(p) at constant pressure. In the isobaric case the heat divides into internal energy and work in a fixed ratio for each type of gas.

Example. Isobaric heating of a diatomic gas: W ÷ Q = 2 ÷ 7 = 28.6% of the heat becomes work; for a monatomic gas 2 ÷ 5 = 40%.

Reading P–V graphs: work, temperature and heat

• T ∝ PV: the point farther from the axes is hotter
• work = area under the curve
• a straight line through the origin (P ∝ V): T ∝ V²
• ΔU depends only on the end temperatures
• Q = ΔU + W for the actual path
• a cycle: ΔU = 0 and Q = W = area

How to apply. Find T at each end from PV, then ΔU from the change in T, then W from the area, and last Q. The same two points joined by different paths have the same ΔU but different W and Q.

Example. P = kV and the volume doubles from V₁: T becomes 4 times, and W = ½(kV₁ + 2kV₁)V₁ = 1.5kV₁².

Polytropic process and molar heat capacity

• PV^x = constant
• C = C(v) + (R)/(1 − x)
• W = (P₁V₁ − P₂V₂)/(x − 1)
• x = 0: isobaric (C = C(p))
• x = 1: isothermal (C = ∞)
• x = γ: adiabatic (C = 0)
• x = ∞: isochoric (C = C(v))

How to apply. A polytropic process is a family: the exponent x tells which process it is. The molar heat capacity along it follows from the first law.

Example. A monatomic gas with PV² = constant (x = 2): C = 1.5R + R ÷ (1 − 2) = 0.5R, which is positive, so the gas warms when heat is added.

Free expansion and mixing of gases

• free expansion into a vacuum (not reversible): Q = 0, W = 0, ΔU = 0, so the temperature of an ideal gas does not change; P₂ = P₁V₁ ÷ V₂
• mixing in a rigid insulated vessel: n₁C(v1)T₁ + n₂C(v2)T₂ = (n₁C(v1) + n₂C(v2))T
• it is not an adiabatic reversible process, so PV^γ is not constant

How to apply. Use conservation of energy for the isolated gas. Because the vessel is rigid, no work is done on or by the gas.

Example. 1 mol of helium at 300 K mixed with 1 mol of helium at 400 K in an insulated rigid vessel: T = 350 K.

Straight-line P–V process: maximum temperature

• P = P₀ − aV
• T = (V(P₀ − aV))/(nR), maximum at V = (P₀)/(2a)
• C = C(v) + (P dV)/(n dT), i.e. Q = ΔU + W on each small step

How to apply. Write T in terms of V, set dT/dV = 0 for the maximum, and test the sign of Q = nC(v)ΔT + PΔV on each part of the line (Q can change sign along a straight line).

Example. P = 4 − V (in suitable units): T ∝ V(4 − V), which is greatest at V = 2, where P = 2.

Irreversible (sudden) adiabatic change against a constant external pressure

• Q = 0
• ΔU = −P(ext)ΔV
• nC(v)(T₂ − T₁) = −P(ext)(V₂ − V₁)
• PV^γ = constant does NOT hold

How to apply. When the gas is released suddenly against a fixed external pressure or a piston is pushed with a fixed load, use the first law with W = P(ext)ΔV, not the reversible power law.

Specific heats

C(p), C(v) and γ

• C(p) − C(v) = R (Mayer’s relation)
• C(v) = (fR)/(2), C(p) = ((f + 2)R)/(2)
• γ = (C(p))/(C(v)) = 1 + (2)/(f)
• monatomic γ = (5)/(3), diatomic (7)/(5), polyatomic (non-linear) (4)/(3)
• C(p) > C(v) because at constant pressure the gas also does work

How to apply. Know C(v) and C(p) for each type of gas and the value of γ. When the type is not stated, a diatomic gas has f = 5.

Example. Diatomic gas: C(v) = 2.5R = 20.8 J/mol K, C(p) = 3.5R = 29.1 J/mol K.

Mixture of gases

• C(v) = (n₁C(v1) + n₂C(v2))/(n₁ + n₂)
• C(p) = (n₁C(p1) + n₂C(p2))/(n₁ + n₂)
• γ(mix) = (n₁C(p1) + n₂C(p2))/(n₁C(v1) + n₂C(v2))
• the pressure of a mixture is the sum of the partial pressures

How to apply. Add the heat capacities in proportion to the number of moles. The γ of a mixture lies between the two values.

Example. 1 mol of monatomic and 1 mol of diatomic gas: C(v) = (1.5R + 2.5R) ÷ 2 = 2R, C(p) = 3R, γ = 1.5.

C(v) and C(p) in terms of γ

• C(v) = (R)/(γ − 1)
• C(p) = (γR)/(γ − 1)
• C(p) − C(v) = R
• U = (nRT)/(γ − 1)

How to apply. Use these when γ is given and f is not. Then ΔU = (nRΔT)/(γ − 1) and W(adiabatic) = (nR(T₁ − T₂))/(γ − 1).

Example. γ = 1.4: C(v) = R ÷ 0.4 = 2.5R, C(p) = 1.4R ÷ 0.4 = 3.5R.

Heat engines

Heat engine

• η = (W)/(Q₁) = 1 − (Q₂)/(Q₁)
• W = Q₁ − Q₂
• Q₁ is the heat absorbed from the source, Q₂ the heat rejected to the sink
• the engine works in a cycle: ΔU = 0
• no engine can have η = 1

How to apply. The work in a cycle is the area of the loop and the net heat. Efficiency is what you get divided by what you pay for: the work divided by the heat taken from the source.

Example. An engine takes 500 J and rejects 400 J: W = 100 J and η = 20%.

Carnot engine and cycle

• four steps: isothermal expansion at T₁, adiabatic expansion, isothermal compression at T₂, adiabatic compression
• η = 1 − (T₂)/(T₁) (temperatures in kelvin)
• (Q₂)/(Q₁) = (T₂)/(T₁)
• all reversible engines between the same two temperatures have the same efficiency, the greatest possible
• lowering T₂ by x raises η more than raising T₁ by the same x

How to apply. Use kelvin. For a change in the source or the sink temperature compute η twice. The efficiency is independent of the working substance.

Example. Source 500 K and sink 300 K: η = 40%. Lowering the sink by 50 K gives 50%; raising the source by 50 K gives only 45.5%.

Otto engine

• η = 1 − (1)/(r^γ − 1) where r is the compression ratio
• two adiabatic and two isochoric processes
• η is higher for a larger r and for a larger γ

How to apply. The cycle is a petrol engine: use the compression ratio r = V₁ ÷ V₂. It is always less than the Carnot efficiency between the extreme temperatures.

Example. r = 8 and γ = 1.4: η = 1 − 1 ÷ 8^0.4 = 1 − 0.435 = 56.5%.

Refrigerator and heat pump

• coefficient of performance β = (Q₂)/(W) = (Q₂)/(Q₁ − Q₂)
• Carnot refrigerator: β = (T₂)/(T₁ − T₂)
• heat pump: COP = (Q₁)/(W) = β + 1
• η = (1)/(1 + β) for the same Carnot cycle run forward
• Q₁ = Q₂ + W is delivered to the hot reservoir

How to apply. A refrigerator is a heat engine in reverse: work W is used to take Q₂ from the cold body and give Q₁ to the hot one. The COP can be greater than 1.

Example. A Carnot refrigerator between 270 K and 300 K: β = 270 ÷ 30 = 9. The same cycle as an engine has η = 1 ÷ 10 = 10%.

Engines in series and a Carnot cycle with ideal-gas data

• two engines in series: Q₂ of the first = Q₁ of the second
• η(total) = 1 − (T₃)/(T₁) for Carnot engines with T₁ > T₂ > T₃
• for the Carnot cycle of an ideal gas (V₂)/(V₁) = (V₃)/(V₄)

How to apply. The sink of the first engine is the source of the second. For equal work output, T₂ = (T₁ + T₃)/(2).

Example. Engines between 600 K, 450 K and 300 K: η(total) = 1 − 300 ÷ 600 = 50%.

Second law and entropy

Second law of thermodynamics

• Kelvin–Planck: no cyclic engine can take heat from a single reservoir and turn it all into work
• Clausius: heat cannot pass from a colder to a hotter body without work being done
• reversible: quasi-static and without friction or dissipation
• irreversible: free expansion, friction, mixing, conduction across a finite ΔT
• the Carnot engine is the most efficient of all engines working between the same two reservoirs

How to apply. The first law counts the energy, and the second law says which direction a process can go. A process that would break either statement is impossible; a reversible process is an ideal limit.

Example. A refrigerator needs a compressor: heat does not flow from a cold to a hot body on its own, so W > 0 is always needed.

Entropy

• dS = (dQ(rev))/(T)
• ΔS = (Q)/(T) for a reversible isothermal change
• S is a state function
• isolated system: ΔS ≥ 0 (= 0 for reversible)
• entropy increases when a solid melts or a gas expands
• for a reversible cycle ΔS = 0

How to apply. Entropy measures the disorder. For a phase change at constant T, ΔS = mL ÷ T. For the universe, the entropy never decreases.

Example. Melting 10 g of ice at 273 K: ΔS = 3360 ÷ 273 = 12.3 J/K. A gas taking 600 J reversibly at 300 K has ΔS = 2 J/K.

Entropy change of an ideal gas and of mixing hot and cold bodies

• ΔS = nC(v) ln(T₂)/(T₁) + nR ln(V₂)/(V₁)
• isothermal: ΔS = nR ln(V₂)/(V₁)
• heating a body of heat capacity C: ΔS = C ln(T₂)/(T₁)
• adiabatic reversible: ΔS = 0
• ΔS(total) > 0 when two bodies at different temperatures exchange heat

How to apply. For the universe add the gas and surroundings. A reversible adiabatic process is isentropic.

Example. 1 mol doubles its volume isothermally: ΔS = 8.31 × ln 2 = 5.76 J/K.

Most asked question types

How do I solve “First law and heat, work and internal energy” questions in JEE Main?

Heat is given to a gas, or a gas does work. Find ΔU, Q or W. 1. Write Q = ΔU + W. 2. Take ΔU = nC(v)ΔT. 3. Watch the signs: work done on the gas is negative. Example: 100 J of heat and 40 J of work give ΔU = 60 J.

How do I solve “Work from a P–V graph” questions in JEE Main?

A P–V graph of a process or a cycle is given. Find the work, the heat or the change in internal energy. 1. Find the area under the curve (or the area of the loop). 2. Take the sign from the direction. 3. Find ΔU from the end temperatures. Example: 2 × 10⁵ Pa, volume 1 L to 3 L: W = 400 J.

How do I solve “Isothermal and adiabatic processes” questions in JEE Main?

A gas expands or is compressed isothermally or adiabatically. Find the work, the final temperature or the final pressure. 1. Isothermal: PV = constant; W = nRT ln(V₂/V₁). 2. Adiabatic: PV^γ = constant; W = (P₁V₁ − P₂V₂)/(γ − 1). 3. Compare the slopes on the graph. Example: Monatomic gas compressed to 1/8 of the volume: T becomes 4 times and P becomes 32 times.

How do I solve “Heat, work and internal energy at constant P or V” questions in JEE Main?

A gas is heated at constant pressure or constant volume. Find the heat used for work and the heat used for internal energy. 1. Isochoric: Q = ΔU, W = 0. 2. Isobaric: Q = nC(p)ΔT, W = nRΔT. 3. Use the ratio 5 : 3 : 2 (monatomic) or 7 : 5 : 2 (diatomic). Example: Isobaric heating of a diatomic gas: 2/7 of the heat becomes work.

How do I solve “Specific heats, γ and degrees of freedom” questions in JEE Main?

A gas with given degrees of freedom or a mixture of gases is considered. Find C(v), C(p), γ or the internal energy. 1. C(v) = fR/2. 2. C(p) = C(v) + R. 3. For a mixture weight by the moles. Example: 1 mol monatomic + 1 mol diatomic: γ = 1.5.

How do I solve “Polytropic process and molar heat capacity” questions in JEE Main?

A gas follows PV^x = constant (P ∝ V is x = −1). Find the molar heat capacity or the work. 1. C = C(v) + R/(1 − x). 2. W = (P₁V₁ − P₂V₂)/(x − 1). 3. Identify x = 0, 1, γ or ∞. Example: Monatomic gas, PV² = constant: C = 0.5R.

How do I solve “Free expansion and mixing of gases” questions in JEE Main?

A gas expands into a vacuum or two gases mix in an insulated vessel. Find the final temperature or pressure. 1. Q = 0 and W = 0, so ΔU = 0. 2. For mixing, add the energies n C(v) T. 3. Use PV = nRT for the pressure. Example: 1 mol of He at 300 K with 1 mol at 400 K gives 350 K.

How do I solve “Cycles: total work and heat” questions in JEE Main?

A cyclic process made up of several steps is given. Find the net work, the heat absorbed or the efficiency. 1. Find the work and heat of each step. 2. The net work is the area of the loop; ΔU = 0. 3. η = W ÷ (heat absorbed). Example: A triangle of ΔP = 2 × 10⁵ Pa and ΔV = 3 × 10⁻³ m³ does W = 300 J per cycle.

How do I solve “Carnot engine and efficiency” questions in JEE Main?

An engine works between two reservoirs. Find the efficiency, the work or the heat rejected. 1. Convert the temperatures to kelvin. 2. η = 1 − T₂/T₁. 3. Use Q₂/Q₁ = T₂/T₁. Example: 500 K and 300 K: η = 40%.

How do I solve “Changing the source or the sink temperature” questions in JEE Main?

The temperature of the source or the sink is changed. Find the new efficiency or the change needed for a given η. 1. Write η in terms of T₁ and T₂. 2. Make the change in one of them. 3. Solve for the unknown. Example: η = 40% at 500 K and 300 K: lowering T₂ by 50 K gives 50%, raising T₁ by 50 K gives 45.5%.

How do I solve “Refrigerator and heat pump” questions in JEE Main?

A refrigerator or heat pump works between two temperatures. Find the coefficient of performance, the work or the heat given out. 1. β = T₂/(T₁ − T₂). 2. W = Q₂/β. 3. Q₁ = Q₂ + W. Example: 270 K and 300 K: β = 9.

How do I solve “Second law and reversibility” questions in JEE Main?

Statements about engines, refrigerators or processes are given. Decide which are possible or reversible. 1. Check the Kelvin–Planck and Clausius statements. 2. Reversible: slow and without friction. 3. Compare any efficiency with the Carnot value. Example: An engine with η above the Carnot limit is impossible.

How do I solve “Entropy change” questions in JEE Main?

Heat flows reversibly at a given temperature or a substance changes state. Find the change of entropy. 1. ΔS = Q/T for an isothermal change. 2. For a phase change ΔS = mL/T. 3. Entropy of an isolated system does not decrease. Example: Melting 10 g of ice: ΔS = 12.3 J/K.

Common myths and the truth

• Myth: Heat and temperature are the same thing. Truth: Temperature measures how hot a body is; heat is the energy that flows because of a temperature difference.
• Myth: A gas must exchange heat to change its temperature. Truth: In an adiabatic process the temperature changes with Q = 0, because work is done on or by the gas.
• Myth: In an isothermal process there is no energy change at all. Truth: The internal energy of an ideal gas is constant, but heat Q = W is exchanged.
• Myth: A perfect engine could convert all the heat into work. Truth: The second law forbids it: some heat must always be rejected to a colder sink.
• Myth: A Carnot engine’s efficiency depends on the gas used. Truth: It depends only on the two temperatures.
• Myth: Raising the source temperature is the best way to improve a Carnot engine. Truth: Lowering the sink temperature by the same amount gains more.
• Myth: The internal energy of a gas depends on its volume. Truth: For an ideal gas it depends on the temperature alone.
• Myth: Work is a property of the state of a gas. Truth: Work and heat depend on the path; only U, P, V, T and S are state functions.
• Myth: A refrigerator with a COP of 5 breaks the laws of physics. Truth: The COP is not an efficiency; it is the ratio of heat removed to work done and can exceed 1.
• Myth: Entropy always increases. Truth: It never decreases for an isolated system, but it can fall in a part of it (a refrigerator interior) if more is produced outside.

Mistakes to avoid

• Use kelvin, not Celsius, in PV = nRT, in adiabatic laws and in η = 1 − T₂/T₁.
• ΔU = 0 for an isothermal process of an ideal gas, not for an adiabatic one.
• Work done on the gas is negative in Q = ΔU + W.
• The adiabatic curve is steeper than the isothermal curve.
• A free expansion does no work and does not change T of an ideal gas; it is not a reversible adiabatic.
• C(p) − C(v) = R applies per mole of an ideal gas.
• Efficiency uses the heat taken from the source, not the net heat.
• The COP of a refrigerator can be greater than 1; the efficiency of an engine cannot.

Simulations for Thermodynamics

• Adiabatic compression and expansion — Move an insulated piston and trade mechanical work with the gas’s internal energy.
• Carnot heat engine — Follow two isotherms and two adiabats around a reversible cycle.
• First law of thermodynamics — Add heat and let the gas do work; track its internal energy.
• Heating gas at constant pressure — Weights hold the piston pressure fixed while you heat the gas. Track Q, W and ΔU.
• Entropy in isothermal expansion — Expand an ideal gas reversibly at constant temperature and connect heat to entropy.
• Petrol engine (Otto cycle) — Compress the fuel–air mixture more and the ideal efficiency rises — watch the piston and crank turn.
• Work from a rectangular P–V cycle — Take a monatomic gas round a rectangular cycle. The enclosed area is the net work per cycle.
• Refrigerator and heat pump — Pump heat from a cold box to a warm room. The ideal coefficient of performance depends only on the…
• Gas under a piston — Compress or heat an ideal gas and watch its pressure respond.

More: Class 11 Physics simulations · all JEE Main chapters.