1. Chapter at a Glance
Thermodynamics studies how heat and work change the internal energy of a system. Its laws decide how efficient an engine can be and why heat flows from hot to cold. The chapter covers thermal equilibrium (zeroth law), the first law, thermodynamic processes (isothermal, adiabatic, isobaric, isochoric), specific heats of gases, the second law, heat engines, refrigerators and the Carnot cycle. JEE Main asks one or two questions, often on P–V diagrams and efficiency.
- Work done in different processes and from P–V graphs.
- First-law bookkeeping: Q, W and ΔU in each step of a cycle.
- Adiabatic relations (PVγ, TVγ−1).
- Carnot efficiency and coefficient of performance of refrigerators.
2. The Zeroth and First Laws
- Zeroth law: if A and B are each in thermal equilibrium with C, they are in equilibrium with each other. This defines temperature.
- Internal energy U is a state function; for an ideal gas it depends only on temperature: ΔU = nCvΔT for any process.
- First law: Q = ΔU + W, where Q is heat given to the system and W is work done by the system.
- Work by a gas: W = ∫P dV = area under the P–V curve. Expansion: W > 0; compression: W < 0. For a cycle, W = enclosed area (clockwise positive), ΔU = 0, so Qnet = Wnet.
3. Thermodynamic Processes
| Process | Condition | Work W | Key relation |
|---|---|---|---|
| Isochoric | V constant | 0 | Q = ΔU = nCvΔT |
| Isobaric | P constant | PΔV = nRΔT | Q = nCpΔT |
| Isothermal | T constant | nRT ln(V₂/V₁) | ΔU = 0, Q = W; PV = const |
| Adiabatic | Q = 0 | (P₁V₁ − P₂V₂)/(γ − 1) = nCv(T₁ − T₂) | PVγ = const; TVγ−1 = const |
| Polytropic | PVn = const | nR(T₁ − T₂)/(n − 1) | C = Cv + R/(1 − n) |
Specific heats of ideal gases
- Mayer's relation: Cp − Cv = R; γ = Cp/Cv.
- Monoatomic: Cv = 3R/2, γ = 5/3; diatomic (no vibration): Cv = 5R/2, γ = 7/5; polyatomic (non-linear): Cv = 3R, γ = 4/3.
4. Second Law, Engines and Refrigerators
- Kelvin–Planck: no engine can convert all heat taken from a source into work. Clausius: heat cannot flow by itself from a colder to a hotter body.
- Heat engine efficiency η = W/Q₁ = 1 − Q₂/Q₁.
- Carnot engine (two isothermals + two adiabatics, reversible): η = 1 − T₂/T₁ — the maximum possible between those temperatures.
- Refrigerator: coefficient of performance β = Q₂/W = T₂/(T₁ − T₂) (Carnot). Relation: β = (1 − η)/η.
- Entropy of an isolated system never decreases; reversible processes are ideal limits.
5. Formula Master Table
| Quantity | Formula |
|---|---|
| First law | Q = ΔU + W |
| Internal energy change | ΔU = nCvΔT |
| Isothermal work | W = nRT ln(V₂/V₁) |
| Adiabatic | PVγ = const; W = nR(T₁ − T₂)/(γ − 1) |
| Slopes | (dP/dV)adi = γ(dP/dV)iso |
| Mayer | Cp − Cv = R |
| Carnot efficiency | η = 1 − T₂/T₁ |
| Refrigerator COP | β = T₂/(T₁ − T₂) |
6. Mind Map
- Zeroth: temperature
- First: Q = ΔU + W
- Second: direction
- Isochoric, isobaric
- Isothermal, adiabatic
- Cyclic
- Cv, Cp, γ
- Mayer's relation
- Carnot engine
- Refrigerator COP
7. Tips and Tricks
- Make a table with columns Q, W, ΔU for each step of a cycle; ΔU over the full cycle must be zero.
- For work from a P–V graph, find the area; for a cycle, the enclosed area (clockwise = work done by gas).
- Use ΔU = nCvΔT for every process — even non-isochoric ones.
- Two Carnot engines in series: equal work ⇒ T₂ = (T₁ + T₃)/2; equal efficiency ⇒ T₂ = √(T₁T₃).
- Using temperatures in °C in efficiency formulas.
- Mixing sign conventions for W (by the gas vs on the gas).
- Thinking temperature stays constant in an adiabatic process — it changes because Q = 0, not ΔT = 0.
8. Important One-liners
- Internal energy of an ideal gas depends only on its temperature.
- In a cyclic process the change in internal energy is zero.
- No work is done in an isochoric process.
- Adiabatic expansion cools a gas; adiabatic compression heats it.
- No engine working between two temperatures is more efficient than a Carnot engine.
- Free expansion of an ideal gas into vacuum: Q = 0, W = 0, ΔU = 0.
9. Solved Numericals: Average to Advanced
Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take R = 8.314 J mol⁻¹ K⁻¹ and ln 2 = 0.693.
Average level
A gas absorbs 500 J of heat and does 200 J of work. Find the change in its internal energy.
Show answer & solution
Concept: Q = ΔU + W.
- ΔU = Q − W = 500 − 200.
- = 300 J (internal energy increases).
Answer: +300 J
One mole of an ideal gas at 300 K expands isothermally from 10 L to 20 L. Find the work done by the gas and the heat absorbed.
Show answer & solution
Concept: W = nRT ln(V₂/V₁); isothermal ⇒ ΔU = 0 ⇒ Q = W.
- W = 1 × 8.314 × 300 × ln 2.
- = 2494.2 × 0.693 ≈ 1729 J.
- Q = W ≈ 1729 J.
Answer: about 1729 J each
Two moles of a diatomic ideal gas are heated at constant pressure from 300 K to 400 K. Find W, Q and ΔU.
Show answer & solution
Concept: isobaric: W = nRΔT, Q = nCpΔT with Cp = 7R/2.
- W = 2 × 8.314 × 100 ≈ 1663 J.
- Q = 2 × 3.5 × 8.314 × 100 ≈ 5820 J.
- ΔU = Q − W ≈ 4157 J (check: nCvΔT = 2 × 2.5 × 831.4).
Answer: W ≈ 1663 J, Q ≈ 5820 J, ΔU ≈ 4157 J
A Carnot engine works between 500 K and 300 K. Find its efficiency. If it absorbs 1000 J per cycle, how much work does it do and how much heat does it reject?
Show answer & solution
Concept: η = 1 − T₂/T₁.
- η = 1 − 300/500 = 0.4 = 40%.
- W = 0.4 × 1000 = 400 J.
- Q₂ = 1000 − 400 = 600 J.
Answer: 40%; 400 J; 600 J
A Carnot refrigerator keeps its inside at 250 K in a 300 K room. Find its coefficient of performance and the work needed to remove 1000 J from inside.
Show answer & solution
Concept: β = T₂/(T₁ − T₂) = Q₂/W.
- β = 250/50 = 5.
- W = Q₂/β = 1000/5 = 200 J.
Answer: 5; 200 J
Advanced level
A monoatomic ideal gas at 300 K is compressed adiabatically to one-eighth of its volume. Find the final temperature and the ratio of final to initial pressure.
Show answer & solution
Concept: TVγ−1 = const and PVγ = const with γ = 5/3.
- T₂ = T₁(V₁/V₂)γ−1 = 300 × 82/3.
- 82/3 = 4, so T₂ = 1200 K.
- P₂/P₁ = (V₁/V₂)γ = 85/3 = 2⁵ = 32.
Answer: 1200 K; 32
For the compression in N6 with 1 mol of gas, find the work done on the gas.
Show answer & solution
Concept: adiabatic ⇒ Q = 0 ⇒ work done on gas = ΔU = nCvΔT.
- ΔT = 1200 − 300 = 900 K; Cv = 1.5R.
- ΔU = 1 × 1.5 × 8.314 × 900 ≈ 11 224 J.
- Work done on the gas ≈ 11.2 kJ (work by the gas ≈ −11.2 kJ).
Answer: about 11.2 kJ on the gas
A gas is taken round a rectangular cycle on a P–V diagram: pressure between 1 × 10⁵ Pa and 3 × 10⁵ Pa, volume between 1 L and 4 L, traversed clockwise. Find the net work done by the gas and the net heat absorbed per cycle.
Show answer & solution
Concept: net work = enclosed area; for a cycle ΔU = 0 so Qnet = Wnet.
- ΔP = 2 × 10⁵ Pa; ΔV = 3 L = 3 × 10⁻³ m³.
- W = 2 × 10⁵ × 3 × 10⁻³ = 600 J (positive, clockwise).
- Qnet = 600 J.
Answer: 600 J; 600 J
One mole of a monoatomic ideal gas starts at state A (P₀, V₀, 300 K). It is heated at constant volume until the pressure doubles (state B), then at constant pressure until the volume doubles (state C). Find Q, W and ΔU for the whole process A → B → C.
Show answer & solution
Concept: use PV = nRT to find the temperatures, then apply each process formula.
- TB = 600 K (P doubled); TC = 1200 K (V doubled at 2P₀).
- A → B: W = 0; Q = CvΔT = 1.5R × 300 = 450R ≈ 3741 J.
- B → C: W = 2P₀V₀ = 2RTA = 600R ≈ 4988 J; Q = CpΔT = 2.5R × 600 = 1500R ≈ 12 471 J.
- Totals: Q = 1950R ≈ 16 212 J; W ≈ 4988 J; ΔU = Cv(1200 − 300) = 1350R ≈ 11 224 J.
- Check: Q − W = 1950R − 600R = 1350R ✓.
Answer: Q ≈ 16.2 kJ, W ≈ 5.0 kJ, ΔU ≈ 11.2 kJ
Two Carnot engines work in series: the first between 800 K and T, the second between T and 200 K. Find T if (a) both do equal work, (b) both have equal efficiency.
Show answer & solution
Concept: for a Carnot engine Q ∝ T. The heat rejected by the first is taken in by the second.
- (a) W₁ = W₂ ⇒ Q₁ − Q = Q − Q₃ ⇒ (800 − T) = (T − 200) in proportion.
- T = (800 + 200)/2 = 500 K.
- (b) 1 − T/800 = 1 − 200/T ⇒ T² = 800 × 200.
- T = √160 000 = 400 K.
Answer: (a) 500 K; (b) 400 K
10. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
In a cyclic process, the change in internal energy of the system is:
- Positive
- Negative
- Zero
- Equal to the work done
Show solution
U is a state function; the system returns to its initial state, so ΔU = 0.
Answer: (C)
For an ideal gas, Cp − Cv equals:
- R/2
- R
- 2R
- γR
Show solution
Mayer's relation: Cp − Cv = R (per mole).
Answer: (B)
The ratio of the slope of an adiabatic curve to that of an isothermal curve at the same point on a P–V diagram is:
- 1
- γ
- 1/γ
- γ − 1
Show solution
Isothermal slope = −P/V; adiabatic slope = −γP/V. Ratio γ.
Answer: (B)
An ideal gas expands freely into a vacuum inside an insulated container. Which statement is correct?
- Temperature falls
- Q = 0, W = 0 and ΔU = 0
- The gas does positive work
- Internal energy increases
Show solution
No heat enters (insulated), no work is done (no opposing pressure), so ΔU = 0 and the temperature of an ideal gas stays the same.
Answer: (B)
A Carnot engine has an efficiency of 50% with its sink at 27 °C. Find the source temperature in kelvin.
Show solution
0.5 = 1 − 300/T₁ ⇒ T₁ = 600 K.
Answer: 600
The work done by an ideal gas in an isochoric process is:
- PΔV
- nRΔT
- Zero
- nCvΔT
Show solution
W = ∫P dV and dV = 0, so W = 0; all heat goes into internal energy.
Answer: (C)
11. Quick Revision Summary
- Q = ΔU + W; ΔU = nCvΔT for any process of an ideal gas.
- W: isochoric 0; isobaric PΔV; isothermal nRT ln(V₂/V₁); adiabatic nR(T₁ − T₂)/(γ − 1).
- Adiabatic: PVγ, TVγ−1 constant; slope γ times isothermal.
- Cp − Cv = R; γ = 5/3, 7/5, 4/3.
- Carnot η = 1 − T₂/T₁; refrigerator β = T₂/(T₁ − T₂).