NEET UG Physics · Chapter 9

Thermodynamics

Zeroth and first laws, work in thermodynamic processes, isothermal and adiabatic changes, second law.

NCERT Class 11NEET: 1–2 Qs per paper 10 min read

1. Chapter at a Glance

Thermodynamics studies how heat and work change the energy of a system. It explains why a bicycle pump warms up, why a gas cools when it expands suddenly, and why heat never flows by itself from a cold body to a hot one. This chapter covers thermal equilibrium, the zeroth and first laws, specific heats of gases, thermodynamic processes on P–V diagrams, and the second law with reversible and irreversible processes. NEET usually asks one or two questions here.

What NEET asks most
  • First law: sign conventions for Q, W and ΔU.
  • Work done in isothermal, isobaric, isochoric and adiabatic processes.
  • Adiabatic relations PVγ and TVγ−1 = constant.
  • Work from the area of a P–V graph and cyclic processes.
  • Cp − Cv = R and γ for different gases.

2. Zeroth and First Laws

  • A system is the part we study; everything else is the surroundings. State variables (P, V, T, U) describe its equilibrium state.
  • Zeroth law: if two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other. This is the basis of temperature and thermometers.
  • Internal energy U is the total molecular energy. It is a state function; for an ideal gas it depends only on temperature, ΔU = nCvΔT.
  • First law (energy conservation): ΔQ = ΔU + ΔW, where ΔQ is heat supplied to the system and ΔW is work done by the system.
  • Work done by a gas: W = ∫P dV = area under the P–V curve. Expansion ⇒ W positive; compression ⇒ W negative.
  • Heat and work are path functions: they depend on how the change happens. U does not.

3. Thermodynamic Processes

ProcessConstantWork by gasFirst-law result
IsochoricV0Q = ΔU
IsobaricPPΔV = nRΔTQ = nCpΔT
IsothermalTnRT ln(V₂/V₁)ΔU = 0 ⇒ Q = W
AdiabaticQ = 0nR(T₁ − T₂)/(γ − 1)W = −ΔU
Cyclicreturns to startarea enclosedΔU = 0 ⇒ Q = W
isobaric isochoric isothermal adiabatic PV
Four processes starting from the same state. The adiabatic curve is steeper than the isothermal (slope ratio γ).
  • Adiabatic: PVγ = constant, TVγ−1 = constant, P1−γTγ = constant. A sudden (fast) process, or one in an insulated container, is adiabatic.
  • Adiabatic expansion cools a gas (it uses internal energy to do work); adiabatic compression heats it (diesel engines, bicycle pumps).
  • Specific heats: Cp − Cv = R (Mayer's relation). Monoatomic γ = 5/3; diatomic γ = 7/5; polyatomic γ = 4/3.

4. Second Law and Reversibility

  • Kelvin–Planck statement: no process can have as its only result the complete conversion of heat into work.
  • Clausius statement: no process can have as its only result the transfer of heat from a colder to a hotter body.
  • A reversible process is quasi-static and has no dissipative effects (friction, viscosity); it can be retraced exactly. All real processes are irreversible (free expansion, heat flow across a finite temperature difference, friction).
  • A Carnot engine works between two temperatures using two isothermal and two adiabatic steps; its efficiency η = 1 − T₂/T₁ is the highest possible between those temperatures.

5. Formula Master Table

QuantityFormula
First lawΔQ = ΔU + ΔW
Internal energy changeΔU = nCvΔT
Isothermal workW = nRT ln(V₂/V₁) = 2.303nRT log(V₂/V₁)
Isobaric workW = PΔV = nRΔT
AdiabaticPVγ = const; TVγ−1 = const
Adiabatic workW = (P₁V₁ − P₂V₂)/(γ − 1)
Mayer's relationCp − Cv = R
Carnot efficiencyη = 1 − T₂/T₁

6. Master Summary Table

ConceptKey pointNEET Note
Zeroth lawDefines temperatureBasis of thermometers
First lawEnergy conservationQ to system, W by system
Internal energyState functionΔU = 0 in isothermal and cyclic
IsochoricW = 0All heat raises U
AdiabaticQ = 0Steeper than isothermal by γ
Second law100% conversion impossibleCarnot gives the upper limit

7. Mind Map

Thermodynamics
Laws
  • Zeroth
  • First: Q = ΔU + W
  • Second
Processes
  • Isochoric, isobaric
  • Isothermal
  • Adiabatic, cyclic
Gases
  • Cp − Cv = R
  • γ values
Reversibility
  • Quasi-static
  • Carnot limit

8. Tips and Tricks

Shortcuts
  • ΔU = nCvΔT is valid for every process of an ideal gas, not only isochoric ones.
  • For a cycle, net work = area enclosed; clockwise on a P–V diagram means positive work by the gas.
  • Remember γ − 1 for the TV relation: monoatomic 2/3, diatomic 2/5. For example, 82/3 = 4 and 322/5 = 4.
  • In free expansion of an ideal gas: Q = 0, W = 0, ΔU = 0, so T is unchanged.
Common NEET traps
  • Using °C in PV = nRT or efficiency formulas — always kelvin.
  • Mixing up "work done on the gas" and "work done by the gas".
  • Thinking temperature stays constant in an adiabatic process.

9. Important One-liners

  • The first law of thermodynamics is the law of conservation of energy.
  • Internal energy of an ideal gas depends only on its temperature.
  • No work is done in an isochoric process.
  • In an isothermal process all the heat supplied is converted into work.
  • A sudden compression of a gas is adiabatic and raises its temperature.
  • All natural processes are irreversible.
  • Cp is always greater than Cv for a gas.

10. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take R = 8.314 J mol⁻¹ K⁻¹ and ln 2 = 0.693. Click to reveal each answer and working.

Average level

N1Average

A gas absorbs 800 J of heat while 200 J of work is done on it. Find the change in internal energy.

Show answer & solution

Concept: ΔQ = ΔU + ΔW, where ΔW is work by the gas; work on the gas makes ΔW = −200 J.

  1. ΔU = ΔQ − ΔW = 800 − (−200).
  2. ΔU = 1000 J.

Answer: +1000 J

N2Average

A gas at a constant pressure of 2 × 10⁵ Pa expands from 2 L to 5 L. Find the work done by it.

Show answer & solution

Concept: isobaric work W = PΔV.

  1. ΔV = 3 L = 3 × 10⁻³ m³.
  2. W = 2 × 10⁵ × 3 × 10⁻³ = 600 J.

Answer: 600 J

N3Average

Two moles of an ideal gas at 400 K expand isothermally to double their volume. Find the work done and the heat absorbed.

Show answer & solution

Concept: W = nRT ln(V₂/V₁); isothermal ⇒ ΔU = 0 ⇒ Q = W.

  1. W = 2 × 8.314 × 400 × 0.693.
  2. W ≈ 4610 J.
  3. Q = W ≈ 4610 J.

Answer: about 4.61 kJ each

N4Average

Find the increase in internal energy when 3 moles of a monoatomic ideal gas are heated through 50 K.

Show answer & solution

Concept: ΔU = nCvΔT with Cv = 3R/2.

  1. ΔU = 3 × 1.5 × 8.314 × 50.
  2. ΔU ≈ 1871 J.

Answer: about 1.87 kJ

N5Average

One mole of a diatomic gas is heated through 100 K (a) at constant pressure, (b) at constant volume. Find the heat needed in each case.

Show answer & solution

Concept: diatomic Cv = 5R/2, Cp = 7R/2.

  1. (a) Qp = 3.5 × 8.314 × 100 ≈ 2910 J.
  2. (b) Qv = 2.5 × 8.314 × 100 ≈ 2079 J.
  3. The extra 831 J (= RΔT) at constant pressure goes into work of expansion.

Answer: about 2910 J; about 2079 J

Advanced level

N6Advanced

A diatomic gas (γ = 1.4) at 300 K is compressed adiabatically to 1/32 of its volume. Find the final temperature and the pressure ratio P₂/P₁.

Show answer & solution

Concept: TVγ−1 and PVγ are constant.

  1. T₂ = T₁(V₁/V₂)γ−1 = 300 × 320.4.
  2. 320.4 = (2⁵)2/5 = 2² = 4 ⇒ T₂ = 1200 K.
  3. P₂/P₁ = 321.4 = 2⁷ = 128.

Answer: 1200 K; 128

N7Advanced

A gas goes round a triangular cycle on a P–V diagram with corners (1 L, 1 × 10⁵ Pa), (3 L, 1 × 10⁵ Pa) and (3 L, 3 × 10⁵ Pa). Find the net work done per cycle.

Show answer & solution

Concept: net work in a cycle = area enclosed.

  1. Base = 3 − 1 = 2 L = 2 × 10⁻³ m³.
  2. Height = 3 × 10⁵ − 1 × 10⁵ = 2 × 10⁵ Pa.
  3. Area = ½ × 2 × 10⁻³ × 2 × 10⁵ = 200 J.

Answer: 200 J

N8Advanced

Two moles of a monoatomic gas expand adiabatically and cool from 600 K to 400 K. Find the work done by the gas.

Show answer & solution

Concept: adiabatic ⇒ Q = 0 ⇒ W = −ΔU = nCv(T₁ − T₂).

  1. W = 2 × 1.5 × 8.314 × 200.
  2. W ≈ 4988 J (done at the expense of internal energy).

Answer: about 4.99 kJ

N9Advanced

A Carnot engine works between 527 °C and 127 °C. Find its efficiency. To what temperature must the source be raised (sink unchanged) to reach 60% efficiency?

Show answer & solution

Concept: η = 1 − T₂/T₁ with kelvin temperatures.

  1. T₁ = 800 K, T₂ = 400 K ⇒ η = 1 − 0.5 = 50%.
  2. For 60%: 400/T₁ = 0.4 ⇒ T₁ = 1000 K = 727 °C.

Answer: 50%; 1000 K

N10Advanced

One mole of a monoatomic ideal gas is heated at constant pressure from 300 K to 600 K. Find Q, W and ΔU and check the first law.

Show answer & solution

Concept: Q = nCpΔT (Cp = 5R/2), W = nRΔT, ΔU = nCvΔT.

  1. Q = 2.5 × 8.314 × 300 ≈ 6236 J.
  2. W = 8.314 × 300 ≈ 2494 J.
  3. ΔU = 1.5 × 8.314 × 300 ≈ 3741 J.
  4. Check: W + ΔU = 2494 + 3741 = 6235 J ≈ Q ✓.

Answer: Q ≈ 6.24 kJ, W ≈ 2.49 kJ, ΔU ≈ 3.74 kJ

11. NEET-pattern Questions with Detailed Answers

These questions are original and modelled on concepts repeatedly asked in NEET UG.

Q1Easy

In an isothermal process for an ideal gas:

  1. ΔQ = 0
  2. ΔU = 0
  3. ΔW = 0
  4. ΔP = 0
Show solution

U depends only on T; T is constant, so ΔU = 0 and Q = W. ΔQ = 0 describes an adiabatic process; ΔW = 0 an isochoric one.

Answer: (B)

Q2Easy

The zeroth law of thermodynamics leads to the concept of:

  1. Internal energy
  2. Temperature
  3. Work
  4. Entropy
Show solution

Systems in mutual thermal equilibrium share a common property — temperature.

Answer: (B)

Q3Medium

A gas expands adiabatically. Its temperature:

  1. Increases
  2. Decreases
  3. Remains constant
  4. First rises then falls
Show solution

Q = 0 and W > 0, so ΔU = −W < 0; the gas cools.

Answer: (B)

Q4Medium

For a monoatomic ideal gas, the ratio Cp/Cv is:

  1. 1.33
  2. 1.40
  3. 1.67
  4. 1.29
Show solution

Cv = 3R/2, Cp = 5R/2 ⇒ γ = 5/3 ≈ 1.67. (1.40 is diatomic, 1.33 polyatomic.)

Answer: (C)

Q5Medium

A system goes through a cycle and does 300 J of net work. The net heat absorbed by it in the cycle is:

  1. 0
  2. 300 J
  3. −300 J
  4. 600 J
Show solution

In a cycle ΔU = 0, so Qnet = Wnet = 300 J.

Answer: (B)

Q6Medium

Which process is represented by a vertical straight line on a P–V diagram?

  1. Isobaric
  2. Isothermal
  3. Isochoric
  4. Adiabatic
Show solution

A vertical line means V is constant while P changes: isochoric. A horizontal line is isobaric.

Answer: (C)

Q7Hard

An ideal gas expands freely into a vacuum inside an insulated container. Which is true?

  1. W > 0, ΔU < 0
  2. Q = 0, W = 0, ΔU = 0
  3. Temperature falls
  4. Internal energy rises
Show solution

No heat enters (insulated) and no work is done (nothing to push against), so ΔU = 0 and T stays the same. Free expansion is irreversible.

Answer: (B)

Q8Hard

At the same point on a P–V diagram, the slope of an adiabatic curve is related to the slope of an isothermal curve by:

  1. Equal
  2. γ times
  3. 1/γ times
  4. (γ − 1) times
Show solution

Isothermal: dP/dV = −P/V. Adiabatic: dP/dV = −γP/V. The adiabatic is γ times steeper.

Answer: (B)

12. Quick Revision Summary

Remember before the exam
  • ΔQ = ΔU + ΔW (Q to system, W by system); ΔU = nCvΔT.
  • Isochoric W = 0; isobaric PΔV; isothermal nRT ln(V₂/V₁); adiabatic Q = 0.
  • PVγ = const; adiabatic is γ times steeper than isothermal.
  • Cp − Cv = R; γ = 5/3, 7/5, 4/3.
  • Second law: no 100% heat-to-work conversion; Carnot η = 1 − T₂/T₁.
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