Thermodynamics for NEET physics

Thermodynamics is a short chapter with a small set of ideas, which is why it rewards careful study. Most NEET questions here test one of four things: the sign convention in the first law, work read off a P–V graph, the difference between isothermal and adiabatic changes, or the efficiency of a Carnot engine.

The chapter also sits on a fault line between physics and chemistry. The two NCERT books write the first law with different signs. Mix them up and a correct method gives a wrong answer.

Zeroth law, heat, work and internal energy

Zeroth law: if A is in thermal equilibrium with C, and B is in thermal equilibrium with C, then A and B are in equilibrium with each other. This is what makes temperature a meaningful, measurable quantity.

Three quantities appear in every question:

• Internal energy U is the total energy of the molecules. It is a state function: its change depends only on the start and end states, not the path. For an ideal gas, U depends only on temperature, so ΔU = nC_vΔT for any process.
• Heat Q is energy transferred because of a temperature difference.
• Work W is energy transferred by a change of volume against pressure.

Q and W are path functions. Two different routes between the same states give the same ΔU but different Q and W.

The first law and its sign convention

The physics NCERT writes the first law as

ΔQ = ΔU + ΔW

• ΔQ is positive when heat is supplied to the gas.
• ΔW is positive when work is done by the gas (expansion).

So heat supplied either raises the internal energy, or is used by the gas to do work, or both.

Work and the P–V diagram

The work done by a gas is W = ∫P dV, which is the area under the P–V curve.

• Expansion (volume increases): W is positive.
• Compression: W is negative.
• Cyclic process: the gas returns to its starting state, so ΔU = 0 and Q = W. The net work is the area enclosed by the loop. A clockwise loop on a P–V graph means net work done by the gas (an engine); anticlockwise means net work done on it.

The four standard processes

A few details that questions hinge on:

• For isothermal work, ln x = 2.303 log x, so W = 2.303 nRT log(V₂/V₁).
• The adiabatic work can also be written (P₁V₁ − P₂V₂)/(γ − 1).
• Other adiabatic forms: TVᵞ⁻¹ = constant, and P¹⁻ᵞTᵞ = constant.
• In an adiabatic expansion, the gas does work out of its own internal energy, so it cools. An adiabatic compression heats the gas. This is why a bicycle pump warms up.

Why the adiabatic curve is steeper. Differentiate PV = constant: slope dP/dV = −P/V. Differentiate PVᵞ = constant: slope = −γP/V. Since γ > 1, the adiabatic through any point is steeper than the isothermal by a factor γ.

Specific heats of a gas

A gas has two molar specific heats: C_v at constant volume and C_p at constant pressure. At constant pressure, some of the heat also does work, so C_p is larger. For an ideal gas, C_p − C_v = R (Mayer's relation), and γ = C_p/C_v.

Where these numbers come from (degrees of freedom and equipartition) is covered in kinetic theory for NEET.

The second law and the Carnot engine

The first law says energy is conserved. The second law says which way energy can flow and how much heat can become work.

• Kelvin–Planck statement: no process can have, as its only result, the absorption of heat from a reservoir and its complete conversion into work. So no engine is 100% efficient.
• Clausius statement: no process can have, as its only result, the transfer of heat from a colder body to a hotter one. A refrigerator needs work to do it.

A reversible process can be run backwards so that both system and surroundings return to their original states. It must be quasi-static and free of dissipation such as friction. Real processes (free expansion, heat flowing across a finite temperature gap) are irreversible.

The Carnot engine is the ideal reversible engine working between a hot reservoir T₁ and a cold reservoir T₂. Its cycle has two isothermal and two adiabatic steps. Its efficiency is

η = W/Q₁ = 1 − Q₂/Q₁ = 1 − T₂/T₁ (temperatures in kelvin)

It depends only on the two temperatures, not on the working substance, and no engine working between the same two temperatures can beat it.

Worked numericals

Example 1: first law with work done on the gas

A gas absorbs 800 J of heat, and 300 J of work is done on it. Find ΔU.

• Work done by the gas: ΔW = −300 J.
• ΔU = ΔQ − ΔW = 800 − (−300) = 1,100 J.

Both the heat and the work add energy to the gas, so the answer must be larger than 800 J. That quick check catches sign errors.

Example 2: isobaric heating of a diatomic gas

Two moles of a diatomic ideal gas are heated by 50 K at constant pressure. Take R = 8.3 J mol⁻¹ K⁻¹. Find Q, W and ΔU.

• Q = nC_pΔT = 2 × (7/2 × 8.3) × 50 = 2 × 29.05 × 50 = 2,905 J.
• W = nRΔT = 2 × 8.3 × 50 = 830 J.
• ΔU = nC_vΔT = 2 × (5/2 × 8.3) × 50 = 2,075 J.
• Check: 2,075 + 830 = 2,905 ✓. And W/Q = 830/2,905 = 2/7, as the key idea predicts.

Example 3: isothermal expansion

One mole of an ideal gas at 300 K expands isothermally from 10 L to 20 L. Take R = 8.314 J mol⁻¹ K⁻¹ and ln 2 = 0.693.

• W = nRT ln(V₂/V₁) = 8.314 × 300 × 0.693 ≈ 1,729 J, about 1.73 kJ.
• ΔU = 0, so the gas absorbs Q = W ≈ 1.73 kJ of heat.

Example 4: adiabatic compression

A monatomic gas at 300 K is compressed adiabatically to 1/8 of its volume. Find the final temperature and the pressure ratio.

• TVᵞ⁻¹ = constant, with γ − 1 = 2/3. So T₂ = T₁ × 8 raised to the power 2/3.
• 8 raised to 2/3 = (cube root of 8)² = 2² = 4. So T₂ = 4 × 300 = 1,200 K.
• PVᵞ = constant: P₂/P₁ = 8 raised to 5/3 = 2⁵ = 32.

Example 5: raising Carnot efficiency

A Carnot engine has an efficiency of 40% with its sink at 300 K. By how much must the source temperature rise to make the efficiency 50%, keeping the sink fixed?

• 0.4 = 1 − 300/T₁, so T₁ = 300/0.6 = 500 K.
• 0.5 = 1 − 300/T₁′, so T₁′ = 600 K.
• The source must rise by 100 K.

Practice MCQs

1. A gas absorbs 500 J of heat and does 200 J of work. The change in internal energy is: (a) 700 J (b) 300 J (c) −300 J (d) 200 J
2. The efficiency of a Carnot engine working between 400 K and 300 K is: (a) 75% (b) 33% (c) 25% (d) 20%
3. In which process does all the heat supplied go into internal energy? (a) isothermal (b) isobaric (c) isochoric (d) adiabatic
4. A monatomic ideal gas is heated at constant pressure. The fraction of the heat used to do work is: (a) 3/5 (b) 2/5 (c) 2/7 (d) 5/7
5. A gas goes round a rectangular cycle on a P–V graph between 1 × 10⁵ Pa and 3 × 10⁵ Pa, and between 1 L and 4 L. The net work per cycle is: (a) 300 J (b) 600 J (c) 900 J (d) 1,200 J
6. An ideal gas in an insulated container expands freely into a vacuum. Its temperature: (a) rises (b) falls (c) stays the same (d) depends on γ
7. At the same point on a P–V graph, the ratio of the slope of an adiabatic to that of an isothermal for a diatomic gas is: (a) 1 (b) 1.4 (c) 1.67 (d) 0.71
8. A Carnot engine takes 1,000 J from a source at 500 K and rejects heat to a sink at 300 K. The heat rejected is: (a) 400 J (b) 500 J (c) 600 J (d) 1,000 J

Answers

1. (b) ΔU = Q − W = 500 − 200 = 300 J.
2. (c) η = 1 − 300/400 = 0.25.
3. (c) At constant volume W = 0, so Q = ΔU.
4. (b) W/Q = 1 − 1/γ = 1 − 3/5 = 2/5.
5. (b) Area = ΔP × ΔV = 2 × 10⁵ × 3 × 10⁻³ = 600 J.
6. (c) Q = 0 (insulated) and W = 0 (no external pressure), so ΔU = 0. For an ideal gas, T is unchanged.
7. (b) The adiabatic slope is γ times the isothermal slope, and γ = 1.4.
8. (c) η = 1 − 300/500 = 0.4, so W = 400 J and Q₂ = 1,000 − 400 = 600 J.

What to do next

• Write the first law in the physics convention on the top of your formula sheet, with the chemistry version beside it and a note on how they differ.
• Redraw the process table from memory, including W, Q and ΔU for each process, until you can do it in two minutes.
• Sketch an isothermal and an adiabatic through the same point and label which is steeper and why.
• Solve ten P–V graph questions, calculating work as an area each time.
• Move on to kinetic theory for NEET, which explains C_v, C_p and γ from molecular motion.

For the thermal properties that come before this chapter, see properties of solids and liquids.