Thermodynamics

In this chapter: thermal equilibrium and the zeroth law, heat, work and internal energy, the first law, specific heats of gases, state variables, isothermal, adiabatic, isochoric, isobaric and cyclic processes, the second law, reversible and irreversible processes, heat engines, refrigerators and the Carnot engine.

Thermal equilibrium and the zeroth law

Two systems in thermal contact are in thermal equilibrium when no heat flows between them, which happens when their temperatures are equal. Zeroth law: if systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. This law is what makes a thermometer meaningful, and it gives temperature its definition as the quantity that is equal for systems in equilibrium.

Heat, work and internal energy

The internal energy U of a system is the sum of the kinetic and potential energies of its molecules, measured in the frame where the system as a whole is at rest. It is a state variable: it depends only on the state of the system (P, V, T), not on how the system got there. For an ideal gas, U depends on temperature alone.

Heat and work are two ways of changing U. Heat flows because of a temperature difference; work is done by some other means, such as moving a piston. Neither heat nor work is a state variable; both depend on the path taken.

When a gas expands against a piston, the work done by the gas is

The first law of thermodynamics

The first law is the law of conservation of energy applied to heat and work. Heat supplied to a gas is partly used to raise its internal energy and partly to do work on the surroundings.

Worked example: A gas absorbs 200 J of heat and does 80 J of work in pushing a piston out. Find the change in its internal energy. What if, instead, 80 J of work had been done on the gas?
Solution: ΔU = ΔQ − ΔW = 200 − 80 = 120 J. If work is done on the gas, ΔW = −80 J, so ΔU = 200 − (−80) = 280 J.

Specific heats of a gas

For a gas the heat needed per mole per kelvin depends on the conditions. At constant volume no work is done, so all the heat raises U; at constant pressure the gas also expands and does work. So C_p > C_v, and for an ideal gas

Thermodynamic processes

A quasi-static process is one carried out so slowly that the system stays in equilibrium with its surroundings at every stage, so P and T are well defined throughout. The standard processes for an ideal gas (μ moles) are these.

• The slope of an adiabatic curve at any point is γ times the slope of the isothermal curve through the same point.
• In an adiabatic process, TV^γ−1 and P^1−γT^γ are also constant. A gas compressed adiabatically heats up; one expanding adiabatically cools.
• Adiabatic conditions need either a perfectly insulated container or a process so fast that there is no time for heat to flow.
• A cycle traced clockwise on the P-V diagram gives positive net work by the gas; anticlockwise gives negative work.

Worked example: 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. (R = 8.31 J mol⁻¹ K⁻¹)
Solution: W = μRT ln(V₂/V₁) = 1 × 8.31 × 300 × ln 2 = 2493 × 0.693 ≈ 1.73 × 10³ J. Since ΔU = 0 in an isothermal process, the heat absorbed is also 1.73 × 10³ J.

Heat engines and refrigerators

A heat engine takes heat Q₁ from a hot reservoir, converts part of it into work W, and rejects heat Q₂ to a cold reservoir, working in a cycle. A refrigerator runs the other way: work W is done on the working substance so that it takes heat Q₂ from the cold body and gives Q₁ to the hot surroundings.

The second law of thermodynamics

The first law allows any process that conserves energy. The second law says which of them can actually happen.

• Kelvin-Planck statement: no process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of that heat into work. So no engine can have an efficiency of 1.
• Clausius statement: no process is possible whose sole result is the transfer of heat from a colder object to a hotter object. A refrigerator needs work to be done on it.

Reversible and irreversible processes

A process is reversible if it can be run backwards so that both the system and the surroundings return exactly to their original states. That requires it to be quasi-static and free of dissipative effects like friction and viscosity. All natural processes, such as free expansion of a gas, heat flowing from a hot body to a cold one, or a block sliding to rest, are irreversible. A reversible process is an idealisation.

The Carnot engine

The Carnot engine is an ideal reversible engine working between a hot reservoir at T₁ and a cold reservoir at T₂. Its cycle has four steps: isothermal expansion at T₁ (absorbing Q₁), adiabatic expansion from T₁ to T₂, isothermal compression at T₂ (rejecting Q₂), and adiabatic compression back to T₁.

Carnot's theorem: no engine working between two given temperatures can be more efficient than a Carnot engine, and the efficiency of a Carnot engine does not depend on the working substance. Since T₂ cannot be zero, even an ideal engine cannot convert all heat into work, which agrees with the second law.

Worked example: A Carnot engine works between 500 K and 300 K and absorbs 1000 J of heat per cycle from the hot reservoir. Find its efficiency, the work done per cycle and the heat rejected.
Solution: η = 1 − 300/500 = 0.4 (40%). W = ηQ₁ = 0.4 × 1000 = 400 J. Q₂ = Q₁ − W = 600 J. Check: Q₂/Q₁ = 0.6 = T₂/T₁.

Common mistakes: (1) Mixing sign conventions: in NCERT, ΔW is work done by the system; some books use work done on the system and write ΔU = Q + W. Pick one and stay with it. (2) Using Celsius temperatures in the Carnot formula. (3) Saying the temperature is constant in an adiabatic process; it is the heat exchange that is zero. (4) Thinking an isothermal process has no heat flow; heat flows in, and all of it becomes work. (5) Using μC_vΔT for the internal energy change only in isochoric processes; for an ideal gas, ΔU = μC_vΔT in every process.

JEE and NEET focus

• First law with correct signs, and ΔU = μC_vΔT for any process of an ideal gas.
• Work done from P-V graphs: area under a curve, area of a loop and its sign.
• Isothermal and adiabatic work, and temperature change in adiabatic compression or expansion.
• Comparing slopes and work of isothermal and adiabatic processes between the same volumes.
• Carnot efficiency, refrigerator coefficient of performance and the two statements of the second law.

Practice questions