Thermodynamics

Thermodynamics studies the relationships between heat, work, internal energy, and the macroscopic state of matter. The PMDC MDCAT 2026 syllabus emphasises the First Law, molar specific heats of gases, and the famous Mayer's relation Cp − Cv = R. Expect 1-2 numerical MCQs.

PMC Table of Specifications. Five subtopics: Thermal Equilibrium & Heat, Work in Thermodynamics, First Law, Molar Specific Heat, and the Cp − Cv = R relation.

Thermal Equilibrium and Heat

Two bodies are in thermal equilibrium when there is no net flow of heat between them — equivalently, they have the same temperature. The Zeroth Law of Thermodynamics: if A is in thermal equilibrium with C and B is in thermal equilibrium with C, then A is in thermal equilibrium with B. This justifies the very concept of temperature.

Heat (Q) is energy in transit between systems due to a temperature difference. SI unit: joule (J). 1 calorie = 4.18 J. Heat is a process quantity (path-dependent), not a state function.

Internal energy

Internal energy U is the total kinetic + potential energy of all the molecules in a system. It is a state function — depends only on the present state, not the history. For an ideal gas U depends only on temperature.

Work in Thermodynamics

When a gas expands against an external pressure P from volume V₁ to V₂, the work done by the gas is:

W = ∫P·dV

Special cases:

Isobaric (constant P): W = P·ΔV = P(V₂ − V₁).
Isochoric (constant V): W = 0 (no volume change).
Isothermal (constant T): W = nRT·ln(V₂/V₁).
Adiabatic (Q = 0): PV^γ = constant; W = (P₁V₁ − P₂V₂)/(γ − 1).

Sign convention: W > 0 when work is done by the gas (expansion); W < 0 when work is done on the gas (compression).

First Law of Thermodynamics

Statement: heat supplied to a system is used to increase the internal energy of the system and/or to do work by the system on its surroundings.

ΔU = Q − W

This is simply conservation of energy. Sign conventions:

Q > 0: heat added to the system; Q < 0: heat lost.
W > 0: work done by the system (expansion); W < 0: work done on the system.

First law in special processes

Thermodynamic processes — what's constant, what's zero
Process	Constant	Q	W (by gas)	ΔU	P-V relation
Isothermal	T	Q = W	nRT·ln(V₂/V₁)	0	PV = nRT (Boyle's law)
Isobaric	P	nC_pΔT	P·ΔV	nC_vΔT	V/T = const (Charles's)
Isochoric	V	nC_vΔT = ΔU	0	nC_vΔT	P/T = const (Gay-Lussac)
Adiabatic	Q = 0	0	(P₁V₁ − P₂V₂) / (γ − 1)	−W	PV^γ = const
Cyclic	Returns to start	Q = W (= area of loop on PV)	Net = area enclosed	0	Closed curve on PV diagram

Common trap. Different textbooks write the First Law as ΔU = Q − W (chemistry/physics convention W = work BY gas) or ΔU = Q + W (some chemistry books, W = work ON gas). Always check which convention is in use; the physics for MDCAT uses ΔU = Q − W.

Molar Specific Heat of Gas

For a gas, the heat needed to raise the temperature of 1 mole by 1 K depends on the path of heating. Two principal molar specific heats are defined:

C_v — molar specific heat at constant volume: Q_v = nC_vΔT. Since W = 0 in an isochoric process, all heat goes to raising U: ΔU = nC_vΔT.
C_p — molar specific heat at constant pressure: Q_p = nC_pΔT. The gas also does PΔV work, so more heat is required for the same ΔT: C_p > C_v always.

Values for ideal gases

Monatomic gas (He, Ar): C_v = (3/2)R, C_p = (5/2)R, γ = 5/3 ≈ 1.67.
Diatomic gas (O₂, N₂): C_v = (5/2)R, C_p = (7/2)R, γ = 7/5 = 1.40.
Polyatomic gas: C_v ≈ 3R, C_p ≈ 4R, γ ≈ 4/3.

Ratio γ = C_p/C_v appears in the adiabatic equation PV^γ = constant.

Relation Cp − Cv = R (Mayer's Relation)

For 1 mole of an ideal gas:

C_p − C_v = R

Where R = 8.314 J/(mol·K) is the universal gas constant. Derivation idea: at constant V, all the heat raises U; at constant P, the same ΔT also requires PΔV = RΔT of work to be done on the surroundings, so more heat is required.

Memory aid. "Pumping a tyre needs more heat than a sealed bottle." Constant-pressure heating requires extra energy to perform expansion work — that extra is exactly R per mole per kelvin.

Heat engines (introduction)

A heat engine takes in heat Q_h from a hot reservoir at T_h, converts part of it into work W, and dumps Q_c to a cold reservoir at T_c. Efficiency:

η = W/Q_h = 1 − Q_c/Q_h. The maximum (Carnot) efficiency is η_max = 1 − T_c/T_h, with temperatures in kelvin.

Worked MCQs

Five MCQs that capture the high-yield testing patterns for this chapter.

Q1. A gas absorbs 500 J of heat and does 200 J of work. The change in its internal energy is:

700 J
300 J
500 J
−300 J

First Law: ΔU = Q − W = 500 − 200 = 300 J.

Q2. For an ideal monatomic gas, the molar specific heat at constant volume is:

R
(3/2)R
(5/2)R
(7/2)R

A monatomic gas has only translational degrees of freedom (3). Equipartition gives U = (3/2)nRT, so C_v = (3/2)R.

Q3. In an isothermal process applied to an ideal gas:

ΔU = Q
Q = W
Q = 0
W = 0

For an ideal gas U depends only on T, so isothermal means ΔU = 0 and the First Law gives Q = W.

Q4. For 1 mole of an ideal gas, C_p − C_v equals:

Mayer's relation: C_p − C_v = R = 8.314 J/(mol·K).

Q5. In an adiabatic process:

Q = 0 and ΔU = 0
Q = 0 and ΔU = −W
W = 0 and Q = ΔU
Q = −W and ΔU = 0

Adiabatic means no heat exchange (Q = 0). First Law ⇒ ΔU = −W. Adiabatic expansion cools the gas; adiabatic compression heats it.

Quick Recap

First Law: ΔU = Q − W (W = work done BY gas).
Isothermal: ΔU = 0; Isochoric: W = 0; Adiabatic: Q = 0; Cyclic: ΔU = 0.
C_v = (3/2)R (monatomic), (5/2)R (diatomic).
C_p − C_v = R (Mayer's relation).
γ = C_p/C_v; PV^γ = constant adiabatic.
Carnot efficiency η = 1 − T_c/T_h.

Test yourself. Take a timed Thermodynamics quiz or browse all Physics MCQs to lock these concepts in.