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Chapter 9 of 16 · Physics
Thermodynamics
Thermodynamics averages 2 MCQs per paper — first law, ideal-gas processes, and Carnot efficiency are the staples.
Thermodynamics is a Physics chapter on the official PMDC MDCAT 2026 syllabus, contributing roughly 2 MCQs to the 36-MCQ Physics section. Mastering the core concepts below typically secures the full chapter weightage.
State variables and the zeroth law
A thermodynamic state is defined by P, V, T, and n. The zeroth law: bodies in thermal equilibrium with a third are in equilibrium with each other — this is what allows temperature to be defined operationally. Ideal-gas equation: PV = nRT, with R = 8.314 J/(mol·K). One mole of any ideal gas occupies 22.4 L at STP (0°C, 1 atm).
The first law and processes
First law: ΔU = Q − W, where W is work done by the gas (sign convention varies — FSc and HRW use Q − W, while some chemistry texts use Q + W; check the problem statement). Internal energy of an ideal monoatomic gas: U = (3/2)nRT. Four key processes: isobaric (P constant, W = PΔV), isochoric (V constant, W = 0), isothermal (T constant, W = nRT·ln(Vf/Vi), ΔU = 0), and adiabatic (Q = 0, PVγ = constant, with γ = Cp/Cv = 5/3 for monoatomic, 7/5 for diatomic).
Heat capacities
Specific heat c = Q/(mΔT). For solids and liquids, c is approximately constant. For ideal gases, Cp − Cv = R (Mayer's relation). Monoatomic ideal gases have Cv = (3/2)R and Cp = (5/2)R; diatomic gases at moderate temperatures have Cv = (5/2)R and Cp = (7/2)R, accounting for two extra rotational degrees of freedom.
Second law, entropy, and Carnot
The second law has multiple equivalent forms: Kelvin-Planck (no engine converts heat entirely to work), Clausius (heat does not flow spontaneously from cold to hot), and the entropy form (ΔSuniv ≥ 0). A Carnot engine operating between TH and TC has efficiency η = 1 − TC/TH (temperatures in kelvin). An engine between 600 K and 300 K has ηmax = 0.5, or 50%. No real engine can exceed this Carnot bound.
Refrigerators and the third law
A refrigerator's coefficient of performance is COP = QC/W = TC/(TH − TC) for the ideal Carnot cycle. The third law: as T → 0 K, the entropy of a perfect crystal approaches zero, and absolute zero cannot be reached in a finite number of steps. The MDCAT often combines this with a numerical Carnot efficiency calculation — always convert °C to K first by adding 273.
Key Concepts
- Laws of thermodynamics
- Heat & internal energy
- Specific heat
- Carnot cycle
- Entropy
Worked MCQs
Q1. A Carnot engine operates between 400 K and 300 K. Its efficiency is:
- A. 25% ✓
- B. 33%
- C. 50%
- D. 75%
Explanation: η = 1 − 300/400 = 0.25 = 25%.
Common trap: Subtracting in Celsius — Kelvin is mandatory for ratios.
Q2. For an isothermal process on an ideal gas:
- A. Q = 0
- B. ΔU = 0 ✓
- C. W = 0
- D. ΔS = 0
Explanation: U depends only on T for an ideal gas, so isothermal means ΔU = 0; Q = W ≠ 0.
Common trap: Choosing Q = 0 — that defines an adiabatic, not isothermal, process.
Q3. C_p − C_v for an ideal gas equals:
- A. 0
- B. R ✓
- C. (3/2)R
- D. (5/2)R
Explanation: Mayer's relation: C_p − C_v = R for one mole of ideal gas.
Common trap: Picking (3/2)R — that is C_v itself for monoatomic.
Q4. 1 mole of ideal gas at STP occupies:
- A. 1 L
- B. 11.2 L
- C. 22.4 L ✓
- D. 44.8 L
Explanation: At STP (0°C, 1 atm) molar volume of any ideal gas = 22.4 L.
Common trap: Choosing 11.2 L — that is the volume at half a mole.
Q5. A heat engine takes 1000 J of heat and rejects 600 J. Its efficiency is:
- A. 20%
- B. 40% ✓
- C. 60%
- D. 100%
Explanation: η = (Q_H − Q_C)/Q_H = 400/1000 = 40%.
Common trap: Choosing 60% — that is Q_C/Q_H, not the efficiency.
Frequently Asked Questions
Why must Carnot efficiency use Kelvin?
Because efficiency depends on the ratio T_C/T_H; ratios are only physically meaningful on an absolute (kelvin) scale.
Can entropy decrease?
It can decrease for a subsystem if the surroundings' entropy increases by at least as much. The total entropy of the universe never decreases.
Is the first law a statement of energy conservation?
Yes. ΔU = Q − W is conservation of energy applied to thermodynamic systems.
What is the difference between heat and temperature?
Temperature is an intensive measure of average kinetic energy of particles; heat is energy in transit due to a temperature difference.
What is an adiabatic process?
One with no heat exchange (Q = 0). Fast compressions like in a piston are nearly adiabatic.
How Thermodynamics Is Tested
MDCAT questions on Thermodynamics are a mix of recall (definitions, classifications), application (predict outcomes, interpret diagrams), and basic numerical/analytical reasoning. PMDC papers from 2020–2025 emphasized the concepts above; older UHS papers (2008–2019) tested them too, with slight variations in question framing.
Practice
Drill Thermodynamics and the rest of Physics — free, no signup.
See the full MDCAT 2026 syllabus or browse all Physics chapters.