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Thermodynamics: MDCAT Notes and MCQs

Heat & thermal equilibrium

Heat is energy transferred from a higher-temperature region to a lower-temperature region. When no net transfer occurs, bodies are in thermal equilibrium (same temperature — zeroth law). Internal energy U of an ideal gas depends only on temperature.

Specific heat vs molar specific heat

  • Specific heat c: heat to raise temperature of 1 kg by 1 K — Q = mcΔT, unit J kg−1 K−1 (water: 4200).
  • Molar specific heat C: heat to raise temperature of 1 mole by 1 K — Q = nCΔT, unit J mol−1 K−1. C = Mc.
  • For gases: Cv (constant volume) and Cp (constant pressure); Cp > Cv because at constant pressure extra heat is needed to do work in expansion.

Work in volume change

At constant pressure: W = PΔV (area under P–V graph). Work done by gas in expansion is positive.

First law of thermodynamics

Q = ΔU + W (Q = heat supplied to system, W = work done by system) — equivalently ΔU = Q − W, or ΔU = Q + Won. It is the law of conservation of energy.

ProcessConditionFirst law
IsothermalΔT = 0 → ΔU = 0Q = W
AdiabaticQ = 0W = −ΔU (expansion cools gas)
IsochoricΔV = 0 → W = 0Q = ΔU
IsobaricΔP = 0Q = ΔU + PΔV

Derivation: Cp − Cv = R

For 1 mole heated through ΔT: at constant volume Qv = CvΔT = ΔU. At constant pressure Qp = CpΔT = ΔU + PΔV. For an ideal gas PΔV = RΔT. So CpΔT = CvΔT + RΔT → Cp − Cv = R (8.314 J mol−1 K−1).

γ = Cp/Cv: monoatomic 5/3 ≈ 1.67; diatomic 7/5 = 1.4.

PMDC learning outcomes for this unit
  • Heat & first law: 7.1 Describe that thermal energy is transferred from a region of higher temperature to lower temperature.
  • Heat & first law: 7.2 Differentiate between specific heat and molar specific heat.
  • Heat & first law: 7.3 Calculate work done by a thermodynamic system during a volume change.
  • Heat & first law: 7.4 Describe the first law of thermodynamics in terms of change in internal energy, heating of the system and work done on the system.
  • Heat & first law: 7.5 Explain that the first law of thermodynamics expresses conservation of energy.
  • Heat & first law: 7.6 Define specific heat and molar specific heats of a gas.
  • Heat & first law: 7.7 Apply the first law of thermodynamics to derive Cp − Cv = R for an ideal gas

Thermodynamics MCQs with answers

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Q1
Heat naturally flows from:
ALarger to smaller body
BHigher to lower temperature
CHigher to lower internal energy always
DLower to higher temperature
Show answer
Answer: B (Higher to lower temperature). Heat flows down a temperature gradient.
Q2
The unit of molar specific heat is:
AJ kg⁻¹ K⁻¹
BJ mol⁻¹
CJ K⁻¹
DJ mol⁻¹ K⁻¹
Show answer
Answer: D (J mol⁻¹ K⁻¹). Molar specific heat refers to one mole: J mol⁻¹ K⁻¹.
Q3
For an ideal gas, Cp − Cv equals:
AZero
BR
Cγ
DR/2
Show answer
Answer: B (R). Cp − Cv = R (from the first law, since PΔV = RΔT for 1 mole).
Q4
In an isothermal process, the change in internal energy of an ideal gas is:
AZero
BMaximum
CEqual to Q
DEqual to −W
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Answer: A (Zero). Internal energy depends only on temperature, which is constant.
Q5
In an adiabatic process:
AΔU = 0
BΔT = 0
CQ = 0
DW = 0
Show answer
Answer: C (Q = 0). No heat enters or leaves the system.
Q6
A gas expands at constant pressure of 2 × 10⁵ Pa from 1 m³ to 1.5 m³. Work done by gas is:
A0.5 × 10⁵ J
B2 × 10⁵ J
C1 × 10⁵ J
D3 × 10⁵ J
Show answer
Answer: C (1 × 10⁵ J). W = PΔV = 2 × 10⁵ × 0.5 = 1 × 10⁵ J.
Q7
The first law of thermodynamics is a statement of conservation of:
AEnergy
BCharge
CMomentum
DMass
Show answer
Answer: A (Energy). Q = ΔU + W expresses energy conservation.
Q8
Cp is greater than Cv because at constant pressure:
AGas does no work
BVolume is constant
CExtra heat is needed for work of expansion
DInternal energy decreases
Show answer
Answer: C (Extra heat is needed for work of expansion). Part of the heat goes into doing work PΔV.
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