This is a 3–4 MCQ-per-year unit on NEET. It is content-rich but formula-heavy. The dominant question types are: (i) calorimetry mixing problems, (ii) thermal conduction in series/parallel rods, (iii) kinetic-theory ratios involving RMS, average, most probable speeds, (iv) first-law calculations on standard processes (isothermal, adiabatic), and (v) Carnot efficiency.
Kinetic Theory of Gases — postulates, pressure formula, kT, RMS, equipartition, degrees of freedom, Cp, Cv, γ, mean free path
First Law of Thermodynamics and processes (isothermal, adiabatic, isobaric, isochoric)
Cyclic processes, heat engines, refrigerator
Carnot cycle and second law
Topic 1: Temperature Scales and Thermal Expansion
Sub-topic A: Temperature Scales
100TC=180TF−32=100TK−273.15,TK=TC+273.15.
Absolute zero: TK=0 K. All thermodynamic temperatures use Kelvin.
Sub-topic B: Thermal Expansion
Type
Relation
Coefficient
Linear
L=L0(1+αΔT)
α
Area
A=A0(1+βΔT)
β=2α
Volume
V=V0(1+γΔT)
γ=3α
For an isotropic solid, α:β:γ=1:2:3.
Anomalous expansion of water: water has maximum density at 4 °C; below this, expansion occurs as T decreases. This is why lakes freeze top-down.
Topic 2: Calorimetry
Sub-topic A: Specific Heat and Heat Capacity
Heat needed to change temperature of mass m by ΔT:
Q=msΔT,
where s is specific heat capacity (J/(kg·K)). For water, sw≈4186J/(kg K)=1cal/(g °C).
Heat capacity C=ms.
Sub-topic B: Latent Heat
For a phase change at constant temperature:
Q=mL,
with Lfusion (solid ↔ liquid) and Lvap (liquid ↔ vapour). For water: Lf=80cal/g, Lv=540cal/g.
Sub-topic C: Mixing (Principle of Calorimetry)
Heat lost by hot body = heat gained by cold body (no external heat loss). Solve for final T:
m1s1(T1−T)=m2s2(T−T2).
For three substances: balance net heat.
Topic 3: Heat Transfer
Sub-topic A: Conduction — Fourier's Law
Rate of heat flow through a rod of cross-section A, length L, conductivity K, temperature difference ΔT:
dtdQ=KALΔT.
Thermal resistance: R=L/(KA). So dQ/dt=ΔT/R — Ohm's-law analogy.
Series rods: Req=R1+R2.
Parallel rods: 1/Req=1/R1+1/R2.
Sub-topic B: Convection
Heat transfer by bulk motion of fluid. Two types:
Natural convection: density-driven (warm air rises).
Forced convection: external pumping (fan).
Quantitatively, dQ/dt=hAΔT (Newton).
Sub-topic C: Radiation
All bodies emit thermal radiation. For a black body:
Stefan-Boltzmann law:
P=σAT4,σ=5.67×10−8W/(m2K4).
For a body with emissivity ϵ: P=ϵσAT4.
Net radiation in surroundings at T0:
dtdQ=ϵσA(T4−T04).
Wien's displacement law: peak emission wavelength
λmT=b,b=2.898×10−3m K.
Hotter bodies emit at shorter wavelengths.
Newton's law of cooling (for small ΔT):
dtdT=−k(T−T0).
This is the linearised version of Stefan's law. Solution: T(t)=T0+(Ti−T0)e−kt.
Topic 4: Kinetic Theory of Gases
Sub-topic A: Postulates
Gas consists of identical molecules in random motion.
Molecules are point-like (volume negligible compared to container).
No intermolecular forces except during elastic collisions.
Collisions with walls are elastic and instantaneous.
Velocity distribution obeys Maxwell-Boltzmann.
Sub-topic B: Pressure of an Ideal Gas
P=31ρv2=31nmv2,
where n is number density and v2 is mean-square speed.
Sub-topic C: Temperature and Kinetic Energy
For monatomic ideal gas, average translational KE per molecule:
⟨K⟩=23kBT,
with Boltzmann's constant kB=1.38×10−23J/K.
For n moles, total translational KE is 23nRT, where R=NAkB=8.314J/(mol K).
Sub-topic D: Speed Distributions
For Maxwell-Boltzmann:
Speed
Formula
Most probable
vp=2kBT/m
Average
vˉ=8kBT/(πm)
RMS
vrms=3kBT/m
Ratio: vp:vˉ:vrms=2:8/π:3≈1:1.128:1.224.
So vrms>vˉ>vp.
Sub-topic E: Equipartition Theorem
Each quadratic degree of freedom contributes 21kBT to the average energy.
Gas
DoF (per molecule)
Cv per mole
Cp
γ=Cp/Cv
Monatomic (He, Ne, Ar)
3 (translational)
23R
25R
5/3≈1.67
Diatomic (N₂, O₂) at moderate T
5 (3 trans + 2 rot)
25R
27R
7/5=1.4
Diatomic (vib. excited, high T)
7
27R
29R
9/7≈1.29
Triatomic (linear)
7
27R
29R
9/7
Triatomic (non-linear)
6 (3 trans + 3 rot)
3R
4R
4/3≈1.33
Mayer's relation: Cp−Cv=R.
Sub-topic F: Mean Free Path
λ=2πd2n1,n=N/V.
At STP, λ≈10−7m.
Topic 5: First Law of Thermodynamics
Sub-topic A: Statement
ΔU=Q−W,
where W is work done BY the gas (some books use Q=ΔU+W). Be consistent with sign convention.
Internal energy U is a state function; Q and W are path functions.
For an ideal gas, U=U(T) only (depends only on temperature, not on V or P).
Sub-topic B: Work Done by a Gas
W=∫PdV.
Sign: W>0 for expansion, W<0 for compression.
Sub-topic C: Specific Processes
Process
Definition
Work
Heat
ΔU
Isothermal
T const
W=nRTln(V2/V1)
Q=W
0
Adiabatic
Q=0
W=(P1V1−P2V2)/(γ−1)=−ΔU
0
−W
Isobaric
P const
W=PΔV=nRΔT
Q=nCpΔT
nCvΔT
Isochoric
V const
0
Q=nCvΔT
Q
Sub-topic D: Adiabatic Relations
For a reversible adiabatic of ideal gas:
PVγ=const,TVγ−1=const,TγP1−γ=const.
Adiabatic curve is steeper than isothermal on a PV diagram.
Sub-topic E: Cyclic Processes
For a cycle: ΔU=0, so Q=W. Net work = area enclosed by the cycle on the PV diagram. Sign: positive if traversed clockwise (heat engine), negative if counter-clockwise (refrigerator).
Topic 6: Heat Engines and the Second Law
Sub-topic A: Heat Engine
Operates between hot reservoir at TH and cold reservoir at TC. Per cycle: absorbs QH, does work W, rejects QC.
Efficiency:
η=QHW=1−QHQC.
Sub-topic B: Refrigerator / Heat Pump
Operates in reverse — work input W extracts QC from cold reservoir and dumps QH=QC+W into hot reservoir.
Coefficient of performance (refrigerator):
βref=QC/W.
For a heat pump: βHP=QH/W=βref+1.
Sub-topic C: Carnot Cycle
Reversible cycle: two isotherms (at TH and TC) + two adiabats. Achieves maximum efficiency between given reservoirs:
ηCarnot=1−THTC.
Carnot refrigerator: β=TC/(TH−TC).
No engine can exceed Carnot efficiency (second law). For 100% efficiency we'd need TC=0 — impossible.
Sub-topic D: Second Law
Several equivalent statements:
Kelvin-Planck: No process can convert heat entirely into work in a cycle.
Clausius: Heat cannot flow spontaneously from a colder to a hotter body.
Entropy of an isolated system never decreases.
NEET Pattern MCQ Tips
Calorimetry mixing: ice + water + warmer water, find final T.
Conduction through composite: series/parallel rods.
Stefan / Newton cooling: P∝T4 or dT/dt=−k(T−T0).
Kinetic theory: ratios of vp,vˉ,vrms; effect of doubling T on vrms (2 times).
γ for ideal gas: 5/3,7/5,9/7.
Carnot efficiency: insert TH and TC in kelvin.
Adiabatic vs isothermal: distinguish work formulas, slope comparisons.
Assertion-Reason: any heat engine efficiency < Carnot.
Common Confusions and Traps
γ=7/5 for diatomic gas at room temperature (5 DoF, vibration frozen).
Cp−Cv=R per mole, not per molecule (per molecule it would be kB).
For monatomic gas Cv=(3/2)R — students sometimes use the diatomic value.
In adiabatic, ΔU=−W (work done by gas equals negative of internal energy change).
PV=nRT: R=8.314J/(mol K); or use the per-molecule kBT.
Stefan's law: rate ∝T4 in kelvin, never centigrade.
For Newton's cooling, the rate dT/dt is proportional to (T−T0) in either kelvin or celsius (only differences matter).
A heat engine never has η=1 — second law.
The most probable speed is the smallest of vp,vˉ,vrms.
Effective conductivity for series: Keff=2K1K2/(K1+K2) (harmonic-like mean).
For parallel arrangement (rods side by side, same length, different areas): Keff=(K1A1+K2A2)/(A1+A2).
Example 3: Black Body Radiation
A black body at 500 K emits at rate P0. At 1000 K it emits at P=P0(1000/500)4=16P0.
Example 4: Carnot Efficiency
Hot reservoir at 400 K, cold at 300 K. Carnot efficiency:
η=1−300/400=0.25=25%.
So per 4 J absorbed at hot, 1 J becomes work, 3 J rejected to cold.
Example 5: RMS Speed of Oxygen
For O₂ at T = 300 K, molecular mass M=32 g/mol = 0.032 kg/mol:
vrms=3RT/M=3×8.314×300/0.032≈2.34×105≈484m/s.
Derivations
Pressure of an Ideal Gas
Consider a cube of side L containing N molecules each of mass m. A molecule with vx component impacts a face, transferring momentum 2mvx. Time between impacts on the same face: 2L/vx. So average force from one molecule: 2mvx/(2L/vx)=mvx2/L.
For N molecules: total force = Nmvx2/L. Pressure on face of area L2:
P=Nmvx2/L3=(N/V)mvx2.
By isotropy, vx2=v2/3:
P=(1/3)(N/V)mv2=(1/3)ρv2.
Connection to Temperature
Compare with ideal gas law PV=nRT=NkBT:
v2=3kBT/m,21mv2=(3/2)kBT.
So average translational KE per molecule = (3/2)kBT.
Adiabatic Process for Ideal Gas
First law: dU=−dW (since dQ=0). For ideal gas dU=nCvdT and dW=PdV.
nCvdT+PdV=0.
From PV=nRT: PdV+VdP=nRdT, so dT=(PdV+VdP)/(nR).
Substituting: Cv(PdV+VdP)/R+PdV=0, so CvPdV+CvVdP+RPdV=0, i.e. (Cv+R)PdV+CvVdP=0, or CpPdV+CvVdP=0.
Rearranging: γdV/V+dP/P=0, giving PVγ= constant.
Carnot Efficiency
Carnot cycle has two isothermals (at TH,TC) and two adiabats. Heat absorbed at TH: QH=nRTHln(V2/V1). Heat rejected at TC: QC=nRTCln(V3/V4). From adiabatic legs: THV2γ−1=TCV3γ−1 and THV1γ−1=TCV4γ−1, so V2/V1=V3/V4. Therefore QC/QH=TC/TH and
η=1−TC/TH.
Indicator Diagrams and Cyclic Work
For any cyclic process, work done by the gas equals the area enclosed on the P-V diagram. Direction:
Clockwise traversal: positive work (heat engine).
Counter-clockwise: negative work (refrigerator).
In a rectangular cycle on P-V plane: ABCDA with corners (V1,P1),(V2,P1),(V2,P2),(V1,P2). Work = (V2−V1)(P1−P2).
Two Specific Heat Relations
Mayer's Relation Derivation
For ideal gas, U=nCvT depends only on T. At constant pressure: Q=nCpΔT, W=PΔV=nRΔT, ΔU=nCvΔT. First law: Q=ΔU+W, so
nCpΔT=nCvΔT+nRΔT⟹Cp−Cv=R.
Effective γ for Mixtures
For a mix of n1 moles of gas with Cv1 and n2 moles with Cv2:
Cvmix=(n1Cv1+n2Cv2)/(n1+n2),
with similar formula for Cpmix. Then γmix=Cpmix/Cvmix. NEET often tests this with mono+diatomic mixtures.
Heat Engine and Pump Efficiencies
Refrigerator COP: β=QC/W=QC/(QH−QC). For Carnot, β=TC/(TH−TC). Higher COP at lower ΔT.
Heat pump COP (for heating): βHP=QH/W=1+βref.
Entropy (Conceptual)
The second law can be phrased: in any process, the total entropy of the universe increases (for irreversible) or stays the same (reversible).
For reversible heat exchange: dS=dQ/T.
For an ideal gas isothermal expansion: ΔS=nRln(Vf/Vi) — positive (irreversible if free expansion, reversible if isothermal but slow).
Phase Diagrams (Qualitative)
A typical P-T diagram for water shows three regions (solid, liquid, gas), separated by phase boundaries meeting at the triple point (273.16 K, 611 Pa). Above the critical point (647 K, 22 MPa), no distinction between liquid and gas.
NEET Unit 8: Thermodynamics and Kinetic Theory — Quiz
15 questions · pick the best answer
Q1
The efficiency of a Carnot engine operating between 400 K and 300 K is:
Q2
Ratio of vrms:vavg:vp for an ideal gas at temperature T is approximately:
Q3
For a diatomic gas at room temperature, γ=Cp/Cv equals:
Q4
Two rods of equal length and cross-section but conductivities K and 2K are joined in series. The effective conductivity is:
Q5
A black body radiates at temperature T. If T doubles, the radiated power becomes:
Q6
100 g of ice at 0 °C is mixed with 100 g of water at 80 °C. Final temperature is (Lf=80cal/g,swater=1 cal/g/°C):
Q7
Wien's displacement law: λmax T = constant. If T of a hot body doubles, λmax:
Q8
Assertion: In an adiabatic process, no heat is exchanged with surroundings. Reason: The system is thermally insulated, so Q = 0.
Q9
For an ideal gas undergoing isothermal expansion, the change in internal energy is:
Q10
Mean free path λ of a gas molecule is inversely proportional to:
Q11
Work done in compressing 2 moles of ideal gas isothermally from V to V/2 at 300 K is:
Q12
If γ = 5/3 for a monatomic gas, the number of degrees of freedom is:
Q13
A body at 100 °C cools to 80 °C in 10 minutes in a room at 20 °C. Using Newton's cooling, time to cool from 80 °C to 60 °C (approximately) is:
Q14
Average translational KE of a gas molecule at temperature T is:
Q15
Assertion: A refrigerator cools its inside by extracting heat from inside and dumping it outside. Reason: This requires external work, in accordance with the second law of thermodynamics.