Thermodynamics · Study · FE Chemical · FE → PE Prep
Thermodynamics
8% of exam
Thermodynamic properties of pure components and mixtures, property data and phase diagrams (steam tables, P-h, T-s, x-y), the first and second laws, isothermal, adiabatic, and isentropic processes, power and refrigeration cycles, phase equilibrium (Raoult's law, fugacity, activity coefficients), chemical equilibrium, and heats of reaction and mixing.
5 concepts
A. Thermodynamic properties of pure components and mixtures
Thermodynamic Properties of Pure Components and Mixtures
The ideal-gas law and real-gas behavior through the compressibility factor and reduced properties, specific heats and property departures, and ideal-mixture property combination.
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B. Properties data and phase diagrams
Property Data and Phase Diagrams (Steam Tables, P-h, T-s, x-y)
Reading the steam tables and computing quality, using P-h and T-s diagrams to follow power and refrigeration cycles, and reading x-y and T-x-y diagrams for binary VLE.
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C. Thermodynamic laws
First and Second Law: Processes, Cycles and Efficiency
Closed- and open-system energy balances, entropy and the second law, the isothermal/adiabatic/isentropic process family, and Carnot ceilings on efficiency and COP.
Almost every thermodynamics point on the FE Chemical is won or lost on two reflexes: getting the energy balance bookkeeping right, and remembering that the second law caps what the first law permits. The first law tells you energy is conserved; it will happily let you write a 100%-efficient engine. The second law is what forbids it, fixes the direction of spontaneous change, and sets the Carnot ceiling that every real machine falls short of. The traps are almost never conceptual — they are sign conventions (Q in positive, W out positive), forgetting that a temperature ·ratio· demands kelvin, and confusing a coefficient of performance with an efficiency. Keep those straight and this entire topic becomes a series of free points. (FE Reference Handbook 10.6 — Thermodynamics, First and Second Law of Thermodynamics.)
Closed-system energy balance
A closed system exchanges energy but no mass with its surroundings. The handbook writes the first law as Q−W=ΔU+ΔKE+ΔPE, with the sign convention that heat ·added to· the system is positive and work done ·by· the system is positive. For most FE problems the kinetic and potential terms vanish and the balance collapses to Q−W=ΔU
Open-system (control-volume) balance
When mass flows across the boundary, each stream carries its internal energy plus flow work Pv — together the enthalpy h=u+Pv — along with kinetic and potential energy. For steady flow the storage term is zero and the rate balance reduces to heat in plus enthalpy in equals work out plus enthalpy out. This single equation specializes to every steady device on the exam: a turbine or compressor (adiabatic, hi=he+w
The ideal-gas process family
Cold-air-standard problems lean on four idealized processes, all with constant heat capacities. A constant-temperature (isothermal) ideal-gas process gives Pv= const and work w=RTln(v2/v1)
Entropy and the second law
Entropy change is defined through the reversible heat transfer, ds=(δq/T)rev. The decisive statement for the exam is the increase-of-entropy principle: for any real (irreversible) process the entropy generated in the system plus surroundings is non-negative, and it is zero only for a reversible process. This is why heat flows spontaneously from hot to cold and never the reverse, and why the rejected heat QL
Power and refrigeration cycles
A heat engine takes in QH at TH, delivers net work W, and rejects QL
The Carnot ceiling
Between two reservoirs the most efficient possible cycle is the reversible Carnot cycle — two isothermals joined by two isentropics, a rectangle on a T–s diagram whose enclosed area is the net work. Its efficiency and COP depend only on the two absolute temperatures, and no real device operating between the same reservoirs can beat them. A computed actual efficiency above the Carnot value is a guaranteed error, almost always Celsius leaking into a temperature ratio.
Convert every temperature to kelvin the instant you read it, then decide whether the device makes work (use η) or spends work to move heat (use COP, and name the useful stream: QL for cooling, QH for heating). For a process, identify the path first — isothermal, isentropic, or polytropic — then pull the matching T
Key equations
Closed-system first lawQ−W=ΔU+ΔKE+ΔPE
Energy balance with no mass flow. Q positive when added to the system,
Worked examples
Isentropic compression of air
Problem. Air (k=1.4, R=0.287kJ/(kg⋅K), cv=0.718kJ/(kg⋅K)
Common pitfalls
•Sign-convention slips: the handbook uses Q positive ·in· and W positive ·out· (Q−W=ΔU). Many textbooks use ΔU=Q+W
References
NCEES FE Reference Handbook — Thermodynamics
Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics — first/second law, reversible work, entropy
Raoult's and Henry's laws, K-values and relative volatility, bubble- and dew-point calculations, and the fugacity/activity framework for non-ideal VLE.
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G. Chemical equilibrium
Chemical Equilibrium and Heats of Reaction/Mixing
The K-ΔG° relationship, the van't Hoff temperature dependence, equilibrium conversion via the extent of reaction, and standard heats of reaction, formation, and mixing.
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. The only work that crosses a closed boundary reversibly is boundary (moving-boundary) work,
wb=∫Pdv
, so an adiabatic compression (
Q=0
) drives
W=−ΔU
— the work you put in shows up entirely as internal energy and a temperature rise.
Q−W=ΔU+ΔKE+ΔPEwb=∫Pdv
), a throttle (
hi=he
), a nozzle (enthalpy converts to kinetic energy), and a boiler or condenser (
. The denominator is always the heat ·purchased·, never the work — a classic slip. Reverse the cycle and you spend work to pump heat from cold to hot: a refrigerator or air conditioner (useful effect
QL
) or a heat pump (useful effect
QH
), rated by coefficient of performance rather than efficiency. Because
QH=QL+W
, the two COPs differ by exactly one, and both routinely exceed unity — you are moving heat, not creating it.
Maximum COP of a reversed-Carnot cycle between the two reservoirs; absolute temperatures only.
Reservoir entropy changeΔSres=TresQ
Entropy change of a thermal reservoir exchanging heat Q at constant absolute temperature Tres (kJ/K).
) is compressed reversibly and adiabatically from
300K
,
100kPa
to
600kPa
. Find the exit temperature, the change in internal energy, and the boundary work per kilogram.
Solution. Exit temperature from the isentropic relation: T2=T1(P2/P1)(k−1)/k=300(6)0.2857=300(1.6685)=500.6K.
Internal-energy change: Δu=cvΔT=0.718(500.6−300)=0.718(200.6)=144.0kJ/kg.
The process is adiabatic, so q=0 and the closed-system first law gives w=−Δu=−144.0kJ/kg (negative = work done ·on· the gas). The formula form agrees: w=R(T2−T1)/(1−k)=0.287(200.6)/(−0.4)=−143.9kJ/kg.
Final: T2=501K, Δu=144kJ/kg, w=−144kJ/kg.
Sanity check: compression raises temperature and pressure, so work is into the gas (negative by the handbook sign convention) and stored entirely as internal energy — exactly what an adiabatic process demands.
T2=300(6)0.2857=501K,w=−Δu=−144kJ/kg
Carnot engine work and rejected heat
Problem. A Carnot heat engine absorbs 500kJ of heat from a reservoir at 800K and rejects heat to a sink at 320K. Find the efficiency, the net work, and the heat rejected.
Solution. Carnot efficiency: η=1−TL/TH=1−320/800=1−0.400=0.600, i.e. 60.0%.
Net work: W=ηQH=0.600(500)=300kJ.
Heat rejected: QL=QH−W=500−300=200kJ.
Final: η=60.0%, W=300kJ, QL=200kJ.
Sanity check: QL/QH=200/500=0.40=TL/TH — for a reversible cycle the heat ratio equals the absolute-temperature ratio, confirming the arithmetic.
η=1−800320=0.600,W=0.600(500)=300kJ
Refrigerator work and the Carnot limit
Problem. A refrigerator removes heat from a cold space at 8.0kW with a rated COPref=4.0. (a) Find the compressor power. (b) The cold space sits at 3∘C and heat is rejected to a 35∘C kitchen — what is the least power thermodynamically possible?
Solution. (a) W˙=Q˙L/COPref=8.0/4.0=2.00kW
COPref,max=308.15−276.15276.15=8.63
; pick one convention and label every term before substituting.
•Celsius in a temperature ratio. ηCarnot=1−TL/TH and every COP ceiling need absolute T (K or °R). Only a ·difference· ΔT is safe in Celsius.
•Dividing by W instead of QH in efficiency — the denominator is the heat purchased, not the work delivered.
•Confusing closed- and open-system isentropic work: boundary work is R(T2−T1)/(1−k) but shaft work is k times larger, kR(T2−T1)/(1−k). Use the right one for a piston versus a turbine.
•Treating COP like an efficiency that must be below 1. Cooling and heating COPs are usually 3–6; a value under 1 signals an error.
•Mixing energy and power. Use kJ for per-cycle problems but kW (kJ/s) for rate problems — never multiply a mass flow rate by a per-cycle energy.
•Getting a real efficiency above ηCarnot, or a real COP above the Carnot COP. Both are ceilings; beating them means you slipped (almost always Celsius in a ratio).