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

Exam areas

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 = \Delta U + \Delta KE + \Delta PE

, 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 = \Delta U

. The only work that crosses a closed boundary reversibly is boundary (moving-boundary) work,

w_b = \int P\,dv

, so an adiabatic compression (

Q = 0

) drives

W = -\Delta U

— the work you put in shows up entirely as internal energy and a temperature rise.

Q - W = \Delta U + \Delta KE + \Delta PE \qquad w_b = \int P\,dv

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,

h_i = h_e + w

), a throttle (

h_i = h_e

), a nozzle (enthalpy converts to kinetic energy), and a boiler or condenser (

h_i + q = h_e

\dot{Q}_{in} - \dot{W}_{net} = \sum_e \dot{m}_e\left(h_e + \tfrac{V_e^2}{2} + gZ_e\right) - \sum_i \dot{m}_i\left(h_i + \tfrac{V_i^2}{2} + gZ_i\right)

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 = RT\ln(v_2/v_1)

. A reversible adiabatic process is isentropic:

Pv^{k} =

const with

k = c_p/c_v

, so

T_2/T_1 = (P_2/P_1)^{(k-1)/k} = (v_1/v_2)^{k-1}

. A polytropic process generalizes this to

Pv^{n} =

const. Throughout,

\Delta u = c_v\Delta T

and

\Delta h = c_p\Delta T

for an ideal gas regardless of the path, and the two specific heats are tied by

c_p - c_v = R

\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = \left(\frac{v_1}{v_2}\right)^{k-1}, \qquad k = \frac{c_p}{c_v}

Entropy and the second law

Entropy change is defined through the reversible heat transfer,

ds = (\delta q/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

Q_L

in any cycle can never be driven to zero. For an ideal gas the property change is

\Delta s = c_p\ln(T_2/T_1) - R\ln(P_2/P_1)

; a reservoir exchanging heat

Q

at fixed

T

changes entropy by

Q/T

\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} \ge 0, \qquad \Delta s_{gas} = c_p\ln\frac{T_2}{T_1} - R\ln\frac{P_2}{P_1}

Power and refrigeration cycles

A heat engine takes in

Q_H

T_H

, delivers net work

W

, and rejects

Q_L

T_L

; over a complete cycle

\Delta U = 0

Q_H = W + Q_L

and the thermal efficiency is

\eta = W/Q_H = 1 - Q_L/Q_H

. 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

Q_L

) or a heat pump (useful effect

Q_H

), rated by coefficient of performance rather than efficiency. Because

Q_H = Q_L + W

, the two COPs differ by exactly one, and both routinely exceed unity — you are moving heat, not creating it.

\eta = \frac{W}{Q_H} = 1 - \frac{Q_L}{Q_H}, \qquad \mathrm{COP}_{ref} = \frac{Q_L}{W}, \qquad \mathrm{COP}_{HP} = \frac{Q_H}{W} = \mathrm{COP}_{ref} + 1

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.

\eta_{Carnot} = 1 - \frac{T_L}{T_H}, \qquad \mathrm{COP}_{HP,\max} = \frac{T_H}{T_H - T_L}, \qquad \mathrm{COP}_{ref,\max} = \frac{T_L}{T_H - T_L}

Exam strategy

Convert every temperature to kelvin the instant you read it, then decide whether the device makes work (use

\eta

) or spends work to move heat (use COP, and name the useful stream:

Q_L

for cooling,

Q_H

for heating). For a process, identify the path first — isothermal, isentropic, or polytropic — then pull the matching

T

–

P

–

v

relation. Keep the per-cycle balance

Q_H = W + Q_L

on hand to recover whichever quantity is missing, and remember

1\,\text{ton refrigeration} = 12{,}000\,\text{Btu/hr} = 3{,}516\,\text{W}

. Finally, sanity-check against the Carnot ceiling: a real

\eta

above

\eta_{Carnot}

, or a real COP above the Carnot COP, means you slipped — usually Celsius in a ratio.

Key equations

Closed-system first law

Q - W = \Delta U + \Delta KE + \Delta PE

Energy balance with no mass flow. $Q$ positive when added to the system, $W$ positive when done by the system; energies in kJ or Btu.

Steady-flow energy equation

\dot{Q}_{in} - \dot{W}_{net} = \sum_e \dot{m}_e\!\left(h_e + \tfrac{V_e^2}{2} + gZ_e\right) - \sum_i \dot{m}_i\!\left(h_i + \tfrac{V_i^2}{2} + gZ_i\right)

Open-system rate balance at steady state. $h$ = specific enthalpy (kJ/kg), $\dot m$ = mass flow (kg/s); flow work is already inside $h = u + Pv$ .

Ideal-gas relations

Pv = RT, \quad \Delta u = c_v\,\Delta T, \quad \Delta h = c_p\,\Delta T, \quad c_p - c_v = R

$R$ = specific gas constant (kJ/kg·K). Property changes depend only on $\Delta T$ for an ideal gas, independent of path.

Isentropic (reversible adiabatic) relations

Pv^{k} = \text{const}, \qquad \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = \left(\frac{v_1}{v_2}\right)^{k-1}

$k = c_p/c_v$ . Use absolute $T$ and $P$ . Valid for an ideal gas with constant specific heats.

Isentropic ideal-gas work

w_{closed} = \frac{R(T_2 - T_1)}{1 - k}, \qquad w_{open} = \frac{kR(T_2 - T_1)}{1 - k}

Closed-system (boundary) work versus open-system (shaft) work for an isentropic process; the two differ by the factor $k$ .

Isothermal ideal-gas work

w = RT\ln\frac{v_2}{v_1} = RT\ln\frac{P_1}{P_2}

Constant-temperature ideal-gas process; for a closed system $q = w$ since $\Delta u = 0$ .

Ideal-gas entropy change

\Delta s = c_p\ln\frac{T_2}{T_1} - R\ln\frac{P_2}{P_1} = c_v\ln\frac{T_2}{T_1} + R\ln\frac{v_2}{v_1}

Entropy change between two states of an ideal gas (kJ/kg·K). Absolute $T$ throughout.

Increase-of-entropy principle

\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} \ge 0

Total entropy never decreases; equality only for a reversible process. Sets the direction of spontaneous change.

Thermal efficiency

\eta = \frac{W}{Q_H} = 1 - \frac{Q_L}{Q_H}

Fraction of supplied heat converted to net work. Denominator is the heat supplied $Q_H$ , never $W$ .

Carnot efficiency

\eta_{Carnot} = 1 - \frac{T_L}{T_H}

Maximum efficiency between reservoirs at absolute temperatures $T_H$ , $T_L$ (K or °R). Upper bound no real engine exceeds.

Coefficients of performance

\mathrm{COP}_{ref} = \frac{Q_L}{W}, \qquad \mathrm{COP}_{HP} = \frac{Q_H}{W} = \mathrm{COP}_{ref} + 1

Cooling values $Q_L$ ; heating values $Q_H$ . The two differ by exactly one because $Q_H = Q_L + W$ .

Carnot COP ceilings

\mathrm{COP}_{HP,\max} = \frac{T_H}{T_H - T_L}, \qquad \mathrm{COP}_{ref,\max} = \frac{T_L}{T_H - T_L}

Maximum COP of a reversed-Carnot cycle between the two reservoirs; absolute temperatures only.

Reservoir entropy change

\Delta S_{res} = \frac{Q}{T_{res}}

Entropy change of a thermal reservoir exchanging heat $Q$ at constant absolute temperature $T_{res}$ (kJ/K).

Worked examples

Isentropic compression of air

Problem. Air ( $k = 1.4$ , $R = 0.287\,\text{kJ/(kg·K)}$ , $c_v = 0.718\,\text{kJ/(kg·K)}$ ) is compressed reversibly and adiabatically from $300\,\text{K}$ , $100\,\text{kPa}$ to $600\,\text{kPa}$ . Find the exit temperature, the change in internal energy, and the boundary work per kilogram.

Solution. Exit temperature from the isentropic relation:

T_2 = T_1\,(P_2/P_1)^{(k-1)/k} = 300\,(6)^{0.2857} = 300\,(1.6685) = 500.6\,\text{K}

. Internal-energy change:

\Delta u = c_v\,\Delta T = 0.718\,(500.6 - 300) = 0.718\,(200.6) = 144.0\,\text{kJ/kg}

. The process is adiabatic, so

q = 0

and the closed-system first law gives

w = -\Delta u = -144.0\,\text{kJ/kg}

(negative = work done ·on· the gas). The formula form agrees:

w = R(T_2-T_1)/(1-k) = 0.287(200.6)/(-0.4) = -143.9\,\text{kJ/kg}

. Final:

T_2 = 501\,\text{K}

\Delta u = 144\,\text{kJ/kg}

w = -144\,\text{kJ/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.

T_2 = 300\,(6)^{0.2857} = 501\,\text{K}, \qquad w = -\Delta u = -144\ \text{kJ/kg}

Carnot engine work and rejected heat

Problem. A Carnot heat engine absorbs $500\,\text{kJ}$ of heat from a reservoir at $800\,\text{K}$ and rejects heat to a sink at $320\,\text{K}$ . Find the efficiency, the net work, and the heat rejected.

Solution. Carnot efficiency:

\eta = 1 - T_L/T_H = 1 - 320/800 = 1 - 0.400 = 0.600

, i.e.

60.0\%

. Net work:

W = \eta\,Q_H = 0.600\,(500) = 300\,\text{kJ}

. Heat rejected:

Q_L = Q_H - W = 500 - 300 = 200\,\text{kJ}

. Final:

\eta = 60.0\%

W = 300\,\text{kJ}

Q_L = 200\,\text{kJ}

. Sanity check:

Q_L/Q_H = 200/500 = 0.40 = T_L/T_H

— for a reversible cycle the heat ratio equals the absolute-temperature ratio, confirming the arithmetic.

\eta = 1 - \frac{320}{800} = 0.600, \qquad W = 0.600(500) = 300\ \text{kJ}

Refrigerator work and the Carnot limit

Problem. A refrigerator removes heat from a cold space at $8.0\,\text{kW}$ with a rated $\mathrm{COP}_{ref} = 4.0$ . (a) Find the compressor power. (b) The cold space sits at $3^\circ\text{C}$ and heat is rejected to a $35^\circ\text{C}$ kitchen — what is the least power thermodynamically possible?

Solution. (a)

\dot{W} = \dot{Q}_L/\mathrm{COP}_{ref} = 8.0/4.0 = 2.00\,\text{kW}

. (b) Convert to absolute temperature:

T_L = 3 + 273.15 = 276.15\,\text{K}

T_H = 35 + 273.15 = 308.15\,\text{K}

. Carnot ceiling:

\mathrm{COP}_{ref,\max} = T_L/(T_H - T_L) = 276.15/32.0 = 8.63

. The minimum power is

\dot{W}_{\min} = \dot{Q}_L/\mathrm{COP}_{ref,\max} = 8.0/8.63 = 0.927\,\text{kW}

. Final:

\dot{W} = 2.00\,\text{kW}

;

\dot{W}_{\min} = 0.927\,\text{kW}

. Sanity check: the real

2.00\,\text{kW}

exceeds the

0.927\,\text{kW}

ideal and the real COP

(4.0)

sits below the Carnot COP

(8.63)

— both as the second law requires.

\mathrm{COP}_{ref,\max} = \frac{276.15}{308.15 - 276.15} = 8.63

Common pitfalls

•Sign-convention slips: the handbook uses $Q$ positive ·in· and $W$ positive ·out· ( $Q - W = \Delta U$ ). Many textbooks use $\Delta U = Q + W$ ; pick one convention and label every term before substituting.
•Celsius in a temperature ratio. $\eta_{Carnot} = 1 - T_L/T_H$ and every COP ceiling need absolute $T$ (K or °R). Only a ·difference· $\Delta T$ is safe in Celsius.
•Dividing by $W$ instead of $Q_H$ in efficiency — the denominator is the heat purchased, not the work delivered.
•Confusing closed- and open-system isentropic work: boundary work is $R(T_2-T_1)/(1-k)$ but shaft work is $k$ times larger, $kR(T_2-T_1)/(1-k)$ . 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 $\eta_{Carnot}$ , or a real COP above the Carnot COP. Both are ceilings; beating them means you slipped (almost always Celsius in a ratio).

References

NCEES FE Reference Handbook — Thermodynamics
Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics — first/second law, reversible work, entropy
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics — Carnot, cycles, COP

F. Phase equilibrium

Vapor-Liquid and Phase Equilibrium

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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