METALLURGICAL ENGINEERING 17, AY '26-'27, First Semester

Thermodynamics - 1st Long Exam Coverage

Dense reference for exam cramming. Formulas, sign conventions, and exam tips and traps (para hindi madale ni sir dave)

01 Intro to Thermo & Concept of State

The Four Laws

LawStatementKey takeaway
ZerothIf A~B and B~C, then A~C (equilibrium)Basis for defining temperature
FirstEnergy can't be created/destroyed, only transformedConservation of energy
SecondEntropy of an isolated system increasesDefines reaction direction / reversibility
ThirdEntropy → constant as T → 0 KCan't reach absolute zero

System Types

Open — exchanges matter + energy  |  Closed — energy only  |  Isolated — neither. Thermally isolated = adiabatic (no heat transfer). Mechanically isolated = no work done on/by system.

Properties

Intensive: independent of amount (T, density, pressure). Extensive: proportional to amount (mass, volume, heat, U, H).

Equilibrium types

Thermal (equal T, no heat flow) · Mechanical (no net force imbalance) · Chemical (forward = reverse rate) → together = Thermodynamic equilibrium.

State function — value depends only on the current state, not the path (U, H, S, P, V, T).
Path function — depends on the path taken (q, w). Not state functions.

02 Equations of State of Gases

$$PV = nRT \quad \text{(Ideal Gas Law)}$$

Gas Laws

LawConditionRelation
Boyle'sconstant T\(P_1V_1 = P_2V_2\)
Charles'constant P\(\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\)
Gay-Lussac'scoefficient of thermal expansion\(\alpha = \dfrac{1}{V_0}\left(\dfrac{\partial V}{\partial T}\right)_P\)

Ideal gas assumptions: no intermolecular forces · gas molecules are point masses (no volume) · perfectly elastic collisions.

Dalton's Law (gas mixtures)

$$p_{total} = \sum p_i \qquad p_i = X_i\,p_{total} \qquad X_i = \frac{n_i}{n}$$

Van der Waals Equation (real / non-ideal gas)

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

a corrects for intermolecular attraction (internal pressure). b corrects for finite molecular volume (excluded volume), \(b = 4 \times\) volume of all particles.

A gas behaves more ideally at low pressure, high temperature, and with a low boiling point.

03 First Law of Thermodynamics

$$\Delta U = q - w \qquad \text{differential form:}\qquad dU = \delta q - \delta w$$

Sign convention

QuantityPositive (+)Negative (–)
q (heat)flows INTO system (endothermic)flows OUT of system (exothermic)
w (work)done BY the system (expansion)done ON the system (compression)

Work of a gas against a piston

$$w = \int_{V_1}^{V_2} P\,dV$$

Expansion (\(V_2 > V_1\)): w is positive. Compression (\(V_2 < V_1\)): w is negative.

Heat capacity

$$q = C\Delta T \qquad \text{differential:} \qquad C = \frac{\delta q}{dT}$$
dU is a state function (path-independent) — q and w are path functions. This is why reversible vs. irreversible processes between the same two states give the same ΔU but different q and w.

04 First Law Applied to Processes

ProcessConditionΔUqwΔH
IsochoricdV=0, w=0\(C_v\Delta T\)\(C_v\Delta T\)0\(\Delta U + V\Delta P\)
IsobaricdP=0\(C_p\Delta T - P\Delta V\)\(C_p\Delta T\)\(P\Delta V\)\(C_p\Delta T\)
IsothermaldT=0, dU=00\(RT\ln\frac{V_2}{V_1}\)= q0
Adiabaticδq=0\(C_v\Delta T\)0\(-\Delta U\)\(C_p\Delta T\)

Enthalpy

$$H = U + PV \qquad \text{at constant pressure:} \qquad \Delta H = q_p$$

Cp − Cv relation

$$C_p - C_v = R \quad \text{(for 1 mole ideal gas)}$$

Adiabatic process equations

$$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} \qquad P_1V_1^\gamma = P_2V_2^\gamma \qquad \gamma = \frac{C_p}{C_v}$$
Sanity check: Isothermal expansion → U constant, all work comes from absorbed heat. Adiabatic expansion → U decreases exactly by the work done (no heat to replace it) → T drops.

05 Heat Capacity & Enthalpy Deep Dive

Estimating heat capacity

MethodUse caseResult
Kinetic theorygasesmonoatomic \(C_v=\frac32R\); diatomic \(C_v=\frac52R\); always \(C_p=C_v+R\)
Dulong-Petitsolids\(C_v \approx 3R \approx 24.9\ J/mol\cdot K\)
Kopp-Neumanncompounds\(C_p\)(compound) ≈ sum of \(C_p\) of constituent elements
Empiricalany, temp-dependent\(c_P = a + bT + cT^{-2}\)

Types of enthalpy change

Heat of Formation (\(\Delta H_f\)) — forming a compound from elements; element in standard state → \(\Delta H_f° = 0\).
Heat of Transformation — phase change (fusion \(L_m\), vaporization \(L_v\), polymorphic \(L_t\)).
Heat of Reaction:

$$\Delta H_{rxn} = \sum n\Delta H_{f,products} - \sum n\Delta H_{f,reactants}$$

(+) = endothermic, (–) = exothermic. Always assumes the reaction goes to completion.

Hess's Law — total ΔH of a reaction equals the sum of ΔH along any stepwise path (ΔH is a state function → path-independent).

Kirchhoff's Law (ΔH at elevated T)

$$\Delta H_T = \Delta H_{T_0} + \int_{T_0}^{T}\Delta C_P\,dT \qquad \Delta C_P = \sum C_{P,products} - \sum C_{P,reactants}$$

Thermodynamic Loop method

Alternative to Kirchhoff when phase changes complicate things: draw reactants/products at T1 (298K) and T2 in a closed loop — sum of ΔH around the loop = 0. Solve by heating reactants up, reacting, then adjusting; same answer, easier bookkeeping.

Adiabatic Flame Temperature (AFT)

Exothermic reaction's heat is fully absorbed as sensible heat by the products (no heat escapes):

$$-\Delta H_{rxn,T_1} = \int_{T_1}^{AFT} n\,C_{p,products}\,dT$$

Solve for AFT (often needs iteration/quadratic since Cp depends on T).

⚠ Common Exam Traps

  1. It is common to confuse \(w=+\Delta U\) vs \(w=-\Delta U\) - check which sign convention the given might use on your exam. This notes deck defines w positive when done BY the system, so \(\Delta U = q - w\).
  2. It's also common to forget units - atm·L vs J vs cal. Convert R consistently: \(0.08206\ L\cdot atm/mol\cdot K = 8.314\ J/mol\cdot K = 1.987\ cal/mol\cdot K\).
  3. Using the ΔH formula (products − reactants) but forgetting stoichiometric coefficients n.
  4. Mixing up isothermal (ΔU=0, but q,w ≠ 0) with adiabatic (q=0, but ΔU,w ≠ 0).
  5. Forgetting that the standard heat of formation of a pure element is zero.
MetE 17 · Metallurgical Thermodynamics · print-friendly (Ctrl/Cmd+P)