Part of PC-04 — Chemical Thermodynamics

Process-Focused Summary

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Classification of Thermodynamic Processes

A thermodynamic process is classified by which variable is held constant (or which quantity is forced to be zero). Each process type has one simplifying constraint that dramatically reduces the number of unknowns in the first law.

Isothermal Process (ΔT=0\Delta T = 0)

In an isothermal process, temperature remains constant. For an ideal gas, internal energy depends only on temperature, so ΔU=0\Delta U = 0 and ΔH=0\Delta H = 0. The entire heat absorbed equals work done by the system (q=wq = -w). For reversible isothermal expansion: w=nRTln(V2/V1)w = -nRT\ln(V_2/V_1). For irreversible against constant PextP_{ext}: w=PextΔVw = -P_{ext}\Delta V.

The key comparison: reversible work >> irreversible work (in magnitude) between the same initial and final states. This is a direct consequence of the second law.

Adiabatic Process (q=0q = 0)

No heat exchange occurs. The entire energy change comes from work alone: ΔU=w\Delta U = w. For adiabatic compression (work done ON system), the gas heats up (ΔU>0\Delta U > 0). For adiabatic expansion, the gas cools. The work done in adiabatic processes equals nCvΔTnC_v\Delta T for ideal gases.

Isochoric Process (ΔV=0\Delta V = 0)

At constant volume, no PV work is performed (w=0w = 0). All heat goes into changing internal energy: ΔU=qv=nCvΔT\Delta U = q_v = nC_v\Delta T. A bomb calorimeter operates at constant volume and therefore measures ΔU\Delta U directly. To get ΔH\Delta H, the correction ΔngRT\Delta n_g RT must be applied.

Isobaric Process (ΔP=0\Delta P = 0)

At constant pressure, heat equals the enthalpy change: qp=ΔH=nCpΔTq_p = \Delta H = nC_p\Delta T. This is why calorimetry at constant pressure (coffee-cup calorimeter) directly measures ΔH\Delta H. The extra heat compared to isochoric heating (nRΔTnR\Delta T per mole) goes into expansion work against constant pressure.

Free Expansion (Pext=0P_{ext} = 0)

When a gas expands into vacuum, no work is done (w=0w = 0) and for an ideal gas, no heat is absorbed (q=0q = 0), so ΔU=0\Delta U = 0. Despite no work or heat, the entropy of the system increases (gas occupies more volume) — demonstrating that entropy generation is possible without heat transfer. This is an irreversible process even though ΔU=0\Delta U = 0.

Summary Table

ProcessConstraintZero quantityKey equation
Isothermal (ideal gas)ΔT=0\Delta T = 0ΔU,ΔH\Delta U, \Delta Hq=wq = -w
Adiabaticq=0q = 0Heat qqΔU=w\Delta U = w
IsochoricΔV=0\Delta V = 0Work wwΔU=qv\Delta U = q_v
IsobaricΔP=0\Delta P = 0qp=ΔHq_p = \Delta H
Free expansionPext=0P_{ext} = 0w,q,ΔUw, q, \Delta UNothing changes

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