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Work in Isothermal, Isobaric and Adiabatic Processes

Each idealised process fixes one variable and gives its own expression for the work, with the adiabatic case following a steeper curve than the isothermal one. Which curve applies decides how much a compressed gas heats up.

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What a learner can do afterwards

  • Derives the work done in isothermal and isobaric expansion by integration
  • Uses the adiabatic relation between pressure and volume and says where the exponent comes from
  • Explains why an adiabatic compression raises the temperature and an isothermal one does not

1 · Read

Work is the area under the pV curve, which is the integral of p dV. At fixed pressure p slides out front, so isobaric work is p times delta V. At fixed temperature the integral gives n R T times the log of V2 over V1, and the ideal gas keeps its internal energy, so every joule of heat becomes work.

Try it together

A car engine squeezes its fuel-air mix from 240 to 40 cubic cm with no heat flow. Pressure jumps from 1.00e5 to 1.23e6 and temperature climbs from 293 K to 600 K, costing about 63 J of compression work. The same squeeze done isothermally would stay at 293 K, with the work leaving as heat instead.

Adiabatic means Q equals zero: p times V to the gamma stays constant, where gamma is Cp over Cv, about 1.4 for diatomic air. Because gamma tops one, the adiabat runs steeper than the isotherm on a pV plot. Compression work then has nowhere to go but inside, so the gas warms; expansion spends internal energy, so it cools.

Good to know

Name the process by what stays fixed: T for isothermal, p for isobaric, Q equals zero for adiabatic. Expanding into a vacuum does zero work, since it pushes against zero pressure. Watch signs: work the gas does counts positive in the integral.

Isobaric work is p delta V, isothermal work is n R T log V2 over V1, and adiabatic work stays inside as temperature change.

More to read

3.6 Adiabatic Processes for an Ideal Gas

An adiabatic change is the lonely one: no heat crosses the boundary, so the gas pays for its own expansion. As it pushes outward it spends internal energy, and the temperature falls. Squeeze it instead and the work you do on it becomes internal energy, so the temperature climbs. This is why a bicycle pump warms in your hands and why spray cans chill as they empty.

The mathematics captures this with the adiabatic condition. Pressure times volume raised to the power gamma stays constant, where gamma is the ratio Cp over Cv. That exponent is not a fudge factor: it comes straight from the degrees of freedom, equal to 5/3 for a monatomic gas. Steeper than an isotherm on the PV diagram, the adiabat is the signature curve of insulated change.

15.2 The First Law of Thermodynamics and Some Simple Processes

Four ideal processes form the complete toolkit, and each holds one variable still. Isobaric keeps pressure fixed, so the work is a simple rectangle of pressure times volume change. Isochoric locks the volume, killing all boundary work so heat maps directly onto internal energy. Isothermal pins the temperature with help from a heat bath, and adiabatic seals off heat entirely.

Comparing them side by side trains your PV diagram instincts. From one starting point, an isothermal expansion glides along its hyperbola while an adiabatic expansion dives more steeply below it, ending colder. Compress instead and the adiabat soars above the isotherm, ending hotter.

3.2 Work, Heat, and Internal Energy

Work is the area under the path, and the three expansion styles give three different areas to integrate. For constant pressure the integral is trivial: pressure steps outside and multiplies the volume change. For isothermal expansion of an ideal gas, pressure falls as one over volume, and integrating gives work proportional to the logarithm of the volume ratio. Each doubling of volume adds the same chunk of work.

Adiabatic work needs the temperature change instead. With no heat allowed, the first law says work out equals internal energy lost, so measuring the temperature drop tells you the work directly. This is the practical payoff: for insulated processes you can skip the integral entirely and read the answer off the thermometer.

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Work in Isothermal, Isobaric and Adiabatic Processes · Science, ages 19 to 20 · LightMySky