Thermodynamics & kinetic theory
The idea
Thermodynamics tracks energy at the macroscopic level, and kinetic theory explains it from below: temperature measures the average translational kinetic energy of molecules, (3/2)kT per molecule for an ideal gas, and pressure is the drumbeat of molecular collisions on the walls. The first law, ΔU = Q − W, is conservation of energy with bookkeeping conventions: Q is heat added to the gas and W is work done by the gas as it expands.
Work is where calculus enters: W is the integral of p dV along the actual path from the initial to the final volume — the area under the process curve on a pV diagram. That is why work and heat are path-dependent while internal energy is not: U of an ideal gas depends only on temperature. An isothermal expansion keeps T (and therefore U) fixed, so every joule of work the gas does must be imported as heat.
Do not conflate heat and temperature. A gas can absorb heat with no temperature change (isothermal expansion) or heat up with no heat input at all (rapid adiabatic compression — a bicycle pump warms because work is done on the gas). Temperature is a state of the gas; heat is energy in transit.
Worked example
Two moles of an ideal gas expand isothermally at 300 K from 10 L to 25 L. Using R = 8.314 J/(mol·K), find the work done by the gas, the change in internal energy, and the heat absorbed.
- Set up the work integral: W = integral of p dV, and on an isotherm p = nRT/V, so W = nRT times the integral of dV/V from V₁ to V₂, which evaluates to nRT ln(V₂/V₁).
- Insert numbers: nRT = 2.0 × 8.314 × 300 = 4988 J, and ln(25/10) = ln 2.5 ≈ 0.916.
- Multiply: W = 4988 × 0.916 ≈ 4.6 × 10³ J — positive, as expected for an expansion pushing on its surroundings.
- Internal energy of an ideal gas depends only on temperature, and T is constant on an isotherm, so ΔU = 0.
- Apply the first law: Q = ΔU + W = 0 + 4.6 kJ, so the gas absorbs 4.6 kJ of heat — every joule of work output was paid for by heat flowing in, none by cooling.
Answer. The gas does about 4.6 kJ of work, its internal energy is unchanged, and it absorbs about 4.6 kJ of heat.
Check your understanding
- Why is work path-dependent on a pV diagram while internal energy change is not?
- How would the work compare if the same expansion happened adiabatically instead — and why must the gas cool in that case?
- What does kinetic theory say doubling the absolute temperature does to the average molecular speed?
- How would you explain to a friend the difference between heat and temperature using the isothermal expansion as the example?
Build the foundations first
Thermodynamics & kinetic theory builds on these concepts. If any feel shaky, start there.