DPST1021 Chap.11 The First Law and Thermodynamic Processes
The First Law and Thermodynamic Processes
Lectures 22 and 23, taught in week 9, turn the kinetic picture of a gas into a working accounting system for energy. Internal energy is stored in molecular motion, and it changes only when heat crosses the boundary or work is done on the gas. The first law states this as ΔEint = Q + W, and the course formula sheet defines W as the negative integral of P dV, so compression counts as positive work.
The set textbook uses work done by the gas instead, so holding one convention is the first skill.
Pressure-volume diagrams carry the rest of the chapter. The area under a path gives the size of the work. Two routes between the same states can enclose different areas, so heat and work depend on the route while internal energy does not.
Molar specific heats follow: CV equals f/2 times R, CP exceeds it by exactly R, and their ratio γ sets the shape of an adiabatic curve through PVγ = constant.
Lecture 23 names four idealised processes: isobaric, isovolumetric, isothermal and adiabatic.
For each you should be able to draw the curve, say what is held fixed, and give Q, W and ΔEint in terms of n, R and the temperature change. Everyday cases such as a petrol engine show which ideal applies, and Test 2 problems often chain the gas law with the first law over a cycle.
What this chapter covers
- 01
First law with course signs: write ΔEint = Q + W with W done on the gas, and convert textbook equations written with work done by it.
- 02
Work as area: compute W = −∫P dV from a PV diagram, using rectangles, trapezia or a logarithm, with the correct sign.
- 03
Path dependence: explain why Q and W differ between routes joining two states while ΔEint depends only on the temperatures at the ends.
- 04
Molar heat capacities: derive CV = fR/2 and CP = CV + R, and infer degrees of freedom from measured heat capacities.
- 05
Four standard processes: state what isobaric, isovolumetric, isothermal and adiabatic changes hold fixed and write Q, W and ΔEint for each.
- 06
Adiabatic curves and engines: apply PVγ = constant to rapid compressions and expansions, such as the strokes of a combustion engine.
Worked example · free
Slow isothermal squeeze of a gas sample
- 1Pressures from P = nRT/V: Pi = 0.200 × 8.314 × 300 / 4.00 × 10−3 = 125 kPa, and halving the volume at fixed T doubles it to 249 kPa.
- 1Work on the gas: W = −∫(nRT/V) dV = nRT ln(Vi/Vf) = 0.200 × 8.314 × 300 × ln 2 = +346 J. Compression makes it positive.
- 1Temperature is unchanged, so ΔEint = (f/2)nRΔT = 0 whatever the value of f.
- 1First law: Q = ΔEint − W = 0 − 346 = −346 J, so 346 J of heat leaves the gas.
Key terms
- Internal energy
- The total kinetic and potential energy of the molecules in a system; for an ideal gas it depends only on temperature.
- State function
- A property fixed by the current state alone, such as P, V, T or Eint, independent of history.
- Isovolumetric process
- A constant-volume change, also called isochoric, in which W = 0 and all heat becomes internal energy.
- Adiabatic process
- A change with no heat transfer, so the work done on the gas equals its change in internal energy.
- Molar specific heat
- The heat needed to raise one mole of a substance by one kelvin, written CV or CP for a gas.
- Adiabatic index
- The ratio γ = CP/CV, equal to 5/3 for a monatomic gas and 7/5 for a diatomic gas near room temperature.
The First Law and Thermodynamic Processes FAQ
Is heat the same thing as internal energy?
No. A system stores internal energy; heat is energy in transit across a boundary because of a temperature difference. A gas can gain internal energy with no heat at all, as in a rapid compression, and it can absorb heat with no internal energy change, as in a slow isothermal expansion.
Why is C_P larger than C_V for a gas?
Heated at constant pressure, a gas expands and pushes back its surroundings, so part of the heat leaves as work. For one mole and one kelvin that work equals R, which is exactly the gap between the two heat capacities. Solids barely expand, so their two values almost coincide.
How can a gas get colder without losing any heat?
In an adiabatic expansion the gas does work on a piston or its surroundings while no heat flows in to replace the energy. That work comes straight out of internal energy, so the temperature falls. Rising air cools this way, which is why clouds form at altitude.
Can a gas change volume with no heat and no work at all?
Yes, in a free expansion. If a partition breaks and gas rushes into an evacuated space, nothing pushes back, so W = 0, and an insulated container gives Q = 0. With both zero, ΔE_int = 0, so an ideal gas ends at its original temperature. No PV curve can represent it.
Exam move
Settle the sign convention first: write the first law at the top of every solution in the course form, and practise converting three or four textbook examples into it until the swap is automatic. Next, sketch all four processes from one starting point on shared PV axes and mark the steepest curve. Build your own table of Q, W and ΔEint by deriving each entry from the first law.
Then work multi-stage cycles: tabulate P, V and T at every corner with the gas law before calculating any energy, and check that ΔEint sums to zero around the loop. Finish by linking CV back to the degrees of freedom from the previous chapter, since questions often hinge on choosing f correctly for a diatomic gas.
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