Thermodynamic processes

Quasi-static and isothermal processes

Narrated video. Transcript:

Slam a piston into a cylinder of gas. Near the piston the gas is squeezed and hot, while the far end has not even noticed yet. For a moment the gas has no single pressure or temperature.

Now push the piston in tiny steps, waiting each time for the gas to settle. At every instant the gas is almost in equilibrium, with a well-defined p and T. This is a quasi-static process.

A truly quasi-static process is an idealisation: it would take forever. But a slow enough real process comes very close, and only quasi-static processes can be drawn as a smooth curve on a P-V diagram.

First, the isothermal process: the temperature stays fixed. Put the cylinder, with conducting walls, on a huge reservoir at temperature T, and let the piston move slowly.

For an ideal gas at constant T, p V equals μ R T is constant. That is Boyle’s law. On the P-V diagram the path is a hyperbola called an isotherm.

The work done by the gas is the area under the isotherm. Adding up p ΔV over tiny steps, calculus gives W equals μ R T times the natural log of V₂ over V₁.

Because T is fixed, the internal energy of an ideal gas does not change: ΔU is zero. So by the first law, ΔQ equals ΔW. Every joule of heat the gas absorbs comes out as work.

Example: one mole at 300 K expands isothermally from 10 litres to 20 litres. W equals 8.314 times 300 times ln 2, about 1729 joules, and the gas absorbs exactly 1729 joules of heat.

Run it backwards, an isothermal compression, and work is done on the gas while it gives out the same amount of heat to the reservoir.

Adiabatic processes

Narrated video. Transcript:

Now wrap the cylinder in thick insulation. No heat can flow in or out. A process with ΔQ equal to zero is called adiabatic.

The first law becomes 0 equals ΔU plus ΔW. If the gas does work as it expands, that energy comes out of its internal energy, so the gas cools down.

Compress it instead, and the work done on the gas raises its internal energy. Pump up a bicycle tyre quickly and feel the pump get warm. A diesel engine squeezes air so hard that it becomes hot enough to ignite the fuel.

For a quasi-static adiabatic process of an ideal gas, p V to the power γ stays constant, where γ is C p over C v. For a monatomic gas, γ is 5 by 3; for air, about 1.4.

Since γ is greater than 1, the adiabat is steeper than the isotherm through the same point. In an adiabatic expansion the pressure falls faster, because the temperature is falling too.

The work done by the gas, from state 1 to state 2, is W equals μ R times T₁ minus T₂, divided by γ minus 1. Equivalently, p₁V₁ minus p₂V₂, over γ minus 1.

Example: one mole of a monatomic gas at 300 K is compressed adiabatically to half its volume. T V to the γ minus 1 is constant, so T₂ equals 300 times 2 to the power two-thirds, about 476 K.

The work done on the gas is 3 by 2 R times 176 kelvin, about 2200 joules. All of it is stored as extra internal energy, since no heat leaves.

Isochoric, isobaric and cyclic processes

Narrated video. Transcript:

Pin the piston so the volume cannot change. Heating at constant volume is an isochoric process. On the P-V diagram it is a vertical line.

There is no area under a vertical line, so no work is done. All the heat goes into internal energy: ΔQ equals ΔU, and the temperature and pressure rise together.

Now put a fixed load on a free piston, so the pressure stays the same. Heating at constant pressure is an isobaric process, a horizontal line on the diagram.

The area under it is a rectangle: W equals p times V₂ minus V₁, which for an ideal gas is μ R times T₂ minus T₁. Part of the heat raises U, and the rest does this work.

Finally, take the gas round a closed loop: expand along one path and come back along another to the starting state. This is a cyclic process.

Back at the start, the state is the same, so ΔU for the whole cycle is zero. The first law says the net heat absorbed equals the net work done.

The net work is the area enclosed by the loop. Going clockwise on the P-V diagram, the gas does more work expanding than is done on it during compression, so the net work by the gas is positive.

Let’s compare them from the same starting point. Isothermal: ΔU is zero. Adiabatic: ΔQ is zero. Isochoric: ΔW is zero. Isobaric: none is zero, but p is fixed.

Try it: the P-V process lab

Interactive simulation. Things to try:

• Pick a process and press Run process (or Space). The shaded area is the work done by the gas.
• Isothermal from 10 L to 20 L gives W = Q ≈ 1729 J and ΔU = 0.
• With all four curves shown, notice the adiabat is steeper than the isotherm.
• Compress (V₂ below 10 L): W becomes negative, meaning work is done on the gas.

Which path does the most work?

An ideal gas expands from the same initial state to the same final volume by three different quasi-static processes: isobaric, isothermal and adiabatic. In which process does the gas do the most work?

1. Isobaric
2. Isothermal
3. Adiabatic
4. The same in all three

Squeeze it fast

One mole of a monatomic ideal gas (γ = 5/3) at 300 K is compressed adiabatically to half its volume. What is its final temperature? (Take 2^(2/3) ≈ 1.59.)

1. 300 K
2. 378 K
3. 476 K
4. 600 K

Summary

• Isothermal (ΔU = 0): pV = const, W = Q = μRT ln(V₂/V₁)
• Adiabatic (ΔQ = 0): pV^γ = const, W = μR(T₁ − T₂)/(γ − 1)
• Isochoric (ΔW = 0): ΔQ = ΔU = μCᵥΔT
• Isobaric: W = p(V₂ − V₁) = μR(T₂ − T₁)
• Cyclic (ΔU = 0): ΔQ = ΔW = area of the loop

• A quasi-static process is infinitely slow, so the system is always in equilibrium; only such processes can be drawn on a P-V diagram.
• Work done by the gas = area under the P-V curve; positive for expansion, negative for compression.
• The adiabat is steeper than the isotherm (γ > 1). Adiabatic expansion cools a gas; compression heats it.
• In a cycle, the net work done by the gas is the area enclosed: positive when traced clockwise.