Adiabatic Processes

No heat in, no heat out

In an adiabatic process no heat crosses the boundary: \(Q = 0\). There are two ways to arrange this. The gas can be wrapped in good insulation, as in a vacuum flask. Or the change can be so fast that heat has no time to flow, as in a quick stroke of a pump. This lesson uses \(R = 8.31\,\text{J/(mol K)}\) and \(\gamma = C_p/C_v = 1.4\) for air.

With \(Q = 0\), the first law says \(W = -\Delta U\). A gas that expands adiabatically does positive work, and it can pay only from its own internal energy. So \(U\) falls and the gas cools. A gas compressed adiabatically has work done on it, and nowhere to lose the energy. So \(U\) rises and the gas warms. That is the pump and the valve.

For a quasi-static adiabatic change of an ideal gas, the pressure and volume follow a definite law. The derivation below combines the first law with the ideal gas equation.

Two more forms follow by putting \(p = nRT/V\) or \(V = nRT/p\) into \(pV^\gamma\). They are \(TV^{\gamma - 1} = \text{constant}\) and \(p^{1-\gamma}T^\gamma = \text{constant}\). The work done also has a neat form. Since \(W = -\Delta U = nC_v(T_1 - T_2)\) and \(C_v = R/(\gamma - 1)\), the work is \(nR(T_1 - T_2)/(\gamma - 1)\). With \(pV = nRT\) this is \((p_1V_1 - p_2V_2)/(\gamma - 1)\).

The adiabat is steeper

The curve \(pV^\gamma = \text{constant}\) is an adiabat. Through any state it falls more steeply than the isotherm. Take the change of \(pV = \text{constant}\) over a small step: \(p\,dV + V\,dp = 0\), so the isotherm's slope is \(dp/dV = -p/V\). The adiabatic law, \(dp/p + \gamma\,dV/V = 0\), gives a slope \(-\gamma p/V\). As \(\gamma > 1\), the adiabat is \(\gamma\) times as steep.

The physics is simple. In an isothermal expansion the pressure falls only because the volume grows. In an adiabatic expansion the gas also cools, and a cooler gas pushes less. Both effects lower the pressure, so it falls faster.

Play the chips of Fig 5.1. The isothermal gas takes in \(1728\,\text{J}\) of heat and turns it all into work. The adiabatic gas takes in nothing, so its \(1509\,\text{J}\) of work comes from its internal energy. Its temperature falls to \(300 \times 0.5^{0.4} = 227\,\text{K}\), and its molecules visibly slow down. In the both chip, the band between the two curves is the extra \(219\,\text{J}\) that the isothermal gas does.

The same two slopes set how hard a gas is to squeeze. The bulk modulus is \(B = -V\,dp/dV\). The isotherm gives \(B = p\), and the adiabat gives \(B = \gamma p\). For air at \(10^5\,\text{Pa}\) these are \(1.0 \times 10^5\,\text{Pa}\) and \(1.4 \times 10^5\,\text{Pa}\). A sound wave squeezes and releases the air so fast that heat cannot flow, so sound uses the adiabatic value. The unit on waves meets this as Laplace's correction.

A molecule meets a moving piston

The first law says a compressed gas warms. Kinetic theory shows how. Think of one molecule bouncing off a piston, like a cricket ball off a moving bat.

In the frame of the piston the bounce is elastic. The molecule arrives at \(400 + 100 = 500\,\text{m/s}\) and leaves at \(500\,\text{m/s}\). Back in the frame of the cylinder, add the piston's velocity: the molecule leaves at \(600\,\text{m/s}\). Each hit on a piston moving in adds twice the piston's speed. A piston moving out takes it away: the molecule returns at only \(200\,\text{m/s}\). The lower half of the figure shows a whole gas. Pushed in, its molecules speed up and its temperature rises; drawn out, they slow down.

Now the pump. A quick stroke squeezes the air from \(1\,\text{atm}\) to about \(3\,\text{atm}\) too fast for heat to escape. By \(p^{1-\gamma}T^\gamma = \text{constant}\), air starting at \(300\,\text{K}\) reaches \(T = 300 \times 3^{0.4/1.4} = 411\,\text{K}\). Between strokes this hot air gives its heat to the barrel. The barrel is far heavier than the air, so it warms only by a few degrees per stroke, but twenty strokes add up.

The valve works the other way. Air at \(3\,\text{atm}\) that rushes out to \(1\,\text{atm}\) expands adiabatically and cools. Ideally it would reach \(300 \times (1/3)^{0.4/1.4} = 219\,\text{K}\). In practice it mixes with room air at once, but it still feels cold on the hand. A punctured tyre hisses cold air for the same reason.

Why the mountains keep their snow

In June the plains of north India bake at \(40\,\text{°C}\), yet the high Himalaya stays white. The air is heated from below, by the ground. A warm parcel of air rises, and as it rises the pressure around it falls, as the barometric formula showed. So the parcel expands. Air is a poor conductor and the parcel is large, so almost no heat flows in: the expansion is adiabatic. The parcel cools as it climbs. This part uses \(g = 9.8\,\text{m/s}^2\), as the atmosphere needs, and \(M = 0.029\,\text{kg/mol}\) for air.

For dry air \(C_p = \tfrac72 R = 29.1\,\text{J/(mol K)}\). So the adiabatic lapse rate is \(Mg/C_p\) \(= 0.029 \times 9.8/29.1\) \(= 0.0098\,\text{K/m}\), about \(9.8\,\text{K}\) per kilometre. Dry air that leaves the plains at \(35\,\text{°C}\) would reach \(0\,\text{°C}\) after \(35/9.8 = 3.6\,\text{km}\).

The real atmosphere cools more slowly, about \(6.5\,\text{K}\) per kilometre on average. Rising air carries water vapour. When it cools enough, the vapour condenses into cloud and gives back its latent heat, which warms the parcel. At \(6.5\,\text{K}\) per kilometre, air at \(35\,\text{°C}\) reaches freezing near \(5.4\,\text{km}\). That is about where the summer snow line lies on the Himalaya.