Laws of Thermodynamics

Lessons

Senior 4 · Physics

Laws of Thermodynamics

5 lessons

Lesson 3 of 5

Adiabatic Changes of an Ideal Gas

25 min

Fire from a push

A fire piston is a narrow tube, closed at one end, with a tight-fitting piston. Put a wisp of cotton wool at the bottom and ram the piston down hard: the cotton bursts into flame, with no match and no spark. People in Southeast Asia made fire this way for centuries, and every diesel engine ignites its fuel the same way. The air is squeezed so fast that no heat can escape, so all the work done on it raises its internal energy, and its temperature shoots up by hundreds of degrees. A change in which no heat flows is called adiabatic. In this lesson you find the laws that adiabatic changes obey and use them to explain diesel engines, cool air from a tyre valve and the clouds that form over hills.

Definition

An adiabatic change is one in which no heat enters or leaves the system: $Q = 0$ . It happens when the system is perfectly insulated, or when the change is so fast that heat has no time to flow. By the first law, $\Delta U = -W$ : a gas that expands adiabatically does work using its own internal energy and cools; a gas compressed adiabatically warms. Example: if a gas does 400 J of work while expanding adiabatically, its internal energy falls by 400 J.

The adiabatic laws for an ideal gas

For a small adiabatic change of $n$ moles of ideal gas, $Q = 0$ and the first law gives

nC_V\,\mathrm{d}T = -p\,\mathrm{d}V = -\frac{nRT}{V}\,\mathrm{d}V \qquad\Rightarrow\qquad \frac{\mathrm{d}T}{T} = -\frac{R}{C_V}\,\frac{\mathrm{d}V}{V} = -(\gamma - 1)\frac{\mathrm{d}V}{V}

using $R = C_{p} - C_{V}$ and $\gamma = \frac{C_p}{C_V}$ . Integrating, $\ln T + (\gamma - 1)\ln V$ is constant, so

TV^{\gamma - 1} = \text{constant} \qquad\text{and, with } pV = nRT, \qquad pV^{\gamma} = \text{constant}

Between two states, $p_1V_1^{\gamma} = p_2V_2^{\gamma}$ and $T_1V_1^{\gamma - 1} = T_2V_2^{\gamma - 1}$ ; in terms of pressure, $\frac{T_2}{T_1} = \left(\frac{p_2}{p_1}\right)^{\frac{\gamma - 1}{\gamma}}$ . The work done by the gas is the fall in its internal energy:

W = -\Delta U = nC_V(T_1 - T_2) = \frac{p_1V_1 - p_2V_2}{\gamma - 1}

Because $\gamma \gt 1$ , the curve $pV^{\gamma} =$ constant, an adiabat, is steeper than the isotherm through the same point (Figure 3.1). In an isothermal expansion heat flows in to keep the gas warm; in an adiabatic one it does not, so the gas cools and its pressure falls further.

Figure 3.1: An isotherm and an adiabat through the same point A, for air (γ = 1.4). The adiabat is steeper.

Worked example 1: compressing air quickly

Air ( $\gamma = 1.40$ ) at $1.0 \times 10^{5}\,\mathrm{Pa}$ and 300 K is compressed adiabatically to one tenth of its volume. Find its new pressure and temperature.

p_2 = p_1\left(\frac{V_1}{V_2}\right)^{\gamma} = 1.0 \times 10^{5} \times 10^{1.40} = 2.5 \times 10^{6}\,\mathrm{Pa}, \qquad T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 300 \times 10^{0.40} = 754\,\mathrm{K}

The pressure rises 25 times, not 10 times as it would at constant temperature, and the air reaches 481 °C. That is hot enough to light cotton wool.

Worked example 2: the diesel engine

A diesel engine compresses air with a compression ratio of 18 ( $\frac{V_1}{V_2} = 18$ ), starting at 300 K. Assuming an adiabatic compression, find the final temperature.

T_2 = 300 \times 18^{0.40} = 953\,\mathrm{K} \approx 680\,^\circ\mathrm{C}

Diesel fuel sprayed into air this hot ignites by itself, so a diesel engine needs no spark plug. Petrol engines use much lower compression ratios, because a hot, compressed petrol–air mixture would ignite too early.

Worked example 3: work done in an adiabatic expansion

1.0 mol of a monatomic ideal gas expands adiabatically, and its temperature falls from 400 K to 300 K. Find the work done by the gas and its change in internal energy.

W = nC_V(T_1 - T_2) = 1.0 \times \tfrac{3}{2} \times 8.31 \times 100 = 1.25 \times 10^{3}\,\mathrm{J}, \qquad \Delta U = -1.25 \times 10^{3}\,\mathrm{J}

All the work comes from the gas's own internal energy, because no heat enters.

Adiabatic changes around us

Tyre valves and aerosol cans. Gas rushing out expands quickly against the air, does work and cools, which is why the valve of a tyre feels cold as it lets air out.
Bicycle pumps. Pumping hard compresses the air quickly, and the end of the pump warms up.
Weather and clouds. Air that rises, over a hill or in a warm updraft, moves into lower pressure and expands. Air conducts heat so poorly that the expansion is very nearly adiabatic, so rising dry air cools by about 10 °C for every kilometre it climbs (Figure 3.2). When it has cooled enough, its water vapour condenses into droplets: a cloud with a flat base forms. This adiabatic cooling is also one reason why high ground is generally cooler than low ground.
Sound. The compressions in a sound wave are too fast for heat to flow between them, so sound travels as a series of adiabatic changes.

Figure 3.2: Air rising up a hillside cools as it expands; a cloud forms where its vapour condenses.

Worked example 4: how fast does rising air cool?

For dry air rising slowly, the cooling rate is $\frac{g}{c_p}$ , where $c_p = 1005\,\mathrm{J\,kg^{-1}\,K^{-1}}$ is the specific heat capacity of air at constant pressure. Find the cooling per kilometre, and the cooling of air that rises 1.5 km.

\frac{g}{c_p} = \frac{9.8\,\mathrm{m\,s^{-2}}}{1005\,\mathrm{J\,kg^{-1}\,K^{-1}}} = 9.8 \times 10^{-3}\,\mathrm{K\,m^{-1}} = 9.8\,\mathrm{K\,km^{-1}}

Rising 1.5 km, dry air cools by about $9.8 \times 1.5 = 15\,\mathrm{K}$ . (Once a cloud forms, condensation releases latent heat and the cooling slows.)

Warning

Common mistakes. (1) Thinking "adiabatic" means "constant temperature": it means no heat flow, and the temperature usually changes a lot. (2) Using $pV^{\gamma} =$ constant with temperature-dependent quantities wrongly: use $TV^{\gamma-1}$ for temperatures and $pV^{\gamma}$ for pressures, with $T$ in kelvin. (3) Using $\gamma = \frac{5}{3}$ for air: air is mainly diatomic, $\gamma \approx 1.4$ . (4) Forgetting that a real compression is only approximately adiabatic: some heat always leaks.

Tip

Fast or insulated means adiabatic. Then $Q = 0$ , so $\Delta U = -W$ : squeeze it and it warms, let it expand and it cools. On a p–V graph the adiabat is steeper than the isotherm by the factor $\gamma$ .

Fire Syringe (Veritasium). Watch the cotton wool ignite when the air is compressed very quickly, and listen for the explanation: the work done on the air raises its temperature because there is no time for heat to escape.

✦

Practice

A monatomic ideal gas at 300 K and $1.0 \times 10^{5}\,\mathrm{Pa}$ expands adiabatically to twice its volume. Find its final temperature and pressure.
Air at 290 K and $1.0 \times 10^{5}\,\mathrm{Pa}$ in a bicycle pump is compressed quickly to $3.0 \times 10^{5}\,\mathrm{Pa}$ . Treating the compression as adiabatic ( $\gamma = 1.40$ ), estimate the final temperature.
A monatomic ideal gas occupying 1.0 L at $2.0 \times 10^{5}\,\mathrm{Pa}$ expands adiabatically to 3.0 L. Find the final pressure and the work done by the gas.
Air at $1.0 \times 10^{5}\,\mathrm{Pa}$ expands to twice its volume. Find the final pressure if the expansion is (a) isothermal, (b) adiabatic, and explain the difference.
Explain why the air coming out of a tyre valve feels cool.
Dry air rises 3.0 km up a mountainside. Using the result of Worked example 4, estimate how much it cools.
Starting from the first law with $Q = 0$ , show that $TV^{\gamma - 1}$ is constant in an adiabatic change of an ideal gas.
A diesel engine has no spark plug. Using the adiabatic laws, explain how its fuel ignites, and evaluate why its compression ratio must be much higher than that of a petrol engine.

Show answersHide answers

$T_2 = 300 \times 0.5^{\frac{2}{3}} = 189\,\mathrm{K}$ ; $p_2 = 1.0 \times 10^{5} \times 0.5^{\frac{5}{3}} = 3.1 \times 10^{4}\,\mathrm{Pa}$ .
$T_2 = 290 \times 3.0^{\frac{0.40}{1.40}} = 397\,\mathrm{K}$ , about 124 °C (in practice less, as some heat leaks to the pump walls).
$p_2 = 2.0 \times 10^{5} \times \left(\frac{1}{3}\right)^{\frac{5}{3}} = 3.2 \times 10^{4}\,\mathrm{Pa}$ ; $W = \frac{p_1V_1 - p_2V_2}{\gamma - 1} = \frac{200 - 96.2}{0.667} = 156\,\mathrm{J}$ .
(a) $5.0 \times 10^{4}\,\mathrm{Pa}$ ; (b) $1.0 \times 10^{5} \times 0.5^{1.40} = 3.8 \times 10^{4}\,\mathrm{Pa}$ . In the adiabatic case no heat enters to replace the work done, so the gas cools as well as expanding, and its pressure falls further.
The compressed air expands rapidly as it escapes, pushing back the atmosphere. There is no time for heat to flow in, so the work comes from its internal energy and its temperature drops.
About $9.8 \times 3.0 = 29\,\mathrm{K}$ , if it stays dry.
$nC_V\mathrm{d}T = -p\mathrm{d}V = -\frac{nRT}{V}\mathrm{d}V \Rightarrow \frac{\mathrm{d}T}{T} = -(\gamma - 1)\frac{\mathrm{d}V}{V}$ ; integrate: $\ln T + (\gamma - 1)\ln V =$ constant, so $TV^{\gamma - 1} =$ constant.
The piston compresses air so quickly that the compression is almost adiabatic: with a ratio of 18, $T$ rises from about 300 K to about 950 K, above the ignition temperature of diesel fuel, which ignites as soon as it is sprayed in. A petrol engine compresses the fuel and air together, so it must keep the temperature below the point where the mixture ignites by itself (which causes "knocking"); it uses a lower ratio and lights the mixture with a spark at the right moment. A high ratio is also what gives the diesel engine its higher efficiency (Lesson 4).

Quick check

Key points

Adiabatic: $Q = 0$ , so $\Delta U = -W$ ; expansion cools a gas, compression warms it.
Ideal gas: $pV^{\gamma} =$ constant and $TV^{\gamma - 1} =$ constant, with $\gamma = \frac{C_p}{C_V}$ ( $\frac{5}{3}$ monatomic, about 1.4 for air).
Work done $W = nC_V(T_1 - T_2) = \frac{p_1V_1 - p_2V_2}{\gamma - 1}$ ; an adiabat is steeper than an isotherm.
Examples: fire pistons and diesel ignition, cool tyre valves, rising air cooling about 10 °C per kilometre and forming clouds.

Sources and credits

Photo: Fire piston by Chocolateoak, Wikimedia Commons, licensed under CC BY-SA 3.0.
Photo: Cumulus clouds, NOAA Photo Library, National Severe Storms Laboratory, Wikimedia Commons, public domain.
Video: Fire Syringe, Veritasium, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagrams, worked examples and practice exercises. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Fire from a push

2The adiabatic laws for an ideal gas

3Adiabatic changes around us

Fire from a push

The adiabatic laws for an ideal gas

Adiabatic changes around us