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Work in reversible and irreversible changes

The same expansion can do very different amounts of work. Against a constant pressure, w = -p_ext ΔV. More stages give more work, and the reversible path gives the most. In compression the reversible path needs the least. An adiabatic change has q = 0, so ΔU = w, and ΔH = ΔU + Δn_g RT links the two for reactions.

This is the first lesson of Chemical Thermodynamics: Numerical Practice, free for everyone to read.

The sign of work

In chemistry, $w$ is the work done on the system. When a gas expands, it pushes the surroundings back and does work on them, so $w$ is negative. When a gas is compressed, the surroundings do work on it, so $w$ is positive.

Work is not a state function. It depends on the path. That is the whole point of this lesson.

One stage against a constant pressure

If the gas pushes against a constant outside pressure $p_{\text{ext}}$ , the work is

\displaystyle w = -p_{\text{ext}}\,\Delta V

A handy conversion: $1\ \text{L bar} = 100\ \text{J}$ .

Worked example. $1.00\ \text{mol}$ of an ideal gas sits at $300\ \text{K}$ in $6.00\ \text{L}$ . It expands at constant temperature to $24.0\ \text{L}$ .

The pressures. $p = \dfrac{nRT}{V}$ with $R = 0.08314\ \text{L bar K}^{-1}\text{mol}^{-1}$ . At the start, $p_1 = 4.157\ \text{bar}$ . At the end, $p_2 = 1.039\ \text{bar}$ .
One stage. Drop the outside pressure at once to $1.039\ \text{bar}$ , and let the gas push against it all the way. $w = -1.039 \times 18.0 = -18.71\ \text{L bar} = -1871\ \text{J}$ .

If the outside pressure were zero, the gas would expand into a vacuum. Then $w = 0$ . This is called a free expansion.

Two stages do more

Now expand in two steps. First, let the gas push against $2.079\ \text{bar}$ from $6.00\ \text{L}$ to $12.0\ \text{L}$ . Then let it push against $1.039\ \text{bar}$ up to $24.0\ \text{L}$ .

Stage 1. $-2.079 \times 6.00 = -12.47\ \text{L bar}$ .
Stage 2. $-1.039 \times 12.0 = -12.47\ \text{L bar}$ .
Total. $-24.94\ \text{L bar} = -2494\ \text{J}$ .

The gas did more work, because for the first half it pushed against a higher pressure.

The reversible path does the most

Make the stages smaller and smaller. In the limit, the outside pressure stays only a tiny bit below the gas pressure at every moment. This is the reversible path. Its work is the whole area under the isotherm:

\displaystyle w_{\text{rev}} = -nRT\ln\dfrac{V_2}{V_1} = -2.303\,nRT\log\dfrac{V_2}{V_1}

For our gas: $w_{\text{rev}} = -1.00 \times 8.314 \times 300 \times \ln 4 = -3458\ \text{J}$ .

Three pressure against volume graphs, side by side, for 1.00 mol of ideal gas at 300 K expanding from 6.00 L to 24.0 L. Each shows the same blue isotherm falling from 4.16 bar to 1.04 bar. In the first, one step drops straight to 1.04 bar and runs across: the orange rectangle under it is the work, w = -1871 J. In the second, two steps, at 2.08 bar to 12.0 L and then at 1.04 bar, shade a larger area: w = -2494 J. In the third, the whole area under the curve is shaded green: the reversible work, w = -3458 J, the largest of the three.

On a graph of pressure against volume, the work done by the gas is the area under the path it pushes against. One stage covers a rectangle. Two stages cover more. The reversible path covers everything under the curve.

Compression turns the rule around

Now push the gas back from $24.0\ \text{L}$ to $6.00\ \text{L}$ at $300\ \text{K}$ .

In one stage, the outside pressure must be at least the final pressure, $4.157\ \text{bar}$ , from the start. $w = -4.157 \times (6.00 - 24.0) = +74.83\ \text{L bar} = +7483\ \text{J}$ .
Reversibly, $w_{\text{rev}} = +3458\ \text{J}$ .

So for an expansion, the reversible path gives the most work out. For a compression, it needs the least work in.

Isothermal changes of an ideal gas

The internal energy of an ideal gas depends only on its temperature. At constant temperature, $\Delta U = 0$ by every path. The first law then gives $q = -w$ .

In the reversible expansion above, the gas took in $q = +3458\ \text{J}$ of heat and did exactly that much work.

Adiabatic changes in outline

An adiabatic change lets no heat in or out: $q = 0$ . The first law then gives

\displaystyle \Delta U = w

When a gas expands adiabatically, it does work, so $w < 0$ . Its internal energy falls, and so does its temperature. That is why air cools as it rises and expands.

For an ideal gas, $\Delta U = nC_V\Delta T$ . A monatomic gas has $C_V = \dfrac{3}{2}R = 12.47\ \text{J K}^{-1}\text{mol}^{-1}$ .

Worked example. $2.00\ \text{mol}$ of a monatomic ideal gas expands adiabatically and does $1.20\ \text{kJ}$ of work. By how much does it cool?

The first law. $w = -1200\ \text{J}$ , so $\Delta U = -1200\ \text{J}$ .
The temperature. $\Delta T = \dfrac{\Delta U}{nC_V} = \dfrac{-1200}{2.00 \times 12.47} = -48.1\ \text{K}$ .

A free expansion into a vacuum is a special case. Then $q = 0$ and $w = 0$ , so $\Delta U = 0$ and an ideal gas keeps its temperature.

Delta H and Delta U for a reaction

A bomb calorimeter keeps the volume fixed, so it measures $\Delta U$ . Most tables list $\Delta H$ , the heat at constant pressure. For a reaction of ideal gases:

\displaystyle \Delta H = \Delta U + \Delta n_g RT

Here $\Delta n_g$ is the moles of gaseous products minus the moles of gaseous reactants. Solids and liquids do not count.

Worked example. Propane burns in a bomb at $298\ \text{K}$ with $\Delta U = -2213\ \text{kJ mol}^{-1}$ :

\displaystyle \text{C}_3\text{H}_8(\text{g}) + 5\,\text{O}_2(\text{g}) \rightarrow 3\,\text{CO}_2(\text{g}) + 4\,\text{H}_2\text{O}(\text{l})

Count the gas. $\Delta n_g = 3 - (1 + 5) = -3$ . The water is liquid.
$RT$ in kilojoules. $8.314 \times 298 \div 1000 = 2.478\ \text{kJ mol}^{-1}$ .
Add. $\Delta H = -2213 + (-3)(2.478) = -2220\ \text{kJ mol}^{-1}$ .

Try it in Python

The program expands the same gas in $1$ , $2$ , $4$ , $10$ and $100$ stages, and compares each with the reversible work. Each stage ends at a volume a fixed factor larger than the last.

Common mistakes

Forgetting the sign. Expansion makes $w$ negative; compression makes it positive.
Using the gas pressure in $w = -p_{\text{ext}}\Delta V$ . It is the outside pressure that counts.
Mixing $\ln$ and $\log$ . $\ln x = 2.303\log x$ .
Leaving $RT$ in joules when $\Delta U$ is in kilojoules. Divide by $1000$ .
Counting liquids in $\Delta n_g$ . Only gases count.
Thinking reversible means fast. A reversible change is infinitely slow, through states that are always at equilibrium.

Remember these

One stage: $w = -p_{\text{ext}}\Delta V$ , with $1\ \text{L bar} = 100\ \text{J}$ . A free expansion has $w = 0$ .
Reversible isothermal: $w = -nRT\ln\dfrac{V_2}{V_1}$ .
Expansion: the reversible path gives the most work out. Compression: it needs the least work in.
Isothermal ideal gas: $\Delta U = 0$ , so $q = -w$ .
Adiabatic: $q = 0$ , $\Delta U = w$ ; an expanding gas cools.
$\Delta H = \Delta U + \Delta n_g RT$ , counting gases only.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

A gas expands from $2.0\ \text{L}$ to $7.0\ \text{L}$ against a constant outside pressure of $3.0\ \text{bar}$ . What is $w$ ?

Which needs less work to compress a gas from $10\ \text{L}$ to $5\ \text{L}$ at constant temperature: one stage, or a reversible path?

For $\text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\,\text{HCl}(\text{g})$ , how do $\Delta H$ and $\Delta U$ compare?

A gas expands adiabatically into a vacuum. What are $q$ , $w$ and $\Delta U$ ?

Liked this lesson? There is more

Chemical Thermodynamics: Numerical Practice is a Standard course. It adds 2 more lessons, practice and tests on top of the free chapter.

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Enthalpy cycles in practice: Hess's law lets you find any enthalpy you cannot measure from ones you can. From combustion enthalpies you get enthalpies of formation. Comparing measured values with bond enthalpies gives the resonance energy of benzene. A Born-Haber cycle gives the lattice enthalpy of a salt, and neutralisation shows what it costs to ionise a weak acid.