The most common work in chemistry is a gas pushing back its surroundings, as when a reaction makes gas in an open flask. The First Law and Sign Conventions needs its sign, and the The Ideal Gas Law gives its size.
Key idea
A gas that changes volume against a constant outside pressure does work $$w = -P_\text{ext}\,\Delta V.$$ Expansion ($\Delta V > 0$) gives $w < 0$: the system spends energy pushing back the surroundings. Compression gives $w > 0$. With $P$ in atm and $V$ in L, $w$ comes out in L·atm; convert with $1\ \text{L·atm} = 101.3\ \text{J}$. Expanding into a vacuum ($P_\text{ext} = 0$) does no work.
Why it works
Work is force times distance. The outside pressure pushes on a piston of area $A$ with force $P_\text{ext}A$; if the piston moves out a distance $d$, the gas does work $P_\text{ext}Ad = P_\text{ext}\Delta V$ on the surroundings. Work done on the gas is the negative of that, hence the minus sign. For a reaction at constant temperature and pressure that changes the moles of gas by $\Delta n_\text{gas}$, the ideal gas law gives $P\Delta V = \Delta n_\text{gas}RT$, so $w = -\Delta n_\text{gas}RT$.
Try it in the Science Lab's First Law chapter. Remember that it writes $\Delta U = Q - W$ with $W$ done by the gas, so its $W$ is our $-w$.
Worked example: An expanding gas
A gas expands from $2.00\ \text{L}$ to $6.00\ \text{L}$ against a constant $1.50\ \text{atm}$. Find $w$ in joules.
- $\Delta V = 6.00 - 2.00 = +4.00\ \text{L}$.
- $w = -P\Delta V = -(1.50)(4.00) = -6.00\ \text{L·atm}$.
- $-6.00 \times 101.3 = -608\ \text{J}$.
Answer: $-608\ \text{J}$
Worked example: Heat and work together
A gas absorbs $500\ \text{J}$ of heat while expanding from $2.0\ \text{L}$ to $5.0\ \text{L}$ against $1.00\ \text{atm}$. Find $\Delta U$.
- $w = -(1.00)(3.0) = -3.0\ \text{L·atm} = -304\ \text{J}$.
- $\Delta U = q + w = 500 - 304 = +196\ \text{J}$.
Answer: $+196\ \text{J}$
Worked example: A reaction that makes gas
At $25^\circ\text{C}$ and constant pressure, $\mathrm{CaCO_3(s)} \to \mathrm{CaO(s)} + \mathrm{CO_2(g)}$ makes 1 mol of gas per mole of reaction. Find $w$.
- $\Delta n_\text{gas} = +1$.
- $$w = -\Delta n_\text{gas}RT = -(1)(8.314)(298.15) = -2479\ \text{J} = -2.48\ \text{kJ}$$
Answer: $-2.48\ \text{kJ}$ per mole of reaction.
Common mistake
Dropping the minus sign, so an expanding gas seems to gain energy from its own expansion, or forgetting to convert L·atm to joules before adding to a heat in joules. Check: expansion always makes $w$ negative; and an answer in L·atm must be multiplied by 101.3 before it meets $q$.