work is equal to negative (pressure times volume change). We write this as W = -PdeltaV.
The system is performing work on the surroundings when the volume of the system expands (meaning deltaV is positive, so work of the system is negative, and the work is leaving the system to go to the surroundings.
The opposite happens when the volume of the system decreases. In this case, the surroundings are putting work into the system for deltaV to be negative.
From the examples, it looks like all of these are chemical equations involving some kind of gas. We might not get volume changes directly, but we can get moles of gas from it.
Recall from the ideal gas law PV = nRT. So V is proportionate to n. Any increase in moles of gas will then correspond to an increase in volume.
Let's start with 1:
Hg(l) -> Hg(g). In this situation, we have the evaporation of liquid helium into a gas. So the volume of the system is increasing.
W = -PdeltaV. If Vfinal > V init, then delta V is positive. So W is negative, and the system is doing work on the surroundings.
2.
3O2(g) —> 2O3(g). Now oxygen is being converted to ozone. Find the moles of gas in the products and the reactants side.
Reactants side (n_gas init) = 3
Products side (n_gas final) = 2
So the number of moles of gas is decreasing, meaning the volume is also decreasing, or deltaV is negative.
If W = -PdeltaV, then a negative times a negative is a positive, and positive work is being done on the system. The surroundings are doing work on the system.
3.
H2(g)+F2(g) —> 2HF(g)
Moles of gas in reactants: 2
Moles of gas in products: 2
There is no change in moles of gas, so no change in volume. W = -P(0) = 0. No work is done by the system or the surroundings.
4.
CuSO4 x 5H2(s) —> CuSO4(s)+5H2O(g)
Work is being done by the system on the surrondings. Using the previous solutions, can you find out why?
