Work is the change in energy that occurs when an object is moved against a force. For example, if you lift an object of mass m against the force of gravity you will do work on the object

w_gravity = mgh

where h is the height you lift the object and g is the acceleration due to the earth's gravitational force (approximately 9.8 m/s²). There are a variety of ways of applying force and, therefore, different types of work.

TYPES OF WORK

In this table, the sign in the column for calculating work ensures that the work will be positive when it is done on the system. The 'd' indicates a small, differential change in the quantity. For example, dV means a very small change in V. To get the work, w, we have to integrate over the change that has occurred. If you know calculus, you will want to ask the starting point (V_initial), the ending point (V_final) and the pathway (how was the work done?). If you don't know calculus, hang on! We will show you some simplifications later in the course module.

There certainly are a lot of different kinds of work! And the table above is not an exhaustive listing! But you are in luck. If this were a physics course on mechanics, we might actually expect you to calculate work for some, if not all, of these processes. But this being a chemistry course on thermodynamics, it will only deal with one kind of work at this point. This is 'PV' or expansion work. The reason that chemistry focuses on 'PV' work is that it is nearly impossible to avoid in chemical reactions and phase transitions, which are naturally of interest to chemists.

To see why 'PV' work is ubiquitous in chemical reactions, consider the simple phase transition

The subscripts (l) and (g) refer to the phase of the water, which starts out as a liquid (l) and ends up as a gas (g). Under most conditions, gases take up a lot more room than do condensed phases (liquids or gases). Let's say we are vaporizing the water by boiling it on a stove. The water vaporizes into the room but the room isn't empty to start! There are air molecules above the pot of water and, to vaporize, the water must push against the air molecules.

If the pressure of the room does not change detectably when we boil the water (usually a pretty good approximation if the room is large relative to volume of water vapor created), we can assume that P is constant and equal to the atmospheric pressure. The constant pressure simplifies our calculation greatly

dw = -PdV

w = - PV = - P (V_gas - V_liquid)

So if we have a barometer or other method of measuring the atmospheric pressure and if we measure the volume of the water to start and the volume of the gas evolved we can find out how much energy was expended in the form of work.

As you can imagine, it might be difficult if not impossible to measure the volume of gas evolved since, being a gas, it does not sit in one nice lump waiting for us to measure its volume. It mixes with the air above it and disperses into the room. To make this measurement more practical, we can also make the assumption that the volume of the gas is much bigger than the volume of the liquid and then use the ideal gas law to calculate V from the temperature and pressure.

w = - PΔV

w = - P (V_gas - V_liquid)

w = - P V_gas

w = - P nRT/P = -nRT

Since pressure and volume are both positive quantities, work is negative. The system (your water) has lost energy because it had to do the work of pushing back the air.

'PV' work does not have to result in an expansion; the system could be compressed [see slide 4] instead. In this case, work would be done on the system because ?V would be negative, which would cancel the negative sign on the formula for 'PV' work. For chemical reactions, the assumption that gas volumes will be much greater than liquid or solid volumes is commonly made. To calculate work for a particular chemical reaction using this assumption, you need only subtract the initial amount of gas present (reactants) from the amount of gas that forms (reactants) So, for example, the following chemical reaction

H_{2 (g)} + 1/2 O₂ (g) ---> H₂O _(g)

we have 1.5 moles of gas reactants (1 mole H₂ + 1/2 mole O₂) forming 1 mole of gas products (1 mole H₂O). If we carry out this reaction at a constant pressure (say, atmospheric pressure on top of a lab bench) and 298.0 K, we can calculate the work in the following manner

Note that work is done on the system (w = +) because the final volume of gases is less than the initial volume. It is always useful to check that sign of the work you calculated agrees with your idea of what is going on. In using this method, make sure that you are carrying out the reaction isobarically (at constant pressure) and that you include only the number of moles of gases in your calculation.

If your reaction is not carried out at constant pressure, you are in for a bit more effort in calculating the work associated with the reaction. You will first need to determine how the pressure changes. This will depend upon how you carry out the process. If the reaction is performed isochorically (no volume change) as might occur in a sealed glass jar for reactions involving gases or for a reaction that occurs entirely in the solid or liquid states, you are in luck. If there is no change in volume, then there is no 'PV' work done and w = 0.

Reversible [slides 18 & 19]changes are ones in which the expansion or compression is carried out very slowly so that the pressure that your system is expanding against (or is being compressed by) always equals your system pressure. The concept of reversible changes will be very important in our discussion of entropy and we will get back to it there. To perform the calculation for the work involved, you can generally still assume solids and liquids don't change volume much and therefore are not involved in 'PV' work. Calculations of irreversible work on or by gases requires calculus. An example calculation.

There are an infinite number of irreversible pathways for carrying out expansions or compressions, and to learn more about them you can visit . One thing to note is that doing something irreversibly always costs you energy. You do less work by the system in an expansion and need to do more work on a system in a compression if you do things irreversibly compared to the reversible case. In the simplest case where the expansion/compression is carried out by expanding/contracting against a constant pressure P_ex

w = -P_exΔV

To find teacher's aids on work, you might try some of the mechanical equivalent of heat demos.