Work done by the System

In thermodynamics work is defined as $~-p_{ex}\Delta V~$ for an ideal gas. If the total work is a negative value, that is in case of expansion of gas, we say that the work is done by the system and is negative.

I don't understand that why we use the pressure exerted on the gas in expression of work that is the external pressure and say that the work is done by the system ? If work is done by the system we should use internal pressure and if we use that in case of expansion the work done by the system will be positive which is contradicting. Please provide an explanation

• There is also an issue here with sign convention. Some developments use the symbol w to represent the work done by the system on the surroundings, and others use the symbol w to represent the work done by the surroundings on the system. In the first case, $\Delta U= q - w$, while in the second case, $\Delta U= q + w$. In working with the first law, one must carefully specify which sign convention is being used. – Chet Miller Sep 14 '15 at 0:00

Well, think of a gas that is expanding against a piston shown below:

We want to find the total work which is given by: $$w = F\Delta d$$ So we need to know what is the force that is applied by the gas onto the piston. To do this, we must realise that, the force that is stopping the gas from expanding freely into the atmosphere is the external atmospheric pressure. It is this pressure which is pushing downwards onto the gas, preventing it from expanding. Therefore, the force that is required for the gas to expand must be equal to the total pressure pushing down it and has nothing to do with its internal pressure.

Consider this example. I want to push a box across my room which is covered in carpet. To move the box, obviously I need to apply a force to it. The force that I need to apply to must be equal or greater than the resisting force which is the friction between the box and carpet. This force that is required has nothing to do with the speed of the box.

So now that we know the force, we get that $$w = -p_{ex}A\Delta d$$ but the area times the displacement is actually equal to the volume, therefore we get: $$w =-p_{ex}\Delta V$$

So, as you can see, we have derived the equation for work done by a gas. The main thing that you need to recognise is that the internal pressure is irrelevant when considering the force required for the gas to expand. All you need to consider is the force that is preventing it from expanding which is the external pressure.

• If force by the gas is equal to the external force how can the gas expand – ashwini abhishek Sep 13 '15 at 5:05
• Why negative sign is added before external pressure? The direction of force by gas is in same direction as the the displacement of the piston – ashwini abhishek Sep 13 '15 at 5:12
• Yes that is true, the direction of the force and displacement are in the same direction. But the actual magnitude of the force is negative. This is because the force applied by the gas is negative of the value of the external pressure (hence a negative number) – Nanoputian Sep 13 '15 at 5:38
• Another way of looking at it is, work can be thought as the change in kinetic energy. So, when a gas (the system expands, it obviously needs to exert a force to do so. Therefore, it is using up its own energy so that it can expand against the external pressure and hence it loses kinetic energy. Therefore, its work will be negative. – Nanoputian Sep 13 '15 at 5:40
• Nanoputian says that "But the actual magnitude of the force is negative" but the magnitude of a vector can't be negative – ashwini abhishek Sep 13 '15 at 6:05

This is an argument I have with the sciences. An expanding system does work onto the surrounding atmosphere. Consider the expanding system as a subsystem of our atmosphere, hence as it expands it does work onto the surrounding atmosphere i.e. it is as if the atmosphere has expanded, hence the work onto the atmosphere is [(Psys)(dVsys)]. Since the system's final pressure is 1 atm and the volume change is such that (dVsys)=(dVatm), we can write that the work done onto the atmosphere is [(Patm)(dV)atm].

The confusion originates from not writing the enthalpy equation correctly because simply writing H=E+PV does not tell us what is happening where. So clarity is given by writing

dH=(dE)sys + (PdV)atm and since the work is done exterior to the system then we can rewrite it as dH=(dE)sys-(PdV)sys But the above can be confusing. Imagine we are heating water bring it to a boil, and dQ is the energy in (due to heating. Now we would write

dQ= (dE)sys + (PdV)atm = (dE)sys + (W)atm

The reason is that the heating must do two things 1) change the system's internal energy 2) do work onto the surrounding atmosphere

The above equation is really latent heat of vaporization with the systems energy change being it changes in the bonding energy i.e. (dE)sys = (dU)sys

Now I argue (this is published in peer review [1,2,3] but remains controversial) that the work done onto the surrounding atmosphere: (W)atm = (PdV)atm is lost work because it is irreversible i.e. in order for the atmosphere to work on a system, then the atmosphere must be at a higher pressure than the system. This is controversial because it can become a challenge to the second law, namely lost work into the atmosphere means that the second law is no longer necessary to explain why we cannot have perpetual motion, or why energy is lost by expanding systems like a steam engine or why the Carnot cycle is ideal rather than realistic.

[1] Mayhew, K.W. "Second Law and Lost Work" Physics Essays 28,1 (2015) pg 152-56 [2] Mayhew, K.W. "Resolving Problematic Thermodynamics" Hadronic Journal vol 41 no 3 (2018) pg259

• Hi Kent, I've removed the link to your website because it seemed like the purpose of the post was advertising for the blog rather than answering the question. – jonsca Mar 28 at 3:06