# Work done on a Gas?

I'm extremely confused about work done on a gas. So if you look at the system pictured,

From what I've read online, the net work done, which is equal to the work done by the objects, is equal to the pressure times the change in volume, $$W=p\,\Delta V$$.

But I'm confused on is how can this be true. Why wouldn't the net work done on the gas be equal to the work done on both the gas and the atmosphere, which is also applying a force on the gas? It doesn't make sense that the net work being done on the gas is equal to only the work being done by the objects.

Thanks, I hope I explained my issue correctly here.

• The pressure doing the work is the sum of the atmospheric pressure plus the effect of the additional masses present on the piston. – Maurice Oct 9 '20 at 14:21
• When you do a force balance on an object A that is acted upon by a force from an object B, do you include both the force that B exerts on A and the equal and opposite force that A exerts on B, or just the force that B exerts on A? – Chet Miller Oct 10 '20 at 11:30

You have differential equations.
Differential equations are ALL about the initial conditions, and A PROCESS.

• So, what’s the initial condition?

You have a piston with a given volume of gas.

• What’s happening then?

I apply an additional force with an object. (The atmospheric pressure already was there)

And what if you want to know the work made by the atmosphere?

No problem. You have to start with a very large volume and start to add the gravitational force. Performing a PdV :-)
Do what it works best to you. (Nerd Joke)

• Yeah, I actually agree with this answer. Your states can only be assigned at equilibrium. Your initial state must be such that p_gas = p_atm (or else not at mechanical equilibrium, states not assigned) . Then you apply external pressure, and it is that extra pressure that does work. If p_atm is to be taken into account, just leaving the gas there doing nothing will have non-zero work done which does not make sense. – TheLearner Oct 10 '20 at 9:20

The total work ( and its value for the constant force case ) for pressure $$p=p_\mathrm{force} + p_\mathrm{atm}$$

$$W_\mathrm{tot} = - \int_{V1}^{V2}{p \cdot \mathrm{d}V}$$

is shared between the source of the explicit mechanical force acting on piston and atmosphere. Atmosphere would do work

$$W_\mathrm{atm} = - \int_{V1}^{V2}{p_\mathrm{atm} \cdot \mathrm{d}V}$$

and the force acting on the piston

$$W_\mathrm{force} = - \int_{V1}^{V2}{p_\mathrm{force} \cdot \mathrm{d} V}$$

If we consider a simple case of constant forces, expressions are simple:

$$W_\mathrm{tot} = -p \cdot \Delta V$$

$$W_\mathrm{atm} = -p_\mathrm{atm} \cdot \Delta V$$

$$W_\mathrm{force} = - p_\mathrm{force} \cdot \Delta V = -\frac FA \cdot \Delta V$$

• Why did you subtract the $p-p_{atm}$ in the final expression? – Buraian Oct 9 '20 at 22:09
• Is not it obvious ? Part of the work does the atmosphere and part is done by the local pressure on the piston. Without proper thinking, only the latter seems to happen – Poutnik Oct 9 '20 at 22:41

Why wouldn't the net work done on the gas be equal to the work done on both the gas and the atmosphere, which is also applying a force on the gas?

Your phrasing is not quite correct: The work done on the gas in the cylinder is equal to the work done by the atmosphere (if any), plus the work done by the lowering of the weights (if any).

I.e., the gas is part of the system, and both the weights and the atmosphere are part of the surroundings. The work done on the system is done by the surroundings. Hence the work done on the gas in the cylinder is the sum of the work done by the weights and the atmosphere.

[It is not, as your wrote, the work done on both the gas and the atmosphere, because the gas is part of the system, and the atmosphere (like the weights) is part of the surroundings.]

Thus:

$$\mathrm{w} = - \int_{V_{i}}^{V_{f}}{p_\text{ ext} \cdot \mathrm{d}V},$$

where $$p_\text{ ext} = p_{\text{ from atmosphere}} + p_\text{ due to weights}.$$