# Work equation in thermodynamics

As known, work done on the system is: $$\mathrm{d}W = -P_{\text{ex}}\mathrm{d}V$$ However, I have some ambiguity about this equation:

1. Why is there $P_{\text{ex}}$? Pressure of a gas contained in a cylinder is by definition the force that gas exerts on a piston divided by a area of a piston. But, by the Newton's third law piston exerts the same force on a gas, and so, why is the infinitesimal work not $\mathrm{d}W = -P\mathrm{d}V$?

2. Some can argue, that we use whatever pressure the thing doing work is pushing against. But I don't accept it. It would be correct if I regarded a gas and a piston as a one system. However, system of interest is only a gas, on which a piston exerts the force with the same magnitude as a gas exerts on a piston, which is $P_{\text{gas}}$.

My calculations:

$$m\frac{\mathrm dv}{\mathrm d t} = F_{\text{ext}} - F_{\text{gas}}=S(P_{\text{ext}}-P_{\text{gas}}) = S\bigg(P_{\text{ext}} - \frac{nRT}{V_\mathrm{i}-Sx} \bigg)$$ $$mv\frac{\mathrm d v }{\mathrm dx} = S\bigg(P_{\text{ext}} - \frac{nRT}{V_i - Sx}\bigg)$$ After intergration: $$\frac{mv^2}{2} = \int_0^x{SP_{\text{ext}}\mathrm dx} - \int_0^x{\frac{SnRT \mathrm dx}{V_i - Sx} } = P_{\text{ext}}\Delta V - nRT\ln\bigg(\frac{V_\mathrm{i}}{V_\mathrm{f}}\bigg)$$ Rearrange, we get: $$P_{\text{ext}}\Delta V = W = \frac{mv^2}{2} + nRT\ln\bigg(\frac{V_\mathrm{i}}{V_\mathrm{f}}\bigg)$$ There, on the right side, first term is just an energy gain of a piston. The second term is familiar, it's work on the system in reversible process.

So work done on the system is the same as it would be if process was reversible. Why then infinitesimal work is $\mathrm d W = P_{\text{ext}}\mathrm d V$ and not just $\mathrm d W = P_{\text{gas}}\mathrm d V$ ?

• because that is how it is defined. This is the convention taken. – Tanuj Mar 3 '18 at 8:13
• Replace the gas by you or some other solid body. The work done by you or a gas depends on the weight you are moving..which is almost the same as the external pressure..; also forces are have not the same magnitude, if it is so, there wouldn't be any movement.. – santimirandarp Mar 3 '18 at 8:20
• This is a good question that deserves a thorough reply. If someone wishes to answer, I recommend reading doi: 10.1021/ed3008704, and 10.1021/ed043p233. – Linear Christmas Mar 3 '18 at 12:26
• Many a times, we are dealing with reversible processes. In such cases, we are dealing with infinitesimal changes - changes so small that we can assume the system to be at equilibrium with its surroundings at all points along the path it is following. This would mean mechanical equilibrium - $P_{ext}=P_{gas}$. – Eashaan Godbole Mar 3 '18 at 12:51

In a reversible process, the gas pressure is spatially uniform within the cylinder, and is described globally by the ideal gas law. However, in an irreversible process, the force per unit area at the piston face is not equal the force per unit area at other locations within the cylinder. Furthermore, the ideal gas law does not describe the behavior of the gas because viscous stresses contribute to the force per unit area for a rapid irreversible deformation. So, even though Newton's 3rd law is satisfied at the piston face, unless we specify the force per unit area externally (e.g., manually), we will get the wrong answer if we try to calculate the pressure at the piston face using the ideal gas law.

In applying the equation $W=\int{P_{ext}dV}$ to calculate the work, $P_{ext}$ is supposed to be the force per unit area exerted by the surroundings on your system, at the interface between your system and the surroundings. So, if the gas is your system, $P_{ext}$ is the force per unit area exerted by the inner face of the piston on your gas (and by your gas on the inner face of the piston). In your example, if the cylinder is vertical and we do a force balance on the piston, we get: $$m\frac{dv}{dt}=P_{ext}A-mg$$assuming there is vacuum on the outer face of the piston. If we multiply this equation by the piston velocity v = dx/dt and integrate, we obtain:$$m\frac{v^2}{2}=\int{P_{ext}dV}-mg\Delta x$$Therefore, the work done by the gas on its surroundings (up until an arbitrary time) is given by: $$W=\int{P_{ext}dV=mg\Delta x}+m\frac{v^2}{2}$$When the final thermodynamic equilibrium state of the system has been realized, the piston will no longer be moving (including any oscillations of the piston, which will have eventually been damped out by viscous stresses) and the work will then be determined by:$$W=\int{P_{ext}dV=\frac{mg}{A}\Delta V}$$ If, instead of vacuum, there is some constant force external to the piston (say, $P_{atm}$ equal to the atmospheric pressure), the previous results change instead to:$$W=\int{P_{ext}dV=mg\Delta x}+m\frac{v^2}{2}+P_{atm}\Delta V$$when the piston is still moving and $$W=\int{P_{ext}dV=(P_{atm}+\frac{mg}{A})\Delta V}$$at final equilibrium when the piston has been damped to rest.

There is no difference and the equation (first law) will yield the same answer to the problems you may come across - particularly based on the first law of thermodynamics. The difference arises in deciding the point of reference for the work done. We have dW = +PdV or -PdV (note that the two Ps are different): you just have to decide where do you look from.

Am I the system and work is done on me, or am I outside and doing work on the system?

Once you choose your point of reference, that is, the convention you desire to follow - you should stick to it and do all your calculations according to it. As in most of these cases, the reason is entirely historical and has no conceptual difference whatsoever.

People who prefer using dW= +PdV, where P is the pressure of the gas, are usually physicists, and engineers – who want to know "what the system can do for us" in practical applications. This is the section of the community that prefers to be in the work-done-by-the-system camp.

On the other hand, work-done-on-the-system seems to foster the view of an experimenter or theoretician operating on a system from outside. Hence, this is usually the view accepted by chemists, i.e. dW = -PdV, where P is the pressure of the surroundings and dV refers to the infinitesimal change in the volume of the system. (Clearly, if dV>0, that is, volume increases, the work done is negative. This is because the surroundings 'oppose' the system's expansion)

Edit: I found this really important to mention -

Pressure due to gas and pressure due to surroundings are not always equal. They are equal, at all times during the process, if and only if the process is quasi static, that is, 'reversible'.

• If I understood the question correctly, it's less about the sign than it is about the proper term for pressure. The usual equation $\delta w = -P_\pu{ex}\pu{d}V$ has some additional requirements that usually go unstated in textbooks. – Linear Christmas Mar 3 '18 at 12:30