# What is the correct expression of pressure-volume work? [duplicate]

My book defines work in pressure-volume work as: $$\mathrm{d}w = - P_\mathrm{ext} \mathrm{d}V$$

However before doing so it mentioned the piston to be massless.

In physics, work is defined as $$\mathrm{d}w = P_\mathrm{in} \mathrm{d}V$$

1. What is the correct definition of pressure-volume work in chemistry?
2. Does the definition of work, $$-P_\mathrm{ext} dV$$ change if the piston was not massless in chemistry?

P.S. My question was mainly pointed towards the mathematical expression of work in Chemistry,that is, 'What is the most general work expression in Chemistry in relation with the First Law'.

Some of the users have been suggesting me to check out answers on sign convention of work. I do appreciate it but my question is different and is perhaps best interpreted in @Chet Miller answer.

• It the case $p_\mathrm{ext} \ne p_\mathrm{int}$ the work done on the piston is not the same as the work done on the system, as the piston itself is not considered as a part of the system. – Poutnik Aug 19 '20 at 12:27
• Similarly, the work the system performs on the piston may not be the same as the work the piston performs on the neighbourgood. Imagine vacuum behind the piston. The work in TD context must be considered in relation to the change of the system internal energy. The massless piston simplifies such system-piston-neighbourhood quicks. – Poutnik Aug 19 '20 at 12:56
• @Poutnik So the definition of work by external pressure is true only for ideal case? – Tony Stark Aug 19 '20 at 13:32
• If you think about it, you may answer yourself. There is no guarantee the energy $p_\mathrm{ext} \cdot \mathrm{d}V$ is passed to the system. It is the work done by the neighbourhood on the piston, but not necesserily the work done on the system by the piston. – Poutnik Aug 19 '20 at 13:46

At the inside piston face, the pressure exerted by the gas on the piston is equal to the pressure exerted by the piston on the gas. This just follows from Newton's 3rd law of motion. But, in an irreversible process, the pressure exerted by the gas on the piston face cannot be obtained from the ideal gas equation (or other real-gas equation of state). This is because the equation of state applies only at thermodynamic equilibrium. Otherwise, the gas pressure on the piston involves viscous stresses which depend not only on the volume but also on the rate of change of gas volume. For a reversible process, the gas pressure throughout the gas, including at the inside piston face can be determined using the equation of state, since a reversible process consists of a continuous sequence of thermodynamic equilibrium states.

In using these work relationships, it is important to specify what you are calling the system. If the system includes only the gas, and the piston is massless and frictionless, the pressure on the outside piston face is equal to the pressure on the inside piston face (and the gas pressure). If the piston is included as part of the system, and the piston is not massless or frictionless, then the external pressure is the pressure on the outside piston face, but this is not equal to the gas pressure on the inside piston face (or, equivalently, the piston pressure on the gas).

• Sir does that mean when we say work done on the system we mean 'Work done by Piston on the gas' keeping the system as gas and when we say work done by the gas we mean 'work done by the gas on the piston' ?Further both of these works are always equal and opposite following Newton's Third Law? – Tony Stark Aug 20 '20 at 2:48
• Yes, that's what it means. – Chet Miller Aug 20 '20 at 14:55

You may notice that there are two conventions for defining the work $$p\Delta V$$. The same work is sometimes positive sometimes negative, for the same process. Let's be more precise.

Usually, the work is positive when it is given or added to the system (syringe), when the system is compressed. Work and heat are both positive when they are added to any system compressed by a piston. But, as $$V$$ decreases in such a compression, $$\Delta V$$ is < $$0$$ and the work $$p\Delta V$$ must be written with a minus sign to stay positive : $$\delta w = - ~p\Delta V$$. So the sum of all energies entering the system is the change of internal energy $$U$$ : $$\Delta U = \delta w + \delta q = + ~ \delta q - p \Delta V$$.

For some scientists, mainly engineers, the approach is different. The whole system is considered as a machine, where you put heat inside, so that it must produce as much work as possible to be useful ($$\delta q > 0$$). So the work done by the machine is positive if it works correctly, which happens if $$\Delta V$$ is positive. For engineers, the work is defined as : $$\delta w = + ~p ~\Delta V$$. As a consequence, the internal energy is the difference of the energy added as heat to the system, and the energy delivered b the machine. It is then $$\Delta U = \delta q - \delta w = \delta q - p\Delta V$$.

The final result is the same all over the world when speaking of change of internal energy, and of course of enthalpy. But the sign of the work is not always the same.