Work in Newtonian Mechanics and Thermodynamics

Well, is the basic difference between the work that we learn in Mechanics and that in Thermodynamics? This is because in Mechanics, whenever work of magnitude $W$ is done on a system $S$, then the system $S$ also does a work of magnitude $W$ on its surroundings. If I take this consideration in Thermodynamics, then according to the sign conventions, the total work in any process would turn $0$ (once positive and once negative), which is not the case. I am really very confused regarding these. Please explain what exactly is work in Thermodynamics, and its differences with Mechanics.

• Related: physics.stackexchange.com/q/37904/2451 and links therein. Commented Feb 26, 2014 at 19:19
• @Qmechanic, the question states only about the sign convention, but I am confused with the concept of work itself. I need an explanation on that. Commented Feb 26, 2014 at 20:05

It may just be the wording of your text that provides the confusion. Work is the same in thermodynamics as it is in mechanics. Work is an energy transfer process. In thermodynamics this often equates to energy in the forms of pressure, temperature, volume, etc. that is transformed into energy associated with position or movement (i.e. kinetic and potential energy). The sign convention that is adopted in thermodynamics is the same as in mechanics. Work done on a system is viewed as energy added to the system. Whereas work, or thermodynamic processes for that matter, that do negative work represent energy being transferred from the system to its environment.

Let me try to attempt to clarify in some direction that I feel comfortable. From usual mechanics, you know the definition of work is given by : $$W = \int \bf{F}.\bf{ds}$$ for simplicity considering the work done to be in the same direction as displacement $$W = \int {F}{ds}$$ Now this is just one version of defining work. In generalising, we redefine force with some other parameter (say $F \rightarrow \Phi$) and the corresponding physical quantity $X$, whose change is induced by $\Phi$. Then work done will be defined as $$W = \pm\int {\Phi}{dX}$$ the $\pm$ sign is to do with the convention if the work is done on the system or by the system (i.e. work extracted from the system).

With this generalisation, we have different definitions,

If $\Phi \rightarrow P$ then $X \rightarrow V$ as application of pressure induces a change in volume.

If $\Phi \rightarrow H$ then $X \rightarrow M$ as application of magnetic field induces a change in magnetisation of the material.

If $\Phi \rightarrow E$ then $X \rightarrow P$ as application of electric field induces a change in polarisation of the material.

Chemical potential ($\Phi \rightarrow \mu$) and number particles($X\rightarrow N$) is one example in this set. There are lot more of them. In general $\Phi$ is called the field variables and $X$ is called the state variable.

In sum, I want to emphasise upon the fact that the mechanical work and thermodynamical work are on the same footing, when a work $W$ is done on system $S$, the system does not do a work, but extracts an amount of work $W$ from the surroundings

Work in classical mechanics is a force acting through a distance, as discussed in earlier responses. Work in thermodynamics is a much broader concept; it is energy that crosses a boundary with no mass transfer due to an intensive property difference other than temperature. Heat is is energy that crosses a boundary with no mass transfer solely due to a difference in temperature. We sometimes consider the "heat" in a system which is sloppy nomenclature, the correct terminology is the "internal energy" in the system. Heat, like work, is energy that crosses a boundary, not energy in the system. To consider mass transfer, engineering texts use the open thermodynamic system and consider the enthalpy of mass flowing into and out of the system boundaries. See any good engineering thermodynamics text. Sometimes, the term "pseudowork" is used for the work in classical mechanics and the term "work" is reserved for the thermodynamics concept of work; you can find articles on this on the web. As an aside, classical mechanics addresses the motion of a rigid body. For a rigid body there is by definition no internal dissipation of energy (no change in internal energy) and for a rigid body the work from friction does not increase the temperature of the body, the work from friction only changes the kinetic energy of the body. Numerous discussions of the motion of a rigid body incorrectly state friction "heats" up the body (changes the internal energy). Of course, in reality bodies are not rigid and the assumption of a rigid body may not always be a good one.