# 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. – Qmechanic Feb 26 '14 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. – Indrayudh Roy Feb 26 '14 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