# Recall that the dot product of two vectors yields a

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is the dot product operator. Recall that the dot product of two vectors yields a scalar. The terms F and x are scalar equivalents valid for one-dimensional systems. The units of force are N , those of distance are m , so the units of work are N m , which have been defined as Joules ( J ). Work is done by a system if the sole effect on the surroundings (i.e. everything external to the system) could be the raising of a weight. We take the following sign convention: CC BY-NC-ND. 2011, J. M. Powers.

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82 CHAPTER 4. WORK AND HEAT + work done by the system, • − work done on the system. This sign convention is not universal. Many physicists use precisely the opposite convention. Probably the reason for this convention is that thermodynamics is a science that was invented by engineers in the nineteenth century. And those engineers wanted to produce work from steam engines. Systems doing work were viewed favorably and endowed with a positive sign. We associate energy with the ability to do work. We define Power : the time rate of doing work = δW/dt . Specific work : the work per unit mass w = W/m . Because work is path-dependent, the intensive quantity w is not a thermodynamic state variable. 4.2.2 Work for a simple compressible substance Consider the scenario sketched in Fig. 4.3. In state 1, we have a material at pressure P P A x state 1 P+dP A x+dx state 2 Figure 4.3: Sketch of piston-cylinder arrangement as work is done as the material expands when weights are removed. confined in a cylinder of cross-sectional area A . The height of the piston in the cylinder is x . The pressure force of the material on the piston is just balanced by weights on top of the piston. Now remove one of the weights. We notice a motion of the piston to a new height x + dx . We let a long time elapse so the system comes to rest at its new equilibrium. We notice there is a new pressure in the chamber, P + dP sufficient to balance the new weight force. Obviously work was done as a force acted through a distance. Let us calculate how much work was done. The differential work is given from Eq. (4.28) as δW = Fdx. (4.29) CC BY-NC-ND. 2011, J. M. Powers.
4.2. WORK 83 Now F varies during the process. At state 1, we have F = PA. At state 2, we have F = ( P + dP ) A. Let us approximate F by its average value: F 1 2 ( PA + ( P + dP ) A ) = PA + dP 2 A. (4.30) So δW = parenleftbigg PA + dP 2 A parenrightbigg dx = PAdx + A 2 dPdx bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright 0 . (4.31) Let us only retain terms which are differential and neglect the square of differential terms, so δW = PAdx. (4.32) Now since Adx = dV , the differential volume, we get the important formula: δW = PdV. (4.33) We can integrate this and get the equally important 1 W 2 = integraldisplay 2 1 PdV. (4.34) Note we employ the unusual notation 1 W 2 to emphasize that the work depends on the path from state 1 to state 2. We are tempted to write the incorrect form integraltext 2 1 δW = W 2 W 1 , but this would imply the work is a state function, which it is not, as shown directly.

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