# An isobaric process is a polytropic process with n 0

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An isobaric process is a polytropic process with n = 0. Thus P ( V ) = C = P 1 . (4.49) We also have P 2 = P 1 . The work is 1 W 2 = integraldisplay 2 1 P 1 dV = P 1 integraldisplay 2 1 dV = P 1 ( V 2 V 1 ) . (4.50) CC BY-NC-ND. 2011, J. M. Powers.

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86 CHAPTER 4. WORK AND HEAT Example 4.8 Find the work for an isochoric process. An isochoric process has dV = 0. Thus 1 W 2 = integraldisplay 2 1 P dV bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright =0 = 0 . (4.51) There is no work for an isochoric process. This also corresponds to a polytropic process with n → ∞ . A family of paths in the P V plane for a set of polytropic processes of varying n is shown in Fig. 4.5. V P 1 n=0 (isobaric) n<1 (polytropic) n>1 (polytropic) n=1 (isothermal) n ∞ (isochoric) 1 2 1 W 2 = ∫ P dV Figure 4.5: P V diagram for various polytropic processes. 4.2.3 Other forms of work We lastly note that there are other forces besides pressure forces, and those forces can also do work. Consider a stretching wire stretched by tension force T through length change dL . The differ- ential work is δW = T dL. (4.52) a surface with surface tension S . The differential work is δW = S dA. (4.53) CC BY-NC-ND. 2011, J. M. Powers.
4.2. WORK 87 a system with electrical work where E is the electrical field strength, q is the particle charge, and x is the distance: δW = q E dx. (4.54) In total, for materials which are more than simple compressible substances, we have δW = PdV T dL S dA q E dx ... (4.55) It can be shown that the more work modes we include, the more independent thermodynamic variables are necessary to specify the state of the system. Lastly we note that a gas expanding into a vacuum has 1 W 2 negationslash = integraltext 2 1 PdV because it is inherently a non-equilibrium process. Example 4.9 An ideal gas undergoes a two-step process. Beginning at state 1, it is isothermally compressed to state 2. Then it is isobarically compressed to state 3. Find the work. The process is sketched in Fig. 4.6. P V 1 2 3 1 W 3 = ∫ P dV < 0 1 3 Figure 4.6: Sketch of two-step, isothermal-isobaric, compression of an ideal gas. We have 1 W 3 = 1 W 2 + 2 W 3 , (4.56) = integraldisplay 2 1 PdV + integraldisplay 3 2 PdV, (4.57) = mRT 1 integraldisplay 2 1 dV V + P 2 integraldisplay 3 2 dV, (4.58) = mRT 1 ln V 2 V 1 + P 2 ( V 3 V 2 ) , (4.59) = P 1 V 1 ln V 2 V 1 + P 2 ( V 3 V 2 ) . (4.60) Note that 1 W 3 < 0, since V 2 < V 1 and V 3 < V 2 . So work is done on the system in compressing it . Note also that even though we do not know T , we can solve the problem. This is because work only requires information about the P - V state of the system and the nature of the process. CC BY-NC-ND. 2011, J. M. Powers.

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88 CHAPTER 4. WORK AND HEAT Example 4.10 Consider the piston-cylinder arrangement sketched in Fig. 4.7. Here m = 2 kg of water is initially water m = 2 kg V 1 = 0.02 m 3 T 1 = 50 ˚C Q water m = 2 kg T 3 = 200 ˚C Q Figure 4.7: Sketch of piston-cylinder arrangement for water heating example.
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