Isothermal vs Adiabatic

sign up log in tour help Take the 2minute tour × Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required. Isothermal vs. adiabatic compression of gas in terms of required energy I am working on an instruction manual of sorts to be used with an introductory course in thermodynamics. As an example of problem solving, I attempt to answer the following question: If you have little energy available, would you rather use an isothermal or an adiabatic process to compress a gas? My analysis is as follows: The work required to compress a gas from volume $V_0$ to volume $V$ is $$W=\int_{V_0}^{V}PdV$$ For the isothermal compression, the ideal gas law, $P=\frac{nRT}{V}$, is used and inserted into the equation above: $$W=\int_{V_0}^{V}\frac{nRT}{V}dV= nRT\int_{V_0}^{V}\frac{dV}{V}=nRT\ln\left(\frac{V}{V_0}\right)$$ For the adiabatic compression, $PV^\gamma=P_0V_0^\gamma \iff P=P_0\left(\frac{V_0}{V}\right)^\gamma$ is valid. Thus, the required work is $$W=\int_{V_0}^{V}P_0\left(\frac{V_0}{V}\right)^\gamma dV= \frac{P_0V_0^\gamma}{\gamma+1}\left(V^{\gamma+1}V_0^{\gamma+1}\right)=\frac{PVP_0V_0}{\gamma1}$$ This result indicates that the magnitude of the required work depends on the properties of the gas in question. Thus, it may vary which of the two compression processes requires the lowest amount of energy. Now, finally, to my question: Intuitively I would assume that the isothermal work is usually lower than the adiabatic work, as compression lowers the volume of the system and therefore usually increases the temperature if heat exchange with the surroundings is not allowed. Isothermal compression requires heat transfer to the surroundings to maintain constant temperature, lowering the pressure of the system and thus lowering the resistance to compression compared to the adiabatic compression (where heat exchange is not allowed). Is the isothermal work actually smaller than the adiabatic one in most cases or is my argument flawed? physicalchemistry thermodynamics gaslaws edited Jan 21 '14 at 10:40 asked Nov 28 '13 at 10:00 Ove Øyås 441 2 13 1 – A better way to express the adiabatic work is $$W=\frac{P_0V_0PV}{\gamma1}$$. Satwik Pasani Nov 28 '13 at 14:06 1 Answer To solve this, try to use what I call the "graphical apparatus". For an isothermal process: $$PV=\text{constant}$$ $$PdV=VdP$$ $$\frac{dP}{dV}=\frac{P}{V}$$ for adiabatic process: $$PV^\gamma=\text{constant}$$ $$\frac{dP}{dV}=\gamma\frac PV$$ Therefore, starting at the same point on a PV graph, the curves for an adiabatic and isothermal processes will diverge and the adiabatic curve will have a steeper slope. For the same reduction in volume ( ), more area will be enclosed by adiabatic, and since the area $\int PdV$, gives the work required, isothermal work is smaller than adiabatic for the same reduction in volume. the graph in the picture is for expansion, not for contraction. In case of contraction, the curves will be reversed, i.e. adiabatic curve will be above the isothermal curve, and will enclose greater area under it for the same reduction in pressure Your argument is correct. To provide more mathematical support to it, you can observe the fact that it is both increase in temperature and reduction in volume which increases the pressure in adiabatic process and only reduction in volume increases pressure in isothermal process. The exponent of

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Isothermalvs.adiabaticcompressionofgasintermsofrequiredenergy

Iamworkingonaninstructionmanualofsortstobeusedwithanintroductorycourseinthermodynamics.Asanexampleofproblemsolving,Iattempttoanswerthefollowingquestion:

Ifyouhavelittleenergyavailable,wouldyouratheruseanisothermaloranadiabaticprocesstocompressagas?

Myanalysisisasfollows:

Theworkrequiredtocompressagasfromvolume$V_0$tovolume$V$is$$W=\int_{V_0}^{V}PdV$$Fortheisothermalcompression,theidealgaslaw,$P=\frac{nRT}{V}$,isusedandinsertedintotheequationabove:$$W=\int_{V_0}^{V}\frac{nRT}{V}dV=nRT\int_{V_0}^{V}\frac{dV}{V}=nRT\ln\left(\frac{V}{V_0}\right)$$Fortheadiabaticcompression,$PV^\gamma=P_0V_0^\gamma\iffP=P_0\left(\frac{V_0}{V}\right)^\gamma$isvalid.Thus,therequiredworkis$$W=\int_{V_0}^{V}P_0\left(\frac{V_0}{V}\right)^\gammadV=\frac{P_0V_0^\gamma}{\gamma+1}\left(V^{\gamma+1}V_0^{\gamma+1}\right)=\frac{PVP_0V_0}{\gamma1}$$

Thisresultindicatesthatthemagnitudeoftherequiredworkdependsonthepropertiesofthegasinquestion.Thus,itmayvarywhichofthetwocompressionprocessesrequiresthelowestamountofenergy.

Now,finally,tomyquestion:

IntuitivelyIwouldassumethattheisothermalworkisusuallylowerthantheadiabaticwork,ascompressionlowersthevolumeofthesystemandthereforeusuallyincreasesthetemperatureifheatexchangewiththesurroundingsisnotallowed.Isothermalcompressionrequiresheattransfertothesurroundingstomaintainconstanttemperature,loweringthepressureofthesystemandthusloweringtheresistancetocompressioncomparedtotheadiabaticcompression(whereheatexchangeisnotallowed).

Istheisothermalworkactuallysmallerthantheadiabaticoneinmostcasesorismyargumentflawed?

physicalchemistry thermodynamics gaslaws

editedJan21'14at10:40 askedNov28'13at10:00Oveys441 2 13

1 Abetterwaytoexpresstheadiabaticworkis$$W=\frac{P_0V_0PV}{\gamma1}$$. SatwikPasani Nov28'13at14:06

1Answer

Tosolvethis,trytousewhatIcallthe"graphicalapparatus".Foranisothermalprocess:$$PV=\text{constant}$$$$PdV=VdP$$$$\frac{dP}{dV}=\frac{P}{V}$$foradiabaticprocess:$$PV^\gamma=\text{constant}$$$$\frac{dP}{dV}=\gamma\fracPV$$Therefore,startingatthesamepointonaPVgraph,thecurvesforanadiabaticandisothermalprocesseswilldivergeandtheadiabaticcurvewillhaveasteeperslope.Forthesamereductioninvolume(

),moreareawillbeenclosedbyadiabatic,andsincethearea$\intPdV$,givestheworkrequired,isothermalworkissmallerthanadiabaticforthesamereductionin

volume.

thegraphinthepictureisforexpansion,notforcontraction.Incaseofcontraction,thecurveswillbereversed,i.e.adiabaticcurvewillbeabovetheisothermalcurve,andwillenclosegreaterareaunderitforthesamereductioninpressure

Yourargumentiscorrect.Toprovidemoremathematicalsupporttoit,youcanobservethefactthatitisbothincreaseintemperatureandreductioninvolumewhichincreasesthepressureinadiabaticprocessandonlyreductioninvolumeincreasespressureinisothermalprocess.Theexponentof
volumeinadiabaticequation($PV^\gamma=K$)is$\gamma>1$,ascomparedto1intheisothermalequation.Hence,pressurechangeismoresensitivetovolumechangeandislargerinmagnitude.

answeredNov29'13at16:19SatwikPasani2,458 1 5 30