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Biology and Chemistry Homework Help
An ideal gas closed system reversible process
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[QUOTE="HethensEnd25, post: 5702787, member: 594242"] [h2]Homework Statement [/h2] an Ideal gas at T = 70 C and 1 bar undergoes following reversible processes: a: Adiabatically compressed to 150 C b: then, cooled from 150 to 70 C at constant pressure c: finally, expanded isothermally to the original state (T=70 C and P = 1 bar) [h2]Homework Equations[/h2] 1-dU=dQ+dW 2-dU=CvdT 3-(T2/T1)=(P2/P1)[SUP](R/Cp)[/SUP] from TP[SUP](1-ϒ)/ϒ[/SUP]=constant 4-CpdT=dH [h2]The Attempt at a Solution[/h2] So far for part a) I have done the equation[B] dU=dQ+dW[/B], since it is an adiabatic process I have [B]dQ=0[/B], thus leaving [B]dU=dW[/B]. I know that[B] dW=-PdV[/B] so I can calculate [B]dU=CvdT[/B] to get my change in internal energy. From there I calculated P2 using equation 3. For b) knowing it is an Cp problem I used equation 4 having [B]CpdT=dH[/B] where my change in temperature is 150 to 70 C I am stuck on part c) I know it is an isotherm that means that there is no change in temperature thus dU and dH are equal to zero since they are temperature dependent. So [B]dQ=dW. [/B]Do I make it so that T is a constant and solve with partial derivatives giving me[B] dT=(V/R)dP+(P/R)dV[/B]. With dT=0 we get[B] -PdV=VdP[/B]. Using the ideal gas equation to give V=RT/P ==> -(RT/P) dP. Integrating gives me[B] -RTln(P2/P1)[/B] My problem is I don't know what pressure to use. Do i just use the pressures calculated in my part b going from there back to the original state? [/QUOTE]
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An ideal gas closed system reversible process
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