Thermodynamics and Integral? I think I was approaching it wrong, and I don't know where to begin.

Question

I started using the idea that delta U = W because it is adiabatic for the first expansion. I was told by a peer to use w = integral of PdV equation, but someone else also suggested that W=nCdeltaT and now I am unsure which method/why to use. I also do not know how to solve out the integral:

Problem: A 1.00 mole sample of an ideal monatomic gas, originally at a pressure of 1.00 atm undergoes a three-step process. It is first expanded adiabatically from T1=550 to T2=389 Kelvin. Then it is compressed at constant pressure until the temperature reaches T3. It then returns to T1 and orginal pressure by a constant volume process. What is T3.

Answer 1

Hi Yogita S.,

No need to agonize. While it's true that pV^(.gamma.) is conserved for an adiabatic expansion, that doesn't really help you, since you are only given T1 and T2, and you don't immediately know the limits for a p.delta.V integration (even if you do know that .gamma.== Cp/Cv = 5/3 for a monatomic gas), though that would correctly give you the work done! And, you can't use an equation using a heat capacity, since you are constant for neither p nor V in an adiabatic expansion! So, use instead the relationship: TV^(.gamma. -1) = constant. That will give you V2 (you know P1 and V1 from the combined gas law pV=nRT, right?), and everything will fall into place from there. (Calculate P2 also, you'll need it in step 2).

Well, almost -- step 2 is tricky, you have to reason it backwards from step 3, so to speak. Since step 3 is isovolume, therefore step 2 must have returned to V1 ! The only way to do that, is to do some heat extraction as you compress in step 2. But, you know the pressure in this step (P2) and V1, so T3 is from the combined gas law. You don't NEED to know how much heat extraction was done, you are calculating from the system variables directly.

The point of this problem is to enable you to see that you've done a "closed" cycle on the gas -- restored it to original conditions -- via work extraction, work supply, heat extraction, and heat supply, variously. The amount of each of these in a step is governed only by the type of process; overall energy changes are per W+Q (signs however you choose to define them for W and Q).

Hope this makes a bit more sense to you now. It's hard, when you're seeing all this for the first time, to try to keep track of all the relationships among state variables. But the more you do it, the easier it will get. Any time you need to refresh, if your notes don't cover it, hit up the Hyperphysics website.

-- Cheers, -- Mr. d.

Answer 2

Hi Yogita S.,

No need to agonize. While it's true that pV^(.gamma.) is conserved for an adiabatic expansion, that doesn't really help you, since you are only given T1 and T2, and you don't immediately know the limits for a p.delta.V integration (even if you do know that .gamma.== Cp/Cv = 5/3 for a monatomic gas), though that would correctly give you the work done! And, you can't use an equation using a heat capacity, since you are constant for neither p nor V in an adiabatic expansion! So, use instead the relationship: TV^(.gamma. -1) = constant. That will give you V2 (you know P1 and V1 from the combined gas law pV=nRT, right?), and everything will fall into place from there. (Calculate P2 also, you'll need it in step 2).

Well, almost -- step 2 is tricky, you have to reason it backwards from step 3, so to speak. Since step 3 is isovolume, therefore step 2 must have returned to V1 ! The only way to do that, is to do some heat extraction as you compress in step 2. But, you know the pressure in this step (P2) and V1, so T3 is from the combined gas law. You don't NEED to know how much heat extraction was done, you are calculating from the system variables directly.

The point of this problem is to enable you to see that you've done a "closed" cycle on the gas -- restored it to original conditions -- via work extraction, work supply, heat extraction, and heat supply, variously. The amount of each of these in a step is governed only by the type of process; overall energy changes are per W+Q (signs however you choose to define them for W and Q).

Hope this makes a bit more sense to you now. It's hard, when you're seeing all this for the first time, to try to keep track of all the relationships among state variables. But the more you do it, the easier it will get. Any time you need to refresh, if your notes don't cover it, hit up the Hyperphysics website.

-- Cheers, -- Mr. d.