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Topic: 1st law of thermodynamics, work/heat problem  (Read 8395 times)

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Offline Jazzified

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1st law of thermodynamics, work/heat problem
« on: October 03, 2008, 02:46:46 AM »

I know the equations I may be able to use are:
adiabatic expansion: q = 0
deltaU = w = Cv (T2-T1)
reversible (ideal gas): P1V1 = P2V2
gamma = Cp/Cv
Cp - Cv = R

And I may need to do:
V1 = nRT/P1
V2 = nRT/P2
Combine into w = -P2 (V2 - V1)

But after that, I am stuck...is anybody able to help me solve this?  Thanks in advance.

Offline Yggdrasil

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Re: 1st law of thermodynamics, work/heat problem
« Reply #1 on: October 03, 2008, 07:47:44 PM »
Start by finding the temperature at each point.  You can use the ideal gas law (PV = nRT) to do this since the graph gives you P and V at each point.

Offline Naumans

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Re: 1st law of thermodynamics, work/heat problem
« Reply #2 on: October 04, 2008, 10:52:14 PM »
Hmm. Let me just give you a little more detailed answer to the problem.

First the question asks you for Temp. so find that using PV=nrT (you have all the values given for points 1,2,3 from the graph)

Then start by finding U for each processes (using deltaU = w = Cv (T2-T1)) (Also, this equation is still valid even if the process is not constant volume). For process 3==>1, since its isothermal U=0 since u is only dependant on T which doesn't change.

Next find work using w=-nrT(ln(V2/V1)) since all process are reversible and in process 2==>3 use this for work since p is changing w=-nrT(ln(p1/p2)

Next use the first law of thermodynamics u=w + q to find q for all process

Then for H use H=Cp(T2-T1) for the first process 1==>2 (again dosn't matter if pressure is constant or not although it is).For process 3 ==>1, H=0 since isothermal

Next to find the total cycle put all the answers you got into a table like this and just add up the values for the total cycle of each. Careful of the signs

 Path   q   w   u   H

Remember to use proper units for each of the equations and values

Good luck>> >  ;D

Offline toffeefan

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Re: 1st law of thermodynamics, work/heat problem
« Reply #3 on: September 02, 2009, 12:58:31 AM »
I don't get it. How do you know the process for all steps are adiabatic? I thought from step 1 to 2, its a constant pressure process, so shouldn't it be qp=Cp(T2-T1) ? As  :delta:H = qp answer for  :delta:H should be the same as qp?

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