Chemical Forums

Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Jazzified on October 03, 2008, 02:46:46 AM

Title: 1st law of thermodynamics, work/heat problem
Post by: Jazzified on October 03, 2008, 02:46:46 AM
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimg404.imageshack.us%2Fimg404%2F1913%2Fcaptureop6.jpg&hash=84d2de906af417dae91ec934f378910489c174e2)

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.
Title: Re: 1st law of thermodynamics, work/heat problem
Post by: Yggdrasil 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.
Title: Re: 1st law of thermodynamics, work/heat problem
Post by: Naumans 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
1-2            
2-3            
3-1            
Cycle            

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

Good luck>> >  ;D


Title: Re: 1st law of thermodynamics, work/heat problem
Post by: toffeefan 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?