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thermo help.......will rate lifesaver

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Date Posted: 7/17/2008 11:27:01 AM  Status: Live
thermo help.......will rate lifesaver
Course Textbook Chapter Problem
Thermodynamics Fundamentals of Engineering Thermodynamics (5th) by Moran, Shapiro 3 N/A
Question Details:
Problem 3.97
Your answer is incorrect.  Try again.
   I got the wrong anwer pls help me figure it out....
Five kg of butane (C4H10) in a piston–cylinder assembly undergo a process from p1 = 5 MPa, T1 = 500 K to p2 = 1 MPa, T2 = 450 K during which the relationship between pressure and specific volume is pvn = constant. Determine the work, in kJ.
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Date Posted: 7/17/2008 12:31:21 PM  Status: Live
Asker's Rating: Helpful   
Response:
 
 in process pv^n    WD=  P1V1 -P2V2/(N-1)
 
BUT PV=nRT
WHERE  n= NO OF MOLES = MASS OF GAS/MOL . MASS
n=5/58
R=8.314
P1=5*10^6 N/M2
P2=1*10^6 N/M2
 
   WD=nR(T1-T2)/(N-1)
PLZ SUBSTITUTE IN THE ABOVE EQ
 
 
 
 
chemlady's Comment:
Thanks for your help....... i had done it but i was getting the wrong N value so yeah....... thanks for all ur help



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