No need to count squares. The integral is easily done.
I get 85.7 kJ
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net http://www.velocity.net/~trebor
----- Original Message -----
From: "SSHS KPHOX" <kphox@MAIL.CCSD.K12.CO.US>
To: <PHYS-L@lists.nau.edu>
Sent: Thursday, December 19, 2002 2:00 PM
Subject: Thermo problem help!
| I can use some help.
|
| I am building a worksheet to show the difference in isothermal and
| adiabatic expansion. Using Q = dU + W as first law. For adiabatic dU
= -W.
| Plotting the process on a PV diagram I used P*V^gamma = constant. For
air
| gamma = 1.40....according to one reference.
|
| Now I calculate dU (really it is deltaU) as = 1.5nR(T2 - T1). The the
work
| should be the area bounded by the PV graph (I hope). The problem is that
| for the numbers I am using dU and W do not agree.
| P1 = 200 kPa ; V1=1 m^3; n = 0.04 mol; T1 = 600K
| P2 = 103 kPa ; V2 = 1.6 m^3; n = 0.04 mol ; T2 = 496K
| delta U = -51 J area under graph (counting squares ) 88J
|
| I am sure I am missing something important or have a gross
misconception.
| Can anyone help?
|
| Thanks in advance!
|
| Ken Fox
| Science Department Coordinator
| IB Physics Teacher
| Smoky Hill High School
| Aurora, CO
| kfox@mail.ccsd.k12.co.us