Re: Thermo problem help!

[Roger Haar <haar@PHYSICS.ARIZONA.EDU>]

Hi,

        My first thought is the 1.5 in
dU = 1.5 n R ( T2-T1)

I think that 1.5 is for monoatomic gasses, but air
is mostly diatomic
The factor before the n is 0.5 X number of degrees
of freedom, f.  For monatomic f = 3 and the factor
is 1.5.
diatomic  f = 5 and the factor is 2.5

        This will increase your dU by 2.5/1.5
or from 51 to 85.

Thanks
Roger Haar U of AZ
**********************************************
SSHS KPHOX wrote:
> 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