T - KTH
T - KTH
T - KTH
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Van der Waals gaslag 2<br />
V<br />
PV = RT<br />
0<br />
4<br />
υ = 4N A πr<br />
3<br />
=ϖ 0 0 =<br />
0.0224<br />
( V − υ ) RT<br />
P =<br />
0<br />
n=1, d.v.s. en mol gas<br />
3<br />
≈<br />
10<br />
liter, anta att Boyles lag gäller.<br />
Inverkan av υ: Viktigt!<br />
http://www.ict.kth.se/courses/2B1116/ 31<br />
−5<br />
m 3 =0.01 liter<br />
I enlighet med resultaten i Tabell 7.3, sid. 246, KP1