Question

Problem 2: Critical behavior of a gas. The pressure P of a gas is related to its density n = N/V and temperature T by the relation b Р%3D КвТп с (12) where b andc are positive constants.

Answer 1

Given equ as p KBTR - but density - N then, KATN but -02 p= 613 22 X CN The Critical isotherm on a pur diagrawn art this point not only has a horizontal tangent, but a point of reflection onwelt so art the critical point ( ) = 0 = CODE So, the con can be written in the form, a (9), -+ KTM Som REAS V2 multiplying by 204 on both sides, - 2431 NV - 26Nr – CH² – (y) from (3) 2KBTN 36N² 2NN² V3 V4 ure Multiplysig by rs on both sides then wegen 2 KBP NVR - 3602 v - 2 CN3 (5)

Composing (4 and 15 we get 26 NV - N² - 3 bNv - 23 CA3 26 Nov ONS a en va eN V man But in Critical point put this value in (4) weget. 2 KBP Nx (CN.) = 26 NX (CN) – en Xb 2 e N3 – CN akan t est 2 KB T C = 66 be T a brc 2kpc 2KBC point T=To. But in Critical hen Tc- QKBC put the values of h. Te in Press meet Pe=KBT ag & ar & 2W2

Реков - - Сове за fcho, Kд те не - Po 1 ... x хувь 3 (4 6 А3 " ь усь ( 4 ча+ с) 944 gb + cЧ