Question

9.4-7. A particular substance satisfies the van der Waals equation of state. The coexistence curve is plotted in the P, † plane, so that the critical point is at (1,1). Calculate the reduced pressure of the transition for T = 0.95. Calculate the reduced molar volumes for the corresponding gas and liquid phases.

Answer 1

solution Jander woals Equdion is (p + q ) (vb) =RT where v= the molas volume uwversel gas of the gas congtant P = pressure y = volume. At a the pr- plane, this T = temperature, vandes wa walls Iso the mos fixed tempertures, the van der Waals qution describes the dependence pu). In dependence is resp resented by as a family of isotherms. Pt 9 ) (urb), = RT (Pr+a) (v - b) = RT v2 pu² tau-pbut-ab- RTV2 = 0 Pv²_(pb + RTJ vt au-ab=o. v3 (b + R) + q v ab 20 ( common P)

It is 1 or 3 a Cubic have an for a fixed value of P the resulting Equation is a third degue equation with expect to the variable v. known that Equation cm real roots. The first Case occurs at high temperature T Corotherm AB, with lowering the temperature, is the iso therm. In this andulating region appears on the Case three roots (Isotherm E F G H I JL) Rection between the two types of isotherms occurs at a certain temperature (TK) , which is called the Critical temp. Pressure there are The E Pk 1 G D V k volume Critical point av- 13 (5+1) + should be written as (V - V) ² - 0

(V-k) ² = 0 v3 - 3y² Vk + 3 vuk - Vic=0 account that the cutiful point P= Pk and Take into T=Tk -b+ RI) = - 3 vk b+RT = 3Vk 32 о Pk 3 Pk Pk ab Рk Divide the third eqution by the second a. b. Pk E V = 36 Pk. a 3 In the problem critical point (1) .عرا b = V₂ a Pk = a = 34 3. (36) 2 276² Pk = a (P+ = 1) from giver problem 911 3 93

Siticed temperture (Tk) = (3 Vk - b) Pk - (3,36-b). a 2752 R 89 arbr TE = a=3 27x1 XR b= Tk = 24 9R on Critical parameters for vander waals gas depend only values a b and the given by the formulas b= 3 Pk- a Vk= 3b . a 2762 Th= 8a 27bR 94 9R reduced Equation of state TET Tk T =P Pk parameter Vk, he and Tk, it is to go the dimension less variables Using the critical sonsert conversient V= Yuk P = TPk T = TTK [ + ] [-] = lett

Ta ta 2762 (36) & 8 a 276R [+] (p-1)-87 = 8 (0.95) 7.6 the is positive liquid phases consider again the van der waals Ernation at temperat above Critical point TETk. In this case the iso therms have an dulding shape. A portion of such iso therms between the minimum and maximum characterized by derivative de which corresponds to unstable substan. lequid-gas phase transition occuss in this the pv-diagram. This trawition is accompained by a bignificant cherage in the volume (V) at constant pressure (p) and constant temp- asutuse (T) such a transition is respresented by Pu diagram (line FJ line) stite dv of the Actually the asea of a straight line in the

Pressure maxwell Equation asean rule ligudt vapons suges coded vapour I rupeshected liquid volume & The set of Criterion de so, these love the FJ- line is called start and end points of the horizontal segment lines at different temperadores T (Where TITL) which is called the binodal Carve, minimum and maximum points of all isotherms at TCT is called Spinodal Curve unstable states (for which do av 0 are within the region bounded by spinodal Culve. between and bi nodal Curves principle satistics the Stability states meta stable stelee The Culve passing through the the А past of the diagram spinodal dv 9 this region are called

The left half of the the super heated liquid, Corresponds to the super indicated and the region respresented right half cooled vapor.

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