Question

For part 1, it says use equations 1.53 and equations 1.54. It really is a typo and meant to say to use equation 7.53 and equation 7.54.

Problem 7.10. The critical point of the van der Waals equation of state (a) Use (1.53) and (1.54) given by to show that the critical point of the van der Waals equation of state is Pe=1 Te 1, (7.55a) (7.55b) (7.55c) Pe 1 Hence, we can write the dimensionless variables T, P, and V as T = T/Te, P = P/Pe, and V V/Vc. We now see why the various numerical factors were included in the definitions (1.48)-(1.50) of P, T, and p. (b) Show that Ve N (7.56a) 3b, = Pe 27b2 8a (7.56b) kTe 27b (7.56c) (c) At what temperature does the compressibility become negative for a given value of the density? (d) What is the value of the compressibility ratio Pe/pckTe according to the van der Waals equation of state? This ratio is close to 0.29 in a wide variety of systems ranging from Ne to CH4. (7.53) = 0, Әр /т,N а?р 0. др? т,N (7.54)

Answer 1

Prob lem 7.10. let us considles a Van dur Waals Equation of state tos seal tos one mole ot the gas ues, . RT V 2 has Her uual meag where the symbol Now, Reanaang ing (1) G RT V-6 -(2) y 2 At Citical temperature, SP 2P function ot V as pur the Van der Waals Equation where P is the So RT 2 0 VJ (v-6)2 RT RT (v-6)2 a -(3) 2v(V-b)2 V 3 the tirct pastial oleuvative again wat V. daneutiahng get Now, we 2P 2 RT 69 (v-b)3 RT > (V-6)& 3a RT v4 3(v-b)3 () (3) B(4) Compaising RT RT 3(V-6)3 3(V-b) 3V-3b=v V.=3b -(5) Ve is the itical volume

Pudhin'g Now the valu &Vc eq' (3), Jat we RT RT (v-b)a v 3 (34-6) a (3b)3 R T 2 T 86a Te =89 Rb (2b)a (3b)3 27 R63 3. The R equivalut to the Boltzmann constant k. as constant Te=8a 8 a kTc 27 6 97 kb is The caitical tempesatuse where Te Calculde witical Phessure, le , Put Ve 8 Te To est() P =RT V-b V 2 P R/8a 27 Rb (36-6) (3b)2 a 49-30 9 62 2구 b2 2762 a which is chitical Pressure