Solutions to theoretical problems

1S This expansion has already been given in the solutions to Exercise 1.20(a) and Problem 1.14 the result is 

The critical point corresponds to a point of zero slope that is simultaneously a point of inflection in a plot of pressure versus molar volume. A critical point exists if there are values of p, V. and T that result in a point that satisfies these conditions. 

Therefore, the limiting slope of a plot of — againstpis ——.From Fig. 1.3 the limiting slope is 

21 The critical temperature is that temperature above which the gas cannot be liquefied by the application of pressure alone. Below the critical temperature two phases, liquid and gas. may coexist at equilibrium, and in the two-phase region there is more than one molar volume corresponding to the same conditions of temperature and pres.sure. Therefore, any equation of state that can even approximately describe this situation must allow for more than one real root for the molar volume at some values of T and p, but as the temperature is increased above Tc, allows only one real root. Thus, appropriate equations of state must be equations of odd degree in Vm. 

The equation of state fw gas A may be rewritten — (RT/p) Vm — (RTb/p) = 0, which is a quadratic and never has just one real root. Thus, this equation can never model critical behavior. It could possibly model in a very crude manner a two-phase situation, since there are some conditions under which a quadratic has two real positive roots, but not the process of liquefaction. 

We note that all terms beyond the second are necessarily positive, so only if 

Thus Z 1 when attractive forces predominate, and Z 1 when size effects (short-range repulsions) predominate. 

16 The Dieterici equation of state is listed in Table 1.7. At the critical point the derivatives ofp with respect 

Each of these equations is evaluated at the critical point giving the three equations  

Solving the middle equation for Tc, substitution of the result into the last equation, and solving for Vc yields the result Vc = 2b or b = l/c/2 (The solution Vc = b is rejected because there is a singularity in the Dieterici equation at the point Vm = b.) Substitution of Vc = 2b into the middle equation and 

The Dieterici equation of state is listed in Table 1.7. At the critical point the derivatives ofp with respect to (wrt) I m equal zero along the isotherm for which T = T. This means that (dp/SV m) - = 0 and (9 p/3V m)r = 0 at the critical point. 

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Solutions to theoretical problems 170 Answers to discussion questions 268... 

Solutions to numerical problems Solutions to theoretical problems 207 ... 

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