Chapter 9: Problem 5
If \(P, v\), and \(T\) are the pressure, molar volume, and temperature of a gas and \(P_{C}, v_{C}\), and \(T_{C}\) are the critical pressure, critical molar volume, and critical temperature, then the reduced pressure \(P_{R}\), the reduced molar volume \(v_{R}\), and the reduced temperature \(T_{R}\) are defined as $$ P_{R}=\frac{P}{P_{C}}, \quad \nu_{R}=\frac{v}{v_{C}}, \quad T_{R}=\frac{T}{T_{C}} $$ (a) Show that, in terms of reduced quantities, the van der Waals equation becomes $$ \left(P_{R}+\frac{3}{v_{R}^{2}}\right)\left(v_{R}-\frac{1}{3}\right)=\frac{8}{3} T_{R} $$ When the van der Waals equation is in this form, the material constants \(a\) and \(b\) do not appear explicitly. Thus, all gases that obey the van der Waals equation may be considered in the same state when the values of \(P_{R}, v_{R}\), and \(T_{R}\) are the same (i.e., each gas is measured in units of its particular values of \(P_{c}, v c\), and \(T_{C}\) ). This is the principle of corresponding states, which is a principle of universal similarity established first by van der Waals.. (b) Plot three curves for \(P_{R}\) as a function of \(v_{R}\), one for \(T=\frac{1}{2} T_{C}\), one for \(T=T_{C}\) and one for \(T=2 T_{c}\). What happens physically when the equation indicates three allowed values of \(v_{R}\) for a single \(P_{R}\) and \(T\) ?
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