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 $T$ ?