Given that the reduced temperature, θ=(T)/(T_C) the reduced pressure, π=(F)/(F_C) the reduced volume, φ=(V)/(V_C) Thus, it can be said that the reduced equation of state may be given as

JIPMER 2015

JIPMER 2015 · Chemistry · States of Matter

Given that the reduced temperature, θ=(T)/(T_C) the reduced...

Given that the reduced temperature,

\theta=\frac{T}{T_C}

the reduced pressure,

\pi=\frac{F}{F_C}

the reduced volume,

\phi=\frac{V}{V_C}

Thus, it can be said that the reduced equation of state may be given as

A $\left(\frac{\pi}{3}+\frac{1}{\varphi}\right)(3 \phi-1)=\frac{8}{3} \theta$
B $\left(\frac{\pi}{4}+\frac{1}{\varphi}\right)(3 \theta-1)=\frac{3}{8} \varphi$
C $\left(\frac{\pi}{3}+\frac{1}{\varphi}\right)(\phi-1)=\frac{3}{8} \theta$
D $\left(\frac{\pi}{3}+\frac{1}{\varphi^2}\right)(3 \phi-1)=\frac{8}{3} \theta$

D. $\left(\frac{\pi}{3}+\frac{1}{\varphi^2}\right)(3 \phi-1)=\frac{8}{3} \theta$

\begin{aligned} p & =p_C \pi=\left(\frac{a}{27 b^2}\right) \pi \\ V & =V_C \phi=(3 b) \phi \\ T & =T_C \theta=\left(\frac{8 a}{27 R b}\right) \theta\end{aligned}

Thus, substituting in the van der Waals' equation,

\left(p+\frac{a}{V_2}\right)(V-b)=R T or \quad\left(\pi+\frac{3}{\varphi^2}\right)(3 \varphi-1)=8 \theta

This is reduced state equation or,

\quad\left(\frac{\pi}{3}+\frac{1}{\varphi^2}\right)(3 \varphi-1)=\frac{8}{3} \theta