Question

The reduced temperature $=\theta =\frac{T}{T_C}$

The reduced pressure $=\pi =\frac{P}{P_C}$

The reduced volume $=\varphi =\frac{V}{V_C}$

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

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

Answer 1

Answer: (A)

$P=P_c\pi =\left(\frac{a}{27b^2}\right)\pi$
$V=V_c\varphi =\left(3b\right)\varphi$
$T=T_c\theta =\left(\frac{8a}{27Rb}\right)\theta$
Hence substituting in the van der Waal's equation
$\Rightarrow \left(\pi +\frac{3}{{\varphi }^2}\right)(3\varphi -1)=8\theta$
This is reduced state equation
$\begin{array}{l}\left(\pi +\frac{3}{{\varphi }^2}\right)\left(\frac{3\varphi }{8\theta }-\frac{1}{8\theta }\right)=1\\ \left(\frac{\pi }{3}+\frac{1}{{\varphi }^2}\right)(3\varphi -1)=\frac{8}{3}\theta \end{array}$