Question

Use the result of the previous problem and the approximate values of a and b to find the value of T_c, P_c, V_c/N for N₂, H₂O and He.

Short answer

	$N_2$	$H_{2}O$	$He$
$V_C/N$ ($m^3$)	$18\times {10}^{-29}$	$18\times {10}^{-29}$	role="math" localid="1647074922910" $3\times {10}^{-29}$
$P_C$ (Pa)	$4.11\times {10}^6$	$1.64\times {10}^7$	$3.7\times {10}^7$
$T_C$(K)	143	572	21.5

Step-by-step solution

Given information

We know from problem 5.48

$$\begin{gathered} V_c=3Nb \\ {P}_c=\frac{1}{27}\frac{a}{b^2} \\ {T}_c=\frac{3}{8}\frac{a}{kb} \end{gathered}$$

Values of a and b are given as

	$N_2$	$H_{2}O$	He
a $(J-m^3)$	$4\times {10}^{-49}$	$1.6\times {10}^{-48}$	$1\times {10}^{-50}$
brole="math" localid="1647074059853" $(m^3)$	$6\times {10}^{-29}$	$6\times {10}^{-29}$	$1\times {10}^{-29}$

substituting the data 

	$N_2$	$H_{2}O$	$He$
$V_c/N$($m^3$)	$18\times {10}^{-29}$	$18\times {10}^{-29}$	$3\times {10}^{-29}$
$P_c$ ($Pa$)	$4.11\times {10}^6$	$1.64\times {10}^7$	$3.7\times {10}^7$
$T_c$($K$)	143	572	21.5