Question

Derivation
relation between critical constants and van der waals equation.​

Answer 1

Derivation

From Vander Waals equation

$(p+ \frac{a}{ {v}^{2} } )(v - b) = rt\\ \\ \\ pv - pb + \frac{a}{v } - \frac{ab}{ {v}^{2} } = rt\\ \\ \\ p {v}^{3} - {v}^{2} pb + av - ab = rt {v}^{2} \\ \\ {v}^{3} - {v}^{2} b + \frac{ab}{p} + \frac{av}{p} - \frac{rt {v}^{2} }{p} = 0 \\ \\ {v}^{3} - (b + \frac{rt}{p}) {v}^{2} + \frac{av}{p} - \frac{ab}{p} = 0 \\ \\ v = v. \\ \\ (v - v.) ^{2} = \\ {v}^{3} - 3 {v}^{2} v. + 3v {v.}^{2} - {v.}^{3} \\ 3v. = b + \frac{rt}{p} .....(1)\\ \\ 3 {v.}^{2} = \frac{a}{p} ......(2) \\ \\ {v.}^{3} = \frac{ab}{p} ........(3)\\ \\ divide \: equ3 \: by \:equ 2 \\ \\ 3 {v.}^{2} = 3b \\ \\ put \: in \: equation \: \: 3\: \: we \: get\\ p = \frac{a}{27 {b}^{2} } \\ \\ putting \: in \: eq \: 1 \\ \\ t = \frac{8a}{27r {b}^{2} }$

$Here~ replace ~t ~by~ T_c \\ v~ by~ V_c \\ p~ by~ P_c \\ rt=RT_c$

They are critical constants.

THIS IS THE RELATION BETWEEN CRITICAL CONSTANTS AND VANDER WAALS EQUATION.

Answer 2

the equation we can derive the values of critical constants Pc, Vc and Tc in terms of a and b, the van der Waals constants. ... substituting the values of Vc and Pc in equation (6.28), The critical constants can be calculated using the values of van der waals constant of a gas and vice versa.

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