By Sunil Bhardwaj

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Vander Waals equation for the real gas is, $$\left( P+\frac { a }{ { V }^{ 2 } } \right) \left( V-b \right) =RT$$ On simplifying the equation, $$PV+\frac { a }{ { V } } -Pb-\frac { ab }{ { V }^{ 2 } } =RT$$ Multilpy with $${ V }^{ 2 }$$ we get, $$P{ V }^{ 3 }+aV-Pb{ V }^{ 2 }-ab=RT{ V }^{ 2 }$$ Devide by P, $${ V }^{ 3 }+\frac { aV }{ P } -b{ V }^{ 2 }-\frac { ab }{ P } =\frac { RT }{ P } { V }^{ 2 }$$ $${ V }^{ 3 }-b{ V }^{ 2 }-\frac { RT }{ P } { V }^{ 2 }+\frac { aV }{ P } -\frac { ab }{ P } =0$$ $${ V }^{ 3 }-\left( b+\frac { RT }{ P } \right) { V }^{ 2 }+\frac { a }{ P } V-\frac { ab }{ P } =0\qquad ....(1)$$ At critical temperature, pressure and volume. $$T={ T }_{ c }, P={ P }_{ c }, \ and \ V=V_{ c },$$ Substituting the values of T and P in equation (1) $${ V }^{ 3 }-\left( b+\frac { R{ T }_{ c } }{ { P }_{ c } } \right) { V }^{ 2 }+\frac { a }{ { P }_{ c } } V-\frac { ab }{ { P }_{ c } } =0 \qquad ....(2)$$ Also $$V=V_{ c }$$ $$\therefore V-V_{ c }=0 \ and \ { \left( V-V_{ c } \right) }^{ 3 }=0$$ Simplifying this equation $${ V }^{ 3 }-3V_{ c }{ V }^{ 2 }+3{ VV_{ c } }^{ 2 }-{ V_{ c } }^{ 3 }=0\qquad ....(3)$$ Hench the coefficients of simirlar power of V must be equal in both the equation (2) and (3) $$3{ V }_{ c }=\left( b+\frac { R{ T }_{ c } }{ { P }_{ c } } \right) \qquad ....(4)$$ $$3{ V_{ c } }^{ 2 }=\frac { a }{ { P }_{ c } } \qquad ....(5)$$ $${ V_{ c } }^{ 3 }=\frac { ab }{ { P }_{ c } } \qquad ....(6)$$ Deviding equation (6) with eq (5) $$\frac { { V_{ c } }^{ 3 } }{ 3{ V_{ c } }^{ 2 } } =\frac { ab }{ { P }_{ c } } \times \frac { { P }_{ c } }{ a }$$ $$\therefore { V_{ c } }=3b\qquad ....(7)$$ Substituting the value of $${ V_{ c } }$$ in equation (5) $$3\times { \left( 3b \right) }^{ 2 }=\frac { a }{ { P }_{ c } }$$ $$\therefore { P }_{ c } = \frac { a }{ 27{ b }^{ 2 } } \qquad ....(8)$$ Putting values of $${ P }_{ c }$$ and $${ V_{ c } }$$ in equation (4) $$3\times 3b=\left( b+\frac { R{ T }_{ c } }{ \frac { a }{ 27{ b }^{ 2 } } } \right)$$ $$9b=b+R{ T }_{ c }\frac { 27{ b }^{ 2 } }{ a }$$ $$\therefore R{ T }_{ c }\frac { 27{ b }^{ 2 } }{ a } =8b$$ $$\therefore { T }_{ c }=\frac { 8 }{ 27 } \frac { a }{ R{ b }^{ 2 } } \qquad ....(9)$$ $$\therefore { V_{ c } }=3b\qquad { P }_{ c } = \frac { a }{ 27{ b }^{ 2 } } \ and \ { T }_{ c }=\frac { 8 }{ 27 } \frac { a }{ b }$$

MCQ on Gaseous State from Physical Chemistry
###### Prof. Gianfranco Coletti

Shared publicly - 2019-08-23 00:00:00

Don’t want your columns to simply stack in some grid tiers? Use a combination of different classes for each tier as needed. See the example below for a better idea of how it all works.

###### Prof. Maheshwar Sharon

Shared publicly - 2019-08-24 00:00:00

For grids that are the same from the smallest of devices to the largest, use the .col and .col-* classes. Specify a numbered class when you need a particularly sized column; otherwise, feel free to stick to

###### sunil

Shared publicly - 2023-02-28 11:09:52

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###### ss

Shared publicly - 2023-02-28 10:48:10

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