By Sunil Bhardwaj

4135 Views


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) $$ 3\times 3b=\left( b+\frac { R{ T }_{ c } }{ \frac { a }{ 27{ b }^{ 2 } } } \right) $$ $$ 9b=b+R{ T }_{ c }\frac { 27{ b }^{ 2 } }{ a } $$ $$ \therefore R{ T }_{ c }\frac { 27{ b }^{ 2 } }{ a } =8b $$ $$ \therefore { T }_{ c }=\frac { 8 }{ 27 } \frac { a }{ R{ b }^{ 2 } } \qquad ....(9) $$ $$ \therefore { V_{ c } }=3b\qquad { P }_{ c } = \frac { a }{ 27{ b }^{ 2 } } \ and \ { T }_{ c }=\frac { 8 }{ 27 } \frac { a }{ b } $$