Question

Why the critical pressure and temperature should increase vander Waals values?​

Answer 1

A Vander Waals equation is given by:

(P+

V

2

a

)(V−b)=RT

Where, a and b are constant

Solving above equation, P=

V−b

RT

−

V

2

a

Taking derivative of P w.r.t volume

∂V

∂P

=0

∂V

2

∂

2

P

=0

So, P becomes:

∂V

∂P

=−

(V−b)

2

RT

+

V

3

2a

=0

V

3

2a

=

(V−b)

2

RT

...........(1)

V

4

a

=

2V(V−b)

2

RT

..............(2)

Taking double derivative again,

∂V

2

∂

2

P

=

(V−b)

3

2RT

−

V

4

6a

=0

or

(V−b)

3

RT

=

V

4

3a

Put equation (2) in above equation

(V−b)

3

RT

=

2V(V−b)

2

3RT

On rearranging,

3V−3b=2V

V

c

=3b

Vc is critical volume

Use this value in equation (1)

4b

2

RT

=

27b

3

2a

T

c

=

27Rb

8a

Tc is critical temperature.