Question

derive the relation between critical constants and Vander waal constants​

Answer 1

Explanation:

From 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.

Answer 2

Answer:

PC = a/27b²

Explanation:

Critical Phenomena and Van der Walls Equation:

For one mole of a gas the Van der Waals equation,

(P + a/V2)(V - b) = RT

V3 - (b + RT/P)V2 + (a/P)V - ab/P = 0

This equation has three roots in V for given values of a, b, P and T. It found that either all the three roots are real or one is real and the other two are imaginary.

Expression of Critical Constants in Terms of Van der Waals Constants:

Again with increase of temperature the minimum and maximum points come close to each other and at the critical point (C) both maximum and minimum coalesce. The slope and curvature both becomes zero at this point.

Thus the mathematical condition of critical point is,

(dP/dV)T = 0 and (d2P/dV2)T = 0

Van der Waals equation for 1 mole gas is,

(P + a/V2)(V - b) = RT

or, P = RT/(V - b) - a/V2

Differentiating Van der Waals equationwith respect to V at constant T,

We get Slope,

(dP/dV)T = - {RT/(V - b)2} + 2a/V3

And Curvature,

(d2P/dV2)T = {2RT/(V - b)3} - 6a/V4

Hence at the critical point,

- {RTC/(VC - b)2} + 2a/VC3 = 0

or, RTC/(VC - b)2 = 2a/VC3

and {2RTC/(VC - b)3} - 6a/VC4 = 0

or, 2RTC/(VC -b)3 = 6a/VC4

Thus, (VC - b)/2 = VC/3

∴ VC = 3b

Putting the value of VC = 3b in

RTC/(VC - b)2 = 2a/VC3.

We have, RTC/4b2 = 2a/27b3

∴ TC = 8a/27Rb

Again the Van der Walls equation at the critical state,

PC = RTC/(VC - b) - a/VC2

Putting the value of VC and TC,

∴ PC = a/27b2