Question

Derive the relationship between van der waals constant and critical constants

Answer 1

Answer:

PC = a/27b²

Explanation:

short and sweet answer

Answer 2

The relationship between van der Waals constant and critical constants is $a=3V_c^{2}P_c$ and $b=\frac{V_c}{3}$.

Given:

Van der Waals constant and critical constants.

To Find:

Derive the relationship between van der Waals constant and critical constants.

Solution:

Here, the van der Waals for $n$ moles is,

$(P+\frac{a n^{2} }{V^{2} } )(V-nb)=nRT$  equation (i)

now, for 1 mole

$(P+\frac{a}{V^{2} })(V-b)=RT$  equation (ii)

now, from the equation, we can derive the value of critical constants $P_c,V_c$ and $T_c$ in the terms of $a$ and $b$, now the van der Waals constants, after expanding the above equation.

$PV+\frac{a}{V} =Pb-\frac{ab}{V^{2} } -RT=0$  equation (iii)

now, multiply the equation (iii) by $\frac{V^{2} }{P}$

here,

$\frac{V^{2} }{P}(PV+\frac{a}{V}-Pb-\frac{ab}{V^{2} } -RT=0$

$V^{3}+\frac{aV}{P}+-bV^{2}-\frac{ab}{P} -\frac{RTV^{2} }{P} =0$  equation (iv)

here, equation (iv) is a cubic equation in V. We will get three solutions after solving this. These three solutions of V are equal to the critical volume $V_c$ at the critical point.

Now, the pressure and the temperature will be $P_c$ and $T_c$.

here,

$V=V_C\\V-V_C=0\\(V-V_C)^{3}=0\\ V^{3}-3V_CV^{3} +3V_C^{2}V-V_C^{3}=0$ equation (v)

here, equation (iv) is identical with equation (v).

now, we can calculate the coefficients of $V^{2} ,V$ and constant terms in equations (iv) and (v).

now,

$-3V_CV^{2} =-[\frac{RT_C}{P_C}+b]V^{2}$

$3V_C=\frac{RT_C}{P_C} +b$  equation (vi)

$3V_C^{2} =\frac{a}{P_C}$  equation (vii)

and,

$V_C^{3} =\frac{ab}{P_C}$  equation (viii)

now, divide the equation (viii) by equation (vii).

we get,

$\frac{V_C^{3} }{3V_C^{2} } =\frac{ab/P_C}{a/P_C}$

$\frac{V_C}{3} =b$

or,

$V_C=3b$ equation (ix)

now, substitute the equation (ix) in equation (viii).

we get,

$3V_C^{2} =\frac{a}{P_C} \\P_C=\frac{a}{3V_C^{2} } =\frac{a}{3(3b^{2}) } =\frac{a}{3\times 9b^{2} }=\frac{a}{27b^{2} }$

$P_C=\frac{a}{27b^{2} }$  equation (x)

now, substitute the value of $V_c$ and $P_c$ in the equation (vii).

we get,

$3V_c=b+\frac{RT_c}{P} \\3(3b)=b+\frac{RT_c}{[a/27b^{2}] } \\9b-b=[\frac{RT_c}{a} ]27b^{2} \\8b=\frac{T_cR27b^{2} }{a} \\T_c=\frac{8ab}{27Rb^{2} } =\frac{8a}{27Rb} \\T_c=\frac{8a}{27Rb}$

By using the values of the van der Waals constant of gas, the critical constants can be calculated.

now,

$a=3V_c^{2}P_c$

and

$b=\frac{V_c}{3}$

Hence, the relationship between van der Waals constant and critical constants is $a=3V_c^{2}P_c$ and $b=\frac{V_c}{3}$.

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