The value of critical temperature in terms of van der Waal's constants $a$ and $b$ is given by

T_{c} = \frac{a}{2 R b}

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T_{c} = \frac{a}{27 R b}

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T_{c} = \frac{8 a}{27 R b}

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T_{c} = \frac{17 a}{8 R b}

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Solution

The correct option is C $T_{c} = \frac{8 a}{27 R b}$
A Vander Waals equation is given by:

$(P + \frac{a}{V^{2}}) (V - b) = R T$

Where, a and b are constant

Solving above equation, $P = \frac{R T}{V - b} - \frac{a}{V^{2}}$

Taking derivative of P w.r.t volume

$\frac{\partial P}{\partial V} = 0$

$\frac{\partial^{2} P}{\partial V^{2}} = 0$

So, P becomes:

$\frac{\partial P}{\partial V} = - \frac{R T}{{(V - b)}^{2}} + \frac{2 a}{V^{3}} = 0$

$\frac{2 a}{V^{3}} = \frac{R T}{{(V - b)}^{2}} . . . . . . . . . . . (1)$

$\frac{a}{V^{4}} = \frac{R T}{2 V {(V - b)}^{2}} . . . . . . . . . . . . . . (2)$

Taking double derivative again,

$\frac{\partial^{2} P}{\partial V^{2}} = \frac{2 R T}{{(V - b)}^{3}} - \frac{6 a}{V^{4}} = 0$

$o r$

$\frac{R T}{{(V - b)}^{3}} = \frac{3 a}{V^{4}}$

Put equation (2) in above equation

$\frac{R T}{{(V - b)}^{3}} = \frac{3 R T}{2 V {(V - b)}^{2}}$

On rearranging,

$3 V - 3 b = 2 V$

$V_{c} = 3 b$

V_c is critical volume

Use this value in equation (1)

$\frac{R T}{4 b^{2}} = \frac{2 a}{27 b^{3}}$

$T_{c} = \frac{8 a}{27 R b}$

T_c is critical temperature.

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