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(A) The Vander Waals equation for one mole of gas is given by:$(P + \frac{a}{V^2})(V - b) = RT$At the critical point,the pressure $P$ as a function of

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$T_c = \frac{8a}{27Rb}$

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(A) The Vander Waals equation for one mole of gas is given by:$(P + \frac{a}{V^2})(V - b) = RT$At the critical point,the pressure $P$ as a function of volume $V$ satisfies the conditions:$\frac{\partial P}{\partial V} = 0$ and $\frac{\partial^2 P}{\partial V^2} = 0$From the equation of state,$P = \frac{RT}{V-b} - \frac{a}{V^2}$.Taking the first derivative with respect to $V$:$\frac{\partial P}{\partial V} = -\frac{RT}{(V-b)^2} + \frac{2a}{V^3} = 0 \implies \frac{RT}{(V-b)^2} = \frac{2a}{V^3}$ ... $(1)$Taking the second derivative with respect to $V$:$\frac{\partial^2 P}{\partial V^2} = \frac{2RT}{(V-b)^3} - \frac{6a}{V^4} = 0 \implies \frac{RT}{(V-b)^3} = \frac{3a}{V^4}$ ... $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{V-b}{1} = \frac{2a/V^3}{3a/V^4} = \frac{2V}{3}$$3V - 3b = 2V \implies V_c = 3b$Substituting $V_c = 3b$ into equation $(1)$:$\frac{RT_c}{(3b-b)^2} = \frac{2a}{(3b)^3}$$\frac{RT_c}{4b^2} = \frac{2a}{27b^3}$$T_c = \frac{8a}{27Rb}$

The value of critical temperature in terms of Vander Waals constants $a$ and $b$ is

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