Van der Waals equation: $$p=\frac{R T}{v-b}-\frac{a}{v^2}$$
I need to calculate $$k_T=-\frac{1}{v}\left(\frac{\partial v}{\partial p}\right)_T$$
Also we know: $$p_c=\frac{a}{27b^2},~v_c=3b,~T_c=\frac{8a}{27bR}$$
So, $$\left(\frac{\partial p}{\partial v}\right)_T=-~\frac{R T}{(v-b)^2}+2\frac{a}{v^3}$$
At $V=V_c$:
$$-~\frac{R T}{(v-b)^2}+2\frac{a}{v^3}=-~\frac{R(T-T_c)}{4b^2}$$
Hence:
$$k_T(T,v=v_c)=-~\frac{1}{v}\left(\frac{\partial v}{\partial p}\right)_T\approx C(T-T_c)^{-\gamma}$$
But I cant calculate $k_T(T=T_c,p)$.
You are asked to calculate the isothermal compressibility, i.e. $ \kappa_T=-\frac{1}{v}\left(\frac{\partial v}{\partial p}\right)_T $.
You can use the fact that $ \left( \frac{\partial v}{\partial p} \right)_T = \left( \frac{\partial p}{\partial v} \right)_T^{-1} $, i.e.
$$ \left(\frac{\partial p}{\partial v}\right)_T = - \frac{R T}{(v-b)^2} + \frac{2a}{v^3} = - \frac{RTv^3}{(v-b)^2v^3} + \frac{2a(v-b)^2}{(v-b)^2v^3} = \frac{2a(v-b)^2 - RTv^3}{(v-b)^2v^3} $$
Substituting $ v = v_c = 3b $ into this equation, we get:
$$ \left(\frac{\partial p}{\partial v}\right)_T = \frac{2a(v_c-b)^2 - RTv_c^3}{(v_c-b)^2{v_c}^3} = \frac{2a(4b^2) - RT(3b)^3}{(2b)^2(3b)^3} = \frac{b^2(8a - 27RTb)}{108b^5} = \frac{8a - 27RTb}{108b^3} $$
Therefore, we get:
$$ \kappa_T = - \frac{1}{v_c} \left(\frac{\partial v}{\partial p}\right)_T = - \frac{1}{v_c} \left( \frac{\partial p}{\partial v} \right)_T^{-1} = - \frac{1}{3b} \times \frac{108b^3}{8a - 27RTb} = \frac{36b^2}{27RTb - 8a} $$