# On Van Der Waals relation: how to derive du/dv = a/v^2 without using the relation dU=TdS−PdV?

I have read several threads and papers which show how to derive $$\left(\frac{\partial u}{\partial v}\right)_T = \frac{a}{v^2}$$ by using the Maxwell's relation $$\left(\frac{\partial{S}}{\partial{V}}\right)_T=\left(\frac{\partial{P}}{\partial{T}}\right)_V$$ with $$dU = T\mathrm dS - P\mathrm dV$$:

$$\left(\frac{\partial{U}}{\partial{V}}\right)_T = T\left(\frac{\partial{P}}{\partial{T}}\right)_V - P \tag{1}$$

and using this and the first derivative of the Van De Walls equation we can arrive at that relation. A good explanation can be found here, here and here

But I'm reading the book Heat and Thermodynamics by W. Zemansky, Richard H., and in Chapter 4 (First Law), in one of it's exercises I need this relation, but at this point in the book the concept of $$S$$ and Maxwell's relations are not yet defined. So at this point, how can I get to Equation 1 using only state equations and the first law?

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