How to Derive Van der Waals Equations from Thermodynamic Relations?

AI Thread Summary
The discussion focuses on deriving Van der Waals equations from thermodynamic relations for a pure substance under constant heat capacity conditions. The main equations involved include Tds = du + pdv and c_{v}dT + T(∂p/∂T)v dv = 0. The user expresses confusion about how to proceed with the derivation, despite successfully manipulating some equations. They have shown relationships between pressure, volume, and temperature but remain uncertain about the next steps. The conversation highlights the challenges of applying thermodynamic principles to derive the desired equations.
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Homework Statement



For any Pure Substance, use the relation (1) to show that for a Van der Waals gas, over a range of conditions where c_{v} is effectively constant, that equation (2) gives equations (3) and (4)

Homework Equations



(1) ~ Tds = du + pdv

(2) ~ c_{v}dT + T \frac{\partial p}{\partial t}_{v}dv = 0

(3) ~ T(v-b)^{\alpha} = constant

(4) ~ p + \frac{a}{v^{2}}\approxT^{\frac{1+\alpha}{\alpha}}

(5) ~ \alpha=\frac{R}{c_{v}}

(6) ~ (p + \frac{a}{v^{2}})(v-b)=RT

The Attempt at a Solution



I really haven't got a clue where to go with this one, this question obviously has something to do with c_{v}.

I was thinking for the first part since its equal to a constant then maybe integration or diferrentiation will come into this at some point. As for the second part i haven't got the slightest clue where to go so any tips would be nice.

I've written down all the equation i think could possibly come into play here, and have tried rearranging them but have not got anywhere with this yet. I can derive equation (2) from equation (1) but i can't see that being much use

I'm new round here and would like to apologise for any thing that I've done wrong

Thanks alot
 
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i have managed to show that

p+\frac{a}{v^{2}}=\frac{RT}{v-b}=T(\frac{\partial p}{\partial T})_{v}

and that

\frac{a}{v^{2}} = (\frac{\partial u}{\partial v})_{T}

but am now unsure where to go or if this is even the right path to take
 
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