Implicit partial differentiation to find $\frac{\partial{V}}{\partial T}$ given $RT=(p+\frac{a}{T(V+c)^2})(V-b)$

Question

Given the gas law

$RT=(p+\frac{a}{T(V+c)^2})(V-b)$

Evaluate $\frac{\partial{V}}{\partial T}$.

Answer 1

This looks like a modification of the vanderWalls equation; I will take it that this is written as intended, since I haven't found this particular variant online (so far). Since we want to differentiate $ \ V \ \ , $ we may wish to find an arrangement of the equation that will create the least misery possible, for instance,

$$ \frac{\partial}{\partial T} \ [ \ RT \ · \ (V-b)^{-1} \ ] \ \ = \ \ \frac{\partial}{\partial T} \ \left[ \ p \ + \ a·T^{-1}·(V+c)^{-2} \ \right]$$

$$ \Rightarrow \ \ R·\frac{\partial T}{\partial T} · (V-b)^{-1} \ + \ RT · \frac{\partial}{\partial T} \ [ \ (V-b)^{-1} \ ] $$ $$ = \ \ \frac{\partial p}{\partial T} \ + \ a · \ \frac{\partial}{\partial T} \ [ \ T^{-1} \ ]·(V+c)^{-2} \ + \ a·T^{-1}·\frac{\partial}{\partial T} \ [ \ (V+c)^{-2} \ ] $$

$$ \Rightarrow \ \ R·1 · (V-b)^{-1} \ + \ RT · [ \ -(V-b)^{-2} \ ] · \frac{\partial V}{\partial T} $$ $$ = \ \ 0 \ + \ a · ( \ -T^{-2} \ )·(V+c)^{-2} \ + \ a·T^{-1}· [ \ -2·(V+c)^{-3} \ ] · \frac{\partial V}{\partial T} $$ [since this particular partial derivative is usually taken at constant pressure]

$$ \Rightarrow \ \frac{\partial V}{\partial T} \ · \ \left[ \ \frac{RT}{(V-b)^2} \ - \ \frac{2a}{T · (V+c)^3} \right] \ \ = \ \ \frac{R}{V-b} \ + \ \frac{a}{T^2 · (V+c)^2} \ \ . $$

You can manipulate the expression from here to whatever purpose you planned for it. I am assuming that $ \ c \ $ is also a constant, as $ \ a \ $ and $ \ b \ $ are. Note that setting $ \ a \ = \ b \ = \ 0 \ $ reduces this to the proper result for an ideal gas, $$ \Rightarrow \ \frac{\partial V}{\partial T} · \frac{RT}{V^2} \ \ = \ \ \frac{R}{V} \ \ \Rightarrow \ \ \frac{\partial V}{\partial T} \ \ = \ \ \frac{V}{T} \ \ . $$

Answer 2

If $f(V,T) = 0,$ then $0 = \partial_T f = f_V \frac{\partial V}{\partial T} + f_T,$ so $\frac{\partial V}{\partial T} = -\frac{f_T}{f_V}.$

We obtain $f_V = p+\frac{a}{T(V+c)^2} - (V-b)\frac{2a}{T(V+c)^3}, f_T = \frac{a(b-V)}{T^2(V+c)^2}-R,$ the simplification is up to you.