Implicit partial differentiation to find $\frac{\partial{V}}{\partial T}$ given $RT=(p+\frac{a}{T(V+c)^2})(V-b)$ Given the gas law
$RT=(p+\frac{a}{T(V+c)^2})(V-b)$
Evaluate $\frac{\partial{V}}{\partial T}$.
 A: 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,
$$ \Rightarrow \ \frac{\partial V}{\partial T}  ·   \frac{RT}{V^2}  \ \
 = \ \ \frac{R}{V}  \ \ \Rightarrow \ \ \frac{\partial V}{\partial T}   \ \
 = \ \ \frac{V}{T} \ \  . $$
A: 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.
