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How do i begin taking the partial derivative of the below equation with respect to T?$$p = \frac{RT}{V}+\left(B_0 RT-A_0-\frac{C_0}{T^2}\right)\frac{1}{V^2}+(bRT-a)\frac{1}{V^3}+\frac{A\alpha}{V^6}+\frac{c\left(1+\frac{\gamma}{V^2}\right)}{T^2}\left(\frac{1}{V^3}\right)e^\frac{-\gamma}{V^2}$$

I need to show that $$\frac{\partial^2p}{\partial T^2} = \frac{6}{V^2T^4}\left({\frac{C}{V}{\left(1+\frac{\gamma}{V^2}\right) e^\frac{-\gamma}{V^2}-C_0}}\right)$$

Thank You

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1 Answer 1

up vote 3 down vote accepted

You hold all variables other than $T$ fixed, and then differentiate with respect to $T$.

To see that taking the first partial is easy, note that your expression has the form

$$ p=a_1 T +a_2 T+a_3+a_4T^{-2}+a_5 T+a_6+a_7 T^{-2} $$ where

$\qquad a_1={R\over V}$,

$\qquad a_2={B_0R\over V^2}$,

$\qquad a_3={-A_0\over V^2}$,

$\qquad a_4={-C_0\over V^2}$,

$\qquad a_5= {bR\over V^3}$,

$\qquad a_6={-a\over V^3}+{A\alpha\over V^6}$,


$\qquad a_7={c(1+{\gamma\over V^2}})\cdot{1\over V^3}\cdot e^{-\gamma/V^2}$.

Note that none of the $a_i$ depend on $T$; thus, when taking the first partial of $p$, we are differentiating a sum whose terms are multiples of powers of $T$. We have $$\tag{1}\eqalign{ {\partial p\over\partial T} &=a_1+a_2+0-2a_4T^{-3}+a_5+0-2a_7T^{-3}\cr &=a_1+a_2+a_5-2a_4T^{-3}-2a_7T^{-3} } $$

Now take the partial of $(1)$ with respect to $T$ above to obtain the second partial of $p$: $$ \eqalign{ {\partial^2 p\over\partial T^2} &={\partial\over\partial T}\bigl(a_1+a_2+a_5-2a_4T^{-3}-2a_7T^{-3}\bigr)\cr &=6a_4T^{-4}+6a_7T^{-4}\phantom{T\over T}\cr &=6\cdot{-C_0\over V^2}\cdot{1\over T^4}+6\cdot{c(1+{\gamma\over V^2})}\cdot{1\over V^3}\cdot e^{-\gamma/V^2}\cdot{1\over T^4}\cr &={6\over V^2T^4}\Bigl( {c\over V}(1+{\gamma\over V^2}) e^{-\gamma/V^2} -C_0\Bigr). } $$

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