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$$\begin{align} p &= kT\left(\frac{\partial \ln Q}{\partial V}\right)_T \\ &= \frac{kT}{Q}\left(\frac{\partial Q}{\partial V}\right)_T \\ &= \frac{NkT}{q}\left(\frac{\partial q}{\partial V}\right)_T \\ &= \frac{NkT\Lambda^3}{V}\cdot\frac{1}{\Lambda^3} \\ &= \frac{NkT}{V} \\ &= \frac{nRT}{V} \end{align}$$

Looking at the first equation in this series I'm really confused how the $\partial \ln Q/\partial V$ has been simplified to $\partial Q / \partial V$.

My hunch is that it's something to do with the derivative of $\ln Q$ being $1/Q$ however I'm not sure how this ties in.

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  • $\begingroup$ The expression $kT \left( \dfrac{\partial \ln Q}{\partial V} \right)_T$ is simply multiplied with $\dfrac{d Q}{d Q}$ (which equals 1, so it is okay). Then you rearrange the denominators. See orthocresol's answer. $\endgroup$ – Yoda Nov 28 '15 at 19:00
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This is more of a maths question, but I can give a quick answer.

As you said,

$$\frac{\mathrm{d}}{\mathrm{d}Q}\ln Q = \frac{1}{Q}$$

You could write this as

$$\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} = \frac{1}{Q}$$

So, by the chain rule,

$$\begin{align} \frac{\partial (\ln Q)}{\partial V} &= \frac{\partial Q}{\partial V}\cdot\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} \\ &= \frac{1}{Q}\frac{\partial Q}{\partial V} \end{align}$$

If you are studying physical chemistry, it is a good idea to remember this trick. It appears everywhere.

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    $\begingroup$ This is called the chain rule in case the OP wants to see an ultra formal Wikipedia page about it. $\endgroup$ – jheindel Nov 28 '15 at 19:08

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