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

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.
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,

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

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