Photosynthetic rate calculation What are the steps in calculating the photosynthetic rate ($\pu{\mu mole}~ \ce{O2} \pu{m2/ min}$) for a leaf that has an area of $\pu{0.0025 m2}$ and a $\ce{O2}$ evolution
rate of $0.096~\ce{O2}/\pu{min}$?

To do so , the first step is to convert it to $\mu L$ i.e. $\pu{960 \mu L} ~\ce{O2}/\pu{min}$.
The second step should be to convert it to $\pu{\mu moles}$. So, how do we do that ?
EDIT : In my textbook, the following calculation has been done but I am not able to understand it...

$\pu{960 \mu L}~\ce{O2}~\pu{/L} / [(273+23)/273) \times 22.423]$

I know that $22.423$ is the volume of a mole of gas and $\pu{273K}$ is $\pu{0^\circ C}$ but I am not able to understand how that fits here.
Using the general case equation
$$\frac{P_1V_1 }{T_1} = \frac{P_2 V_2} {T_2}$$
and given that pressure is constant at $\pu{1  atm}$ in your experiment, the equation simplifies to
$$\frac{V_1 }{T_1} = \frac{ V_2} {T_2}$$
We know $V_1=\pu{960 \mu L}$ and it appears that the experiment is being run at $\pu{(273+23)K}$, so we need to calculate what the volume ($V_2$) would be at standard T ($\pu{273 K}$).  This is given by
$$V_2 = \frac{V_1 T_2}{T_1}$$
if we next divide $V_2$ by $\pu{22.423E6 \mu~L}$, we now know how many moles of $\ce{O2}$ we have.