# Photosynthetic rate calculation

What are the steps in calculating the photosynthetic rate ($\mu\text{mole}\ce{O2}/$ $\text{m}^2 / \text{min}$) for a leaf that has an area of $0.0025 \text{m}^2$ and a $\ce{O2}$ evolution rate of $0.096$% $\ce{O2}/\text{min}$?

To do so , the first step is to convert it to $\mu L$ i.e. $960 \mu L\text{ }\ce{O2}/\text{min}$.

The second step should be to convert it to $\mu \text{moles}$. So, how do we do that ?

EDIT : In my textbook, the following calculation has been done but I am not able to understand it...

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

I know that $22.423$ is the volume of a mole of gas and $273K$ is $0^\circ C$ but I am not able to understand how that fits here.

Using the general case equation $[P(1) * V(1) / T(1)] = [P(2) * V(2) / T(2)]$ , and given that pressure is constant at $1\text{ atm}$ in your experiment, the equation simplifies to $[V(1) / T(1)] = [V(2) / T(2)]$
We know $V(1)$ is $960 \mu l$ and it appears that experiment is being run at $(273+23)K$, so we need to calculate what the volume $(V(2))$ would be at standard $T (273 K)$. This is given by
$$V(2) = V(1) * T(2)/T(1)$$
if we next divide $V(2) \mu L$ by $22.423 * 10^6 \mu L$, we now know how many moles of $\ce{O2}$ we have.