# 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.

• You did beat me by seconds. Sometimes, mathjaxing the ideal gas law is a burden :D Jan 28, 2014 at 18:32
• It's too bad that SE doesn't permit the OP to select more than 1 correct answer.
– ron
Jan 28, 2014 at 18:53
• I think that upvoting several answers is possible, but only one can be accepted. You were faster and explained in more detail than I would have done! Jan 28, 2014 at 19:00