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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.

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1 Answer 1

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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.

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    $\begingroup$ You did beat me by seconds. Sometimes, mathjaxing the ideal gas law is a burden :D $\endgroup$ Jan 28, 2014 at 18:32
  • $\begingroup$ It's too bad that SE doesn't permit the OP to select more than 1 correct answer. $\endgroup$
    – ron
    Jan 28, 2014 at 18:53
  • $\begingroup$ 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! $\endgroup$ Jan 28, 2014 at 19:00

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