A $54.0\ \mathrm{mL}$ sample of oxygen is collected over water at $20\ \mathrm{^\circ C}$ and $770\ \mathrm{Torr}$ pressure. What is the volume of the dry gas at STP?

By using the combined gas law, I came up with the answer $\pu{51.2 mL}$.

$770 - 21.1 = 748.9\ \mathrm{Torr}$ pressure for oxygen

$21.1$ = vapor pressure of water at $23\ \mathrm{^\circ C}$

$\frac{P_1V_1}{ T_1 }= \frac{P_2V_2 }{ T_2}$

$\frac{748.9V_1}{273 }= \frac{770\times54}{296}$

$V_1 = \pu{51.2 mL}$

Is this how it's supposed to be done?

  • $\begingroup$ No. What are the conditions for STP? $\endgroup$ – MaxW Mar 10 at 22:24
  • $\begingroup$ @MaxW ohh I see now. STP would be 760 torr and 273 K so P1 should be 760 and P2 should be 748.9. Would that be right? $\endgroup$ – user75030 Mar 10 at 22:48
  • $\begingroup$ That is correct. However I'll point out that STP is a bit different now than 760 mm Hg and O C. It is now about 750 mm Hg at 0 C. $\endgroup$ – MaxW Mar 10 at 22:54
  • $\begingroup$ So $$\frac{P_{STP}V_{STP}}{T_{STP}} = \frac{PV}{T}$$ $\endgroup$ – MaxW Mar 10 at 23:01
  • 1
    $\begingroup$ @CharlieCrown - I wasn't trying to answer the question per se but rather I was trying to lead the OP to the answer. $\endgroup$ – MaxW Mar 11 at 8:01

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