# Volume of dried gas at STP

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?

• No. What are the conditions for STP? – MaxW Mar 10 '19 at 22:24
• @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? – user75030 Mar 10 '19 at 22:48
• 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. – MaxW Mar 10 '19 at 22:54
• So $$\frac{P_{STP}V_{STP}}{T_{STP}} = \frac{PV}{T}$$ – MaxW Mar 10 '19 at 23:01
• @CharlieCrown - I wasn't trying to answer the question per se but rather I was trying to lead the OP to the answer. – MaxW Mar 11 '19 at 8:01