Volume of dried gas at STP

Asked Mar 10 '19 at 20:37

Active Mar 11 '19 at 06:57

Viewed 1,913 times

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?

edited Mar 11 '19 at 06:57

asked Mar 10 '19 at 20:37

user75030

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
I like your comment about not closing it MaxW, but my own comment is why don't you put your answer in the answer section :P – B. Kelly Mar 11 '19 at 01:05
@user75030 see this post for details on how the definition of STP has changed. – Tyberius Mar 11 '19 at 01:06
1

@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 08:01

Volume of dried gas at STP

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