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Volume of wet hydrogen collected at room temperature: $\pu{40.48 mL}$, ambient pressure: $\pu{29.7 inHg}$ ($\pu{754 torr}$), ambient temperature: $\pu{295 K}$ ($\pu{22^\circ C}$), vapor pressure of water at $\pu{22^\circ C}$: $\pu{19.83 torr}$. Partial pressure of dry hydrogen: $\pu{9.87 inHg}$.

Find the volume of hydrogen under standard conditions.

The equation of combined gas law is $p_1V_1/T_1 = p_2V_2/T_2$.

Say if the volume of wet hydrogen gas is $\pu{40.48 mL}$, does this equation still work if part of that volume is water vapor? Or, in other words, does the same volume of water vapor and hydrogen gas exert the same pressure?

One of the correct way to do this problem is find the moles of dry hydrogen gas using the room temperature and volume of wet hydrogen gas, gas constant and room temp. Right here, I am kinda confused why the volume of wet hydrogen if we are trying to find the moles of dry hydrogen gas. Then rest of this solution is plugging the number of moles into the equation along with STP to find the volume of dry hydrogen gas.

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most venerable sir
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  • Look up Dalton's Law of Partial Pressures: it states that the total pressure is equal to the sum of the pressures of the individual gases. So, any amount of moisture in the hydrogen gas will have an influence on the pressure. – LDC3 Oct 19 '14 at 18:45
  • That is what I mean, because the water vapor added, the equation is not proportional – most venerable sir Oct 19 '14 at 18:55
  • If water vapor is considered an ideal gas, then the vapor pressure of the water is only dependent on temperature, and not the gas that it is dissolved in. So, from your experiment, you should have the atmospheric pressure reading at the time of the experiment, and the temperature of the experiment (usually room temperature). Now, you just need to look up the vapor pressure of water at the temperature recorded and determine the pressure of the hydrogen. – LDC3 Oct 19 '14 at 19:02
  • Something I find contradicting is at constant temperature if the dry hydrogen has a lower pressure than than the wet hydrogen, which is equal to the air pressure, the the volume of dry hydrogen has to much bigger than that of wet hydrogen. It is impossible for dry hydrogen to have a larger volume – most venerable sir Oct 19 '14 at 19:06
  • It doesn't have a larger volume. When you mix gases, the volume of the 1 atm of the mix (which is determine from the amount of each gas) is made up of the volumes of the individual gases. In your statement, you have different amount of hydrogen (wet and dry) in the same volume so it looks like you need a larger volume when it is cause by the amount you have. – LDC3 Oct 19 '14 at 19:26
  • But the dry hydrogen is part of the wet hydrogen( dry hydrogen+ water vapor) shouldn't smaller amount = smaller volume – most venerable sir Oct 19 '14 at 19:31
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    You're still confused. Dry hydrogen (less pressure, some volume < 40.48mL) + water vapor (pressure and some volume) equals 1 atm and 40.48mL. Wet hydrogen (water is included) = 1 atm and 40.48mL. I believe that you don't see that there are different volumes involved for the same amount of hydrogen. – LDC3 Oct 19 '14 at 19:39
  • LDC3 is all your saying that the volume of water in wet hydrogen gas is negligible? – most venerable sir Oct 19 '14 at 21:16
  • Because the volume of wet hydrogen could be seen as the volume of dry hydrogen?? – most venerable sir Oct 19 '14 at 21:17
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    @user1.618 No, I'm not saying that. But the amount of water vapor is small. We know we have (754.38mm Hg - 19.83mm Hg) for the pressure of dry hydrogen and 19.83mm Hg for water vapor. For wet hydrogen we have 754.38mm Hg and 40.48mL. The pressure for water vapor is about 2.6% that of hydrogen, that is not negligible. – LDC3 Oct 19 '14 at 23:37
  • But then why do we use the volume of wet hydrogen gas in the equation?? – most venerable sir Oct 19 '14 at 23:40
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    Because we can use $PV=nRT$ to determine the amount of hydrogen. $754.38mm Hg \times 40.48mL = n \times R \times 295K$. You will need to look up R with the correct units. This give the amount of wet hydrogen and since the law of partial pressure ($P_T=P_H+P_W$) infers that amount of the gases is $n_T=n_H+n_W$. When we get the correct amount of hydrogen, then we can determine the volume at STP. – LDC3 Oct 20 '14 at 00:16
  • @user1.618: The volume of the wet gas is used because that's the volume all the gas is in. As LDC3 says, the pressure of hydrogen is ambient pressure - water vapor pressure. An alternative way of thinking about the problem is: let both water vapor and hydrogen be at ambient pressure. Then together they must take up the observed volume. – Abel Friedman Oct 20 '14 at 00:18
  • We don't have the moles for water! – most venerable sir Oct 20 '14 at 00:24
  • @user1.618: If you like you can calculate moles for the amount of water, you know the pressure and the volume of the water vapour. But you don't have to, this number isn't needed. – Abel Friedman Oct 20 '14 at 00:41
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    @user1.618 We can figure it out both, after all we have Dalton's Law of Partial Pressure. Read the first part here: http://www.chm.davidson.edu/vce/gaslaws/daltonslaw.html – LDC3 Oct 20 '14 at 00:42

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Consider the composition of the gas in your eudiometer. Most of it is hydrogen, but some of it is water vapour. In other words, not all the pressure is exerted by hydrogen, part of it is due to water. You could ask yourself: what would happen to the volume of gas produced, had you passed it through a drying tube.

You should have been given vapour pressure of water in a table. You should also have taken down value of air pressure and ambient temperature.

Abel Friedman
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  • The temperature is kept the same, I know the pressure and the vaole of wet H2. But because part of the hydrogen is water vapor, the proportion cannt be established – most venerable sir Oct 19 '14 at 17:55
  • How would you determine the proportion of water vapour in your hydrogen? Because that is what you need to know to solve the problem. – Abel Friedman Oct 19 '14 at 18:05
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WAIT! I got it; the "volume of the wet hydrogen gas" is ALSO the volume of the DRY hydrogen gas. That's because Dalton's Law of partial pressure assumes/ is for CONSTANT volume. IE * at constant volume *: the total pressure of wet hydrogen at x mL = vapor pressure at x ML + dry hydrogen pressure at x mL. It's x mL for both.

Another way to think about it is IF the volume of the dry gas were less, then that would mean the pressure of the dry gas is more than the one we calculated for using Dalton's law of partial pressure * at constant volume * (Boyle's Law) and we are keeping the pressure that we calculated.

Alexander Ye
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