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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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  • $\begingroup$ 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. $\endgroup$ – LDC3 Oct 19 '14 at 18:45
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    $\begingroup$ 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. $\endgroup$ – LDC3 Oct 19 '14 at 19:39
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    $\begingroup$ @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. $\endgroup$ – LDC3 Oct 19 '14 at 23:37
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    $\begingroup$ 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. $\endgroup$ – LDC3 Oct 20 '14 at 0:16
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    $\begingroup$ @user1.618 We can figure it out both, after all we have Dalton's Law of Partial Pressure. Read the first part here: chm.davidson.edu/vce/gaslaws/daltonslaw.html $\endgroup$ – LDC3 Oct 20 '14 at 0:42
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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.

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  • $\begingroup$ 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 $\endgroup$ – most venerable sir Oct 19 '14 at 17:55
  • $\begingroup$ How would you determine the proportion of water vapour in your hydrogen? Because that is what you need to know to solve the problem. $\endgroup$ – 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.

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