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Topic: Dissociation of Water and Gas Expansion  (Read 3556 times)

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Offline halpmeh

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Dissociation of Water and Gas Expansion
« on: December 14, 2007, 03:38:45 PM »
I'm having some difficulty understanding this problem:

Inside an expandable tube with a current volume of 1L, and an internal pressure of 1atm that is at equilibrium with the outside and an internal temperature of 273K, a perfectly regulated electric current is used to dissociate 500 mL of water into its constituent parts.   If the resultant gases were allowed to expand until the system reaches a new equilibrium, at what volume would such a state occur? 

I assumed that the way to do this would be the following.
First I calculated how many moles of gas result from the dissociation by calculating the number of grams of water from its density and then using the formula for the reaction. H2O(l) --> H2(g) + 1/2O2(g)
500mL water = 500g water * 1mol water/18g water = 27.777mol which becomes 27.777 moles of H2(g) + 13.8888 moles of O2(g).
Next I simply plugged this into PV = nRT for conditions of 1 atm pressure and 273K, but this came out with an answer of over 90000L, which doesn't seem to make much sense to me.
Does anyone know if this is correct or if there is some other way to go about doing this problem? The teacher put the problem under the category of Work, which I know is the Integral of PdV, but I don't see how this relates to the problem in any way.
Any input is much appreciated.

Offline Yggdrasil

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Re: Dissociation of Water and Gas Expansion
« Reply #1 on: December 15, 2007, 02:07:20 PM »
Your approach looks right, but I get an answer that's closer to 900L.  Check your calculations on the ideal gas law (everything before then looks ok).

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