Posted by **mel** on Monday, November 26, 2007 at 1:24pm.

One mole of H2O(g) at 1.00 atm and 100.°C occupies a volume of 30.6 L. When one mole of H2O(g) is condensed to one mole of H2O(l) at 1.00 atm and 100.°C, 40.66 kJ of heat is released. If the density of H2O(l) at this temperature and pressure is 0.996 g/cm3, calculate E for the condensation of one mole of water at 1.00 atm and 100.°C.

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**DrBob222**, Monday, November 26, 2007 at 11:03pm
Delta E = q + w

q = -40.66 kJ.

w = -p(delta V) = -p(V2-V1)

p = 1.00 atm.

V1 = 30.6 L

V2 = calculated from density. 1 mol H2O has mass of 18.015 g, density is 0.996, V2 = ??

Don't forget to change w from L*atm units to J by multiplying w*101.3 J/L*atm. Also remember w is in Joules and q is in kJ so make them the same unit.

Post your work if you get stuck.

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**Anonymous**, Wednesday, November 28, 2007 at 4:59pm
i don't get how to convert the density to V2...

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**Anonymous**, Friday, February 13, 2009 at 12:24pm
density=mass/volume so volume=mass/density

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**Anonymous**, Saturday, April 10, 2010 at 7:44pm
how do you convert cm^3 to L so you can subtract V1 from V2

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**Anonymous**, Friday, October 22, 2010 at 12:25am
100 Cm^3=1 L

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**Anonymous**, Tuesday, September 25, 2012 at 10:14pm
1000 cm^3 = 1 L since 1 cm^3 = 1 mL and since there are 1000 mL in one L.

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