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.
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.
i don't get how to convert the density to V2...
density=mass/volume so volume=mass/density
how do you convert cm^3 to L so you can subtract V1 from V2
100 Cm^3=1 L
1000 cm^3 = 1 L since 1 cm^3 = 1 mL and since there are 1000 mL in one L.
To calculate the energy (E) for the condensation of one mole of water at 1.00 atm and 100.°C, we can use the formula:
E = q / n
Where:
E = Energy
q = Heat released
n = Number of moles of substance
From the given information, we know that 40.66 kJ of heat is released when one mole of H2O(g) is condensed to H2O(l). So, q = -40.66 kJ (negative sign because heat is released).
We also know that one mole of H2O(g) occupies a volume of 30.6 L at 1.00 atm and 100.°C.
Now, we can calculate the number of moles (n) of water using the ideal gas law:
PV = nRT
Where:
P = Pressure
V = Volume
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature
Rearranging the equation, we have:
n = PV / RT
Substituting the given values, we have:
n = (1.00 atm * 30.6 L) / (0.0821 L.atm/mol.K * 373 K)
Simplifying the equation further, we have:
n = 1.00 mol
Now, we can substitute the values of q and n into the formula for energy:
E = (-40.66 kJ) / (1.00 mol)
Calculating this, we find:
E = -40.66 kJ/mol
Therefore, the energy for the condensation of one mole of water at 1.00 atm and 100.°C is approximately -40.66 kJ/mol.