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 ÄE for the condensation of one mole of water at 1.00 atm and 100.°C.
To calculate the change in internal energy (ΔE) for the condensation of one mole of water at 1.00 atm and 100.°C, we need to use the equation:
ΔE = q + w
where q is the heat transfer and w is the work done.
In this case, the heat transfer (q) is given as 40.66 kJ (kilojoules).
The work done (w) can be calculated using the equation:
w = -PΔV
where P is the pressure and ΔV is the change in volume.
Given that the pressure (P) is 1.00 atm and the change in volume (ΔV) can be calculated using the initial and final volumes of the gas and liquid:
ΔV = V_final - V_initial
In this case, the initial volume (V_initial) is given as 30.6 L.
To find the final volume (V_final), we need to convert the density of water (0.996 g/cm3) to volume. This can be done using the molar mass of water.
The molar mass of water (H2O) is 18.01528 g/mol, so one mole of water has a mass of 18.01528 g.
The volume of one mole of water (V_one_mole) can be calculated using the density formula:
density = mass/volume
rearranging the formula, we get:
volume = mass/density
Substituting the mass (18.01528 g) and density (0.996 g/cm3), we can calculate the volume of one mole of water.
Finally, substitute the calculated values of q and w into the equation ΔE = q + w to find the change in internal energy (ΔE) for the condensation of one mole of water at 1.00 atm and 100.°C.