Assume at exactly 100.0°c and 1.00 atm total pressure, 1.00 mole of liquid water and 1.00 mole of water vapor occupy 18.80 ml and 30.62 l, respectively.

Answer 1) : When we calculate the ΔQ = -40.66 kJ (heat of vaporisation).

Then, we can relate the work done with this equation,

ΔW = -PΔV

First we need to calculate the volume, which will be,

Volume of water vapor = (nRT) / (P)

Then, V = (1 mole) X (0.0821 L X atm/mole X K) X (373.15K) / (1atm) = 30.6 liters So, it is given, and verified.

Volume of Water can be used for finding the; Density = M / V

D = ( 1 mole) X (18.02 g/mole) / (18.80 mL) = 0.9585 g/mL

On, substituting the values, we get,

ΔW = -PΔV

Work = - (1 atm) X (0.00188L - 30.62L) X 0.1013kJ/L X atm = 3.10 kJ

hence the work done is 3.10 kJ

Answer 2) Now, for calculating the internal energy, we can use the formula as,

ΔE = ΔQ + ΔW

Here, we have the values of ΔQ = -40.66 kJ and ΔW as = 3.10 kJ

So, we get, ΔE = (-40.66) + (3.10) = - 37.55 kJ

Therefore, the change in the internal energy will be -37.55 kJ.

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