At 373.15K and 1 atm, the molar volume of liquid water and steamare 1.88 X 10-5 m3 and 3.06 X 10-2m3, respectively. Given that the heat of vaporization ofwater is 40.79 kJ/mol, calculate the values of ?H and ?Ufor 1 mole in the following process:
H2O (l, 373.15 K, 1 atm) ---> H2O(g, 373.15 K, 1 atm)

Answer The value of and is, 40.79 kJ and 37.7 kJ respectively.

Explanation :

Heat released at constant pressure is known as enthalpy.

The formula used for change in enthalpy of the gas is:

Now we have to calculate the work done.

Formula used :

where,

w = work done  = ?

P = external pressure of the gas = 1 atm

= initial volume =

= final volume =

Now put all the given values in the above formula, we get:

Now we have to calculate the change in internal energy.

Thus, the value of and is, 40.79 kJ and 37.7 kJ respectively.