Question

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 1

Answer The value of $\Delta H$ and $\Delta U$ 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:

$\Delta Q_p=\Delta H=40.79kJ/mol$

Now we have to calculate the work done.

Formula used :

$w=-P\Delta V\\\\w=-P\times (V_2-V_1)$

where,

w = work done  = ?

P = external pressure of the gas = 1 atm

$V_1$ = initial volume = $1.88\times 10^{-5}m^3=1.88\times 10^{-5}\times 10^3L=1.88\times 10^{-2}L$

$V_2$ = final volume = $3.06\times 10^{-2}m^3=3.06\times 10^{-2}\times 10^3L=3.06\times 10^{1}L$

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

$w=-(1atm)\times (3.06\times 10^{1}-1.88\times 10^{-2})L$

$w=-30.5812L.atm=-30.5812\times 101.3J=-3097.87556J=-3.09\times 10^3J=-3.09kJ$

Now we have to calculate the change in internal energy.

$\Delta U=q+w$

$\Delta U=40.79kJ+(3.09kJ)$

$\Delta U=37.7kJ$

Thus, the value of $\Delta H$ and $\Delta U$ is, 40.79 kJ and 37.7 kJ respectively.