Question

The molar enthalpy of vaporization of water at 373K is 41.16kJ/mol. What fraction of this energy is used to change the internal energy of the water, and what fraction is used to do work against the atmosphere? (Assume that water vapor is an ideal gas.)

Short answer

92.47% of energy is used to change the initial energy of water and 7.53% is used to do work against the atmosphere.

Step-by-step solution

Determine the change in volume ∆V.

$$\begin{gathered} The volume of water=\frac{18.2g}{1g/mL} \\ =18.2 mL \end{gathered}$$

For calculating the volume of gas, we will use theideal gas law:

$$\begin{gathered} PV=nRT \\ {V}_{gas}=\frac{1mol\times 0.08206Latm/molK\times 373K}{1atm} \\ =30.61L \end{gathered}$$

Change in volume:

$$\begin{gathered} \Delta V=30.61 L-0.1802 L \\ =30.59 L \end{gathered}$$

Determine the fraction of energy used to change the internal energy of water and the fraction used to work against the atmosphere.

First, we will determine the energy used to work against the atmosphere:

$$\begin{gathered} p\Delta V=0.101325kJ\times 30.59 \\ =3.10kJ \end{gathered}$$

The $molar enthalpy of vaporisation=41.16kJ$

Now the fraction of energy used to work against the atmosphere

$$\begin{gathered} The fraction ofenergy used to work against atmosphere=\frac{3.10kJ}{41.16kJ}\times 100 \\ =7.53\% \end{gathered}$$.

The fraction of energy used to change the internal energy of water

$$\begin{gathered} The fraction of energy=\left(100\%-7.53\%\right) \\ =92.47\% \end{gathered}$$.