Question

What is $\Delta U$ when $2.0$ a mole of liquid water evaporates at ${100^0}C$? The heat of evaporation $\left( {\Delta {H_v}} \right)$ of water at ${100^0}C$ is $40kJmo{l^{ - 1}}$.

Answer 1

The standard enthalpy of evaporation is equal to the sum of the change in internal energy and the product of change in moles of gaseous reactants and products, universal gas constant and temperature. The heat of vaporisation and the product of moles gives the standard enthalpy of evaporation.
Formula used:
$\Delta {H_{vap}} = \Delta U + \Delta {n_g}RT$
$\Delta {H_{vap}}$is standard enthalpy of vaporization
$\Delta U$is change in internal energy has to be determined
$\Delta {n_g}$ change in moles of gaseous products and gaseous reactants
R is universal gas constant is $8.314 \times {10^{ - 3}}kJ{\left( {Kmol} \right)^{ - 1}}$
T is temperature is ${100^0}C = 100 + 273 = 373K$

Complete answer:
The moles of liquid given is $2$
Vaporization is the process of converting the reactants present in liquid phase to the products present in vapor state. The number of moles were converted into gaseous phase from liquid phase. The entire moles were converted. So, the number of moles in liquid phase will be zero and the number of moles in gas phase is $2$
The reaction for vaporization of $2$ moles of water will be
$2{H_2}{O_{\left( l \right)}} \to 2{H_2}{O_{\left( g \right)}}$
The standard enthalpy of vaporisation is $\Delta {H_{vap}} = 2 \times \Delta {H_v} = 2 \times 40.66 = 81.32kJmo{l^{ - 1}}$
The change in moles of gaseous products and gaseous reactants $\Delta {n_g} = 2 - 0 = 2$
Substitute all the values in the above equation
$81.32 = \Delta U + 2 \times 8.314 \times {10^{ - 3}} \times 373$
By simplifying the above values, we will get $\Delta U = 74.12kJmo{l^{ - 1}}$
The change in internal energy of a given chemical reaction is $74.12kJmo{l^{ - 1}}$

Note:
The change in the number of moles must be determined from the number of moles of reactants and products. The standard enthalpy of vaporization is equal to the heat of vaporization and number of moles. The change in internal energy can be calculated from standard enthalpy of vaporization.