- #1
NoahCygnus
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Homework Statement
If water vapor is assumed to be a perfect gas, molar enthalpy change for vaporization of 1 mol of water at 1 bar and 100 C is 41 kJ/mol . Calculate the internal energy, when 1 mol of water is vaporized at 1 bar pressure and 100 C.
Homework Equations
$$\Delta U = \Delta H = \Delta n_{g}RT$$
The Attempt at a Solution
The chemical equation is;
$$H_{2}O_{l} \longrightarrow H_{2}O_{g}$$
Then change in internal energy is related to change in enthalpy by;
$$\Delta U = \Delta H - \Delta n_{g}RT$$
Putting in the values;
$$\Delta U = 41 - (0)RT = 41 kJ/mol$$
However in the book they have given, $$\Delta n_{g} = 1$$. If the number of moles of water in liquid and gaseous phase are same , why isn't $$\Delta n_{g} = 0$$. ?