Question

Calculate $w$ and $\Delta U$ for the conversion of $1mol$ of water at $100^\circ C$ to steam at $1atm$ pressure. Heat of vaporization of water at $100^\circ C$ is $40.670kJmo{l^ -1 }$ . Assume ideal gas behavior.

Answer 1

As it shows ideal gas behavior the volume of the gas is negligible, the particles do not attract or repulse with other particles and the particles move in random motion as per the Newton’s Laws of Motion. The work done and internal energy can be calculated from the mathematical expression of the First Law of Thermodynamics.
Formulas used: $w = P \times V$ here $w$ is the work done, $P$ is the pressure and $V$ is the volume
$PV = nRT$ where $n$ is the number of moles, $R$ is the universal gas constant and $T$ is the absolute temperature.
$\Delta U = \Delta {H_{vap}} + w$, here $\Delta U$ is the change in internal energy, $\Delta {H_{vap}}$ is the heat of vaporization

Complete step by step answer:
We have to calculate the work done ($w$) and the internal energy change ( $\Delta U$ ).
The equation for work done is $w = P \times V$ , here the pressure ( $P$ ) is $1atm$ and the volume ( $V$ ) is negligible according to ideal gas behavior.
From the ideal gas equation $PV = nRT$
Therefore, equating these two equations we get, $w = nRT$
where $n$ is the number of moles, $R$ is the universal gas constant and $T$ is the absolute temperature that is $373K$.
on substituting the values of universal gas constant and temperature ,
Then the equation becomes, $w = 1 \times 8.314 \times 373$
$\Rightarrow w = 3101.122J$
$\Rightarrow w = 3.1kJ$, the unit should be in kilojoules and is achieved by dividing by $1000$
Now we have to calculate the change in internal energy change $\Delta U$
According to first law of thermodynamics,
$\Delta U = \Delta {H_{vap}} + w$, here $\Delta {H_{vap}}$is the heat of vaporization that is $40.670kJ\,mo{l^ - }\,$
Substituting value of $w = 3.1kJ$
$\Rightarrow \Delta U = 40.67kJ + 3.1kJ$
$\Rightarrow \Delta U = 43.77kJ$
Therefore the Internal energy change is $43.77kJ$ and the work done is $3.1kJ$ in converting $1mol$ of water to steam at ${100^ \circ }C\,$ at $1atm$ .

Note: Note that the first law of thermodynamics is also a restatement of the law of conservation of energy, and says that heat supplied to a gas is used in increasing the internal energy of the gas and in doing work. Here the heat supplied is in the form of latent heat of vapourization. The units should be converted to kilojoules in the final answer. The other important aspect is to clearly understand the ideal gas behaviours and that can help in the assumption of the volume in the question.