Question

One mole of $H_{2}O\left(g\right)$ at 1.00 atm and 100 degrees Celsius occupies a volume of 30.6 L. When one mole of $H_{2}O\left(g\right)$ is condensed to one mole of $H_{2}O\left(l\right)$ at 1.00 atm and 100 degrees Celsius, 40.66 kJ of heat is released. If the density of $H_{2}O\left(l\right)$ at this temperature and pressure is 0.996 g/cm${}^3$, calculate $\Delta E$ for the condensation of one mole of water at 1.00 atm and 100 degrees Celsius.

Answer 1

Given Data:

• The initial pressure of the water gas is 1.00 atm.

• The volume of the water gas is 30.6 L.

• The temperature is 100 degrees Celsius.

• The heat released is 40.66 kJ.

• The density of the liquid water is 0.996 g/centimetre cube.

The internal energy is determined by the following formula:

$\Delta U=q+w...\left(I\right)$

Here,

• U is the internal energy.

• q is the heat.

• w is the work.

Use the density formula to determine the volume of liquid water as follows:

The mass of one mole water is 18 g/mol.

$Density=\frac{Mass}{Volume}$

Substituted the known values.

$\begin{array}{cc}0.996g/c{m}^3& =\frac{18g}{Volume}\\ Volume& =\frac{18g}{0.996g/c{m}^3}\\ & =18.07c{m}^3\end{array}$

Convert the volume of liquid water from centimetre cube to liter as follows:

$\begin{array}{cc}1000c{m}^3& =1L\\ 18.07c{m}^3& =0.01807L\end{array}$

The work is determined as follows:

$w=-P\left(V_{liq.water}-V_{gas.water}\right)$

Substitute the known values.

$\begin{array}{cc}w& =-1.00atm\left(0.01807L-30.6L\right)\\ & =30.58atmL\end{array}$

Convert the work form atm L to joule as follows:

$\begin{array}{cc}0.0821atmL& =8.314J\\ 30.58atmL& =\frac{8.314J}{0.0821atmL}\times 30.58atmL\\ & =3096.73J\end{array}$

Convert the work form joule to kilojoule as follows:

$\begin{array}{cc}1000J& =1kJ\\ 3096.73J& =3.09673kJ\end{array}$

Use the equation (I) to calculate the internal energy change.

$\begin{array}{cc}\Delta U& =-40.66kJ+3.09673kJ\\ & =-37.56kJ\end{array}$

Therefore, the internal energy change for the condensation is -37.56 kJ.