Question

How would you use the van der Waals equation of state to calculate the pressure of $3.60$mol of ${H_2}O$ at $4253K$ in a $5.90L$ vessel?

Answer 1

We have to know that an equation that deals with the relationship between the pressure, volume, temperature, and amount of real gases is defined as van der Waals equation of gases. In the modifications which are made related to equation of ideal gas for real gases, the reduction in pressure is because of forces of attractions among the molecules is straightly proportional to $\dfrac{{{n^2}}}{{{V^2}}}$

Complete step by step answer:
Given data contains,
The number of moles = $3.60mol$
Gas constant = $0.08314lbar \cdot L \cdot {K^{ - 1}}mo{l^{ - 1}}$
Temperature = $453K$
Volume = $5.90L$
a = $5.536{\text{ }}bar \cdot {L^2}mo{l^{ - 2}}$
b = $0.030{\text{ }}49{\text{ }}L \cdot mo{l^{ - 1}}$
We could substitute the values of ideal volume and ideal pressure in ideal gas equation $PV = nRT$, the modified expression could be written as,
$\left( {P + a\dfrac{{{n^2}}}{{{V^2}}}} \right)\left( {v - nb} \right) = nRT$
We can rearrange the above expression as,
$\left( {P + \dfrac{{a{n^2}}}{{{V^2}}}} \right) = \dfrac{{nRT}}{{\left( {v - nb} \right)}}$
$P = \dfrac{{nRT}}{{\left( {v - nb} \right)}} - \dfrac{{{n^2}a}}{{{V^2}}}$
Here the pressure is indicated by P.
The molar volume is indicated by V.
The temperature of the given sample of the gas is indicated by T.
The gas constant is indicated by R.
van der waals constant is indicated by a and b.
n indicated the moles of the gas.
Let us now substitute the given values in the above expression of Vander Waals equation as
$P = \dfrac{{nRT}}{{\left( {v - nb} \right)}} - \dfrac{{{n^2}a}}{{{V^2}}}$
Now we can substitute the known values we get,
$P = \dfrac{{\left( {3.60mol} \right)\left( {0.08314\,bar\,L \cdot {K^{ - 1}}mo{l^{ - 1}}} \right) \times \left( {453K} \right)}}{{\left( {5.90L - 3.60mol \times 0.03049\,L \cdot mo{l^{ - 2}}} \right)}}$
$P = \dfrac{{135.6bar}}{{5.90 - 0.1098}} - 2.061bar$
$P = 23.42bar - 2.061bar$
On simplification we get,
$P = 21.4bar$

The pressure is calculated as $21.4bar$.

Note: Now we can discuss the certain of the advantages of van der Waals equation are,
Determines the behaviour of gases better than the ideal gas equation.
Applicable for gases and fluids.
Certain of the disadvantages of van der Waals equation are,
We could get better results of all real gases only above critical temperature using van der Waals equation.
Vander Waals completely fails in the transformation phase of gas to the liquid below a critical temperature.