The Ideal Gas Law
1. Computing Number of Molecules in an Ideal Gas.
Follow the link: Example 21.2.
2. Fraction of Volume of a Gas Occupied by Molecules.
Follow the link: Example 21.3.
Section 21.5 Ideal Gas Bootcamp
Exercises Exercises
State Change Processes
Mechanical Work on Gas
3. Path Dependence of Mechanical Work by Gas.
Follow the link: Example 21.13.
4. (Calculus) Work in an Adiabatic Process.
Follow the link: Exercise 21.3.3.1.
5. (Calculus) Work in an Isobaric Process.
Follow the link: Exercise 21.3.3.2.
Real Gas
6. Amount of Gas According to Ideal versus Real Gas Assumptions.
Follow the link: Example 21.15.
7. Pressure in Real Gas is Less Than Pressure in Ideal Gas.
Follow the link: Example 21.16.
8. Ideal Gas Versus van der Waals Gas.
Follow the link: Exercise 21.4.2.1.
Miscellaneous
9. Number of Moles of the Same Gas Treated as Ideal Gas and van der Waals Gas.
A \(16\)-liter gas cylinder has the Oxygen gas at the room temperature of \(20^{\circ}\text{C}\text{.}\) The gauge pressure shows a pressure of \(200\, \text{kPa}\text{.}\) The amount of gas in the cylinder is some definite amount, but the calculated value will depend upon the model you use. (a) What is the calculated value of number of moles if you assume the ideal gas behavior? (b) What is the calculated value of number of moles if van der Waals gas behavior is assumed?
Solution 1. a
The pressure in the ideal gas law is the absolute pressure and not the gauge pressure. The gauge pressure is the difference of the pressure inside to the pressure outside. Since the outside pressure is just the atmospheric pressure, the pressure in the gas will be
\begin{equation*}
p = p_{\textrm{atm}} + p_{\textrm{gauge}} = 1.013\times 10^{5}\ \textrm{Pa} + 200\ \textrm{kPa} = 3.013\times 10^{5} \textrm{Pa}.
\end{equation*}
This pressure in atm is
\begin{equation*}
p = 3.013\times 10^{5} \textrm{Pa}/1.013\times 10^{5}\ \textrm{Pa/atm} = 2.97\ \textrm{atm}
\end{equation*}
The other variables in the ideal gas law are given as
\begin{align*}
\amp V = 16\ \textrm{L}\\
\amp T = 20^{\circ}\textrm{C} + 273.15 = 293.15\ \textrm{K}.
\end{align*}
Therefore, the gas the the following number of moles, where all quantities are expressed in atm, L and K units.
\begin{equation*}
n = \frac{pV}{RT} = \frac{2.97\times 16}{0.082\times 293.15} = 1.98 \ \textrm{mol.}
\end{equation*}
Solution 2. b
We need \(a\) and \(b\) for Oxygen. They are
\begin{equation*}
a = 1.382\ \textrm{atm.L}^2/\textrm{mol}^2;\ \ b = 0.0319\ \textrm{L/mol}.
\end{equation*}
The equation for \(n\) to solve now is
\begin{equation*}
\left( 2.97+ \frac{1.382\ n^2}{16} \right) \left( 16-0.0319\ n\right) = 0.082\times 293.15\ n,
\end{equation*}
where \(p\text{,}\) \(V\text{,}\) and \(T\) have been expressed in atm, L and K units. This equation is a cubic equation in \(n\) and is left for the student to solve. Mathematica gave me three answers: \(n= 480\) mol, \(15.8\) mol, and \(2.26\) mol. The physical answer should be close to the ideal gas behavior since the van der Waals is a correction to the ideal gas behavior. We expect the root \(n = 2.26\) mol to be the physical value of \(n\) here.
10. Thermal Equilibrium Between Gases in Two Containers Seprated by a Freely Movable Wall.
A metallic container of fixed volume of \(35\text{ L}\) immersed in a large tank of temperature \(27^{\circ}\text{C}\) contains two compartments separated by a freely movable wall. There are \(1.2\text{ moles}\) of nitrogen on one side and \(1.5\text{ moles}\) of oxygen on the other side of the wall. Find the temperature, pressure and volume of the two sides when equilibrium has reached. Assume ideal gas behavior.
Hint.
At equilibrium, the pressure on the two sides will be same. Also, the volumes of the two add up to the unchanging total volume.
Answer.
\(1.9\text{ atm} \text{.}\)
Solution.
Since the separating wall is freely movable, the two sides will come to the same pressure at the equilibrium in addition to being at the same temperature. Therefore the ratio of volume to number of moles on the two sides will be equal. Denoting the Nitrogen side by a subscript 1 and the Oxygen side by the subscript 2 we have
\begin{equation*}
\frac{V_1}{n_1} = \frac{V_2}{n_2} = \frac{RT}{p}=\text{ Same}.
\end{equation*}
In addition the sum of the volumes is equal to the total volume, \(V_0\text{.}\)
\begin{equation*}
V_1 + V_2 = V_0.
\end{equation*}
Therefore,
\begin{align*}
\amp V_1 = \left( \frac{n_1}{n_1 + n_2} \right) V_0, \\
\amp V_2 = \left( \frac{n_2}{n_1 + n_2} \right) V_0.
\end{align*}
Putting in the numbers we get
\begin{align*}
\amp V_1 = \left( \frac{1.2}{1.2 + 1.5} \right)\times 35 \ \text{L} = 16\ \text{L}, \\
\amp V_2 = \left( \frac{1.5}{1.2 + 1.5} \right) \times 35 \ \text{L} = 19 \ \text{L}.
\end{align*}
Now, since we know the volume, and the temperature, which is 300.15 K, we can find the common value of the pressure by using the properties of either side. For instance, using the properties of the nitrogen side, we have
\begin{align*}
p \amp = \frac{n_1 RT}{V_1} = \frac{1.2\ \text{mol}\times 0.082\ \text{L.atm/mol.K}\times 300.15\ \text{K}}{16\ \text{L}}\\
\amp = 1.9\ \text{atm}.
\end{align*}
11. A Metallic Container with Two Compartments Separated by Movable Wall.
A metallic container of fixed volume of \(35\ \textrm{L}\) immersed in a large tank of temperature \(27^{\circ}\text{C}\) contains two compartments separated by a freely movable wall. There are \(1.2\) moles of nitrogen on one side and \(1.5\) moles of oxygen on the other side of the wall. Find the temperature, pressure and volume of the two sides when equilibrium has reached. Assume ideal gas behavior.
Solution.
Since the separating wall is freely movable, the two sides will come to the same pressure at the equilibrium in addition to being at the same temperature. Therefore the ratio of volume to number of moles on the two sides will be equal. Denoting the Nitrogen side by a subscript 1 and the Oxygen side by the subscript 2 we have
\begin{equation*}
\frac{V_1}{n_1} = \frac{V_2}{n_2} = \frac{RT}{p}=\textrm{ Same}.
\end{equation*}
In addition the sum of the volumes is equal to the total volume, \(V_0\text{.}\)
\begin{equation*}
V_1 + V_2 = V_0.
\end{equation*}
Therefore,
\begin{align*}
\amp V_1 = \left( \frac{n_1}{n_1 + n_2} \right) V_0.\\
\amp V_2 = \left( \frac{n_2}{n_1 + n_2} \right) V_0.
\end{align*}
Putting in the numbers we get
\begin{align*}
\amp V_1 = \left( \frac{1.2}{1.2 + 1.5} \right)\times 35 \ \textrm{L} = 16\ \textrm{L}\\
\amp V_2 = \left( \frac{1.5}{1.2 + 1.5} \right) \times 35 \ \textrm{L} = 19 \ \textrm{L}
\end{align*}
Now, since we know the volume, and the temperature, which is 300.15 K, we can find the common value of the pressure by using the properties of either side. For instance, using the properties of the nitrogen side, we have
\begin{equation*}
p = \frac{n_1 RT}{V_1} = \frac{1.2\ \textrm{mol}\times 0.082\ \textrm{L.atm/mol.K}\times 300.15\ \textrm{K}}{16\ \textrm{L}} = 1.9\ \textrm{atm}.
\end{equation*}
12. Virital Expansion Corresponding to van der Walls Equation.
Find the virial expansion corresponding to the van der Waals equation of state.
Solution.
Divide out the volume part from both sides and rearrange to arrive at
\begin{equation*}
p = - \frac{n^2a}{V^2} + \frac{nRT}{V-nb}.
\end{equation*}
Now, we multiply both sides by $V/nRT$.
\begin{equation*}
\frac{pV}{nRT} =- \left( \frac{na}{RT}\right)\frac{1}{V} + \frac{V}{V-nb}.
\end{equation*}
The first term on the right side is already in the form we want. The second term can be expanded in powers of $V$ as follows.
\begin{equation*}
\frac{V}{V-nb} = \frac{1}{1-\frac{nb}{V}} = \sum_{i=0}^{\infty} \left(- \frac{nb}{V}\right)^i.
\end{equation*}
Now, putting all together we will write the Virial expansion as
\begin{equation*}
\frac{pV}{nRT} = 1 + \left(nb- \frac{na}{RT}\right)\frac{1}{V} - \sum_{i=2}^{\infty} \left(- \frac{nb}{V}\right)^i.
\end{equation*}
