Section 21.5 Ideal Gas Bootcamp
Subsection 21.5.1 The Ideal Gas Law
Problem 21.5.1. Computing Number of Molecules in an Ideal Gas.
Follow the link: Checkpoint 21.1.2.
Problem 21.5.2. Fraction of Volume of a Gas Occupied by Molecules.
Follow the link: Checkpoint 21.1.3.
Subsection 21.5.2 State Change Processes
Subsection 21.5.3 Mechanical Work on Gas
Problem 21.5.3. Path Dependence of Mechanical Work by Gas.
Follow the link: Checkpoint 21.3.5.
Problem 21.5.4. (Calculus) Work in an Adiabatic Process.
Follow the link: Checkpoint 21.3.6.
Problem 21.5.5. (Calculus) Work in an Isobaric Process.
Follow the link: Checkpoint 21.3.7.
Subsection 21.5.4 Real Gas
Problem 21.5.6. Amount of Gas According to Ideal versus Real Gas Assumptions.
Follow the link: Checkpoint 21.4.2.
Problem 21.5.7. Pressure in Real Gas is Less Than Pressure in Ideal Gas.
Follow the link: Checkpoint 21.4.3.
Subsection 21.5.5 Miscellaneous
Problem 21.5.8. 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.
At equilibrium, the pressure on the two sides will be same. Also, the volumes of the two add up to the unchanging total volume.
\(1.9\text{ atm} \text{.}\)
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
In addition the sum of the volumes is equal to the total volume, \(V_0\text{.}\)
Therefore,
Putting in the numbers we get
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