PV-Diagrams chart the change in pressure and volume of some gas as it undergoes some thermodynamic process. They are typically a plot with pressure on the vertical axis and volume on the horizontal. They show transitions from one state ($P_i, V_i$) to another state ($P_f, V_f$).
Here is a video to illustrate an object undergoing non-Uniform Circular Motion.
Watch the pre-lecture videos and read through the OpenStax text before doing the pre-lecture homework or attending class.
PV-Diagrams are a useful tool for understanding thermodynamic processes. These diagrams are typically shown with pressure along the vertical axis and volume along the horizontal.
f you had an ideal gas for instance ($PV=Nk_BT$), and the number of molecules was known, then for a given pressure and volume, there would be a unique temperature. So a point on a PV diagram tells you a great deal about the state of the system. Similarly a change from one point to another point tells you a great deal about the changes in state variables and the energy transfers during that transition.
In the above diagram the pressure and volume both decreased. Assuming the number of particles in your system remained constant (an assumption always made here), with P and V both decreasing, than so must T according the ideal gas law. So this diagram helps you understand that the temperature decreased during this transition. Since internal thermal energy is related to temperature, you also know that $E_{th}$ has decreased. You can also know information about the work during this process.
Work is equal to the area under the PV-diagram curve
work is positive if the volume decreases and negative if the volume increases
So the you know the sign of the work is positive in this case, meaning the environment did work on your system. Think about compressing a gas, you'd have to push the gas into a smaller volume. You would lose energy doing the compression, which means you (the environment) is losing energy (Wenvironment is negative) so the system (the gas) is gaining that energy and thus Wsystem is positive. Put all that together with the first law of thermo $\Delta E_{th}=Q+W$: you know that $\Delta E_{th}$ is positive, and so is W, so what is the sign of the heat Q and how large is it? In this case Q could be positive or negative and to find the magnitude, you'd have to calcuted the change in thermal energy and the work, then use the first law to solve for Q. If $\Delta E_{th}$ is a larger positive than W, then Q must be postive. If $\Delta E_{th}$ is a smaller positive than W, then Q must be negative.
It's important to note that there are other important thermodynamic curve diagrams, pressure vs. temperature or temperature vs. volume among others, but we have decided to focus only on the PV-diagrams here.
Isobaric: Pressure is sometimes measured in a non-SI unit called Bars. Iso is a prefix meaning constant. So, isobaric refers to a constant pressure process.
Work = area under PV-curve, for isobaric: $W=P \Delta V$
With pressure constant and volume increasing, the temperature must also increase. Since the temperature increases, so does the thermal energy. In an expansion, like that shown above, the work is negative. With the first law ($\Delta E_{th}=Q+W$), heat (Q) is equal to the the change in thermal energy minus the work. With a positive $\Delta E_{th}$ and negative work (W), heat (Q) must be positive and greater in magnitude than the work. So in an isobaric expansion, energy goes in via heat more quickly than it goes out via work and thus the thermal energy increases.
Isochoric: A process where the volume remains constant is considered an isochoric process. From the ideal gas law, if the volume remains constant but the pressure increases, like in the diagram below, then the temperature must increase.
Work = area under curve, w/ $W = 0$, $Q = \Delta E_{th}$
Since work is the area under the PV-curve, the work in an isochoric process is zero. With work equal to zero, the heat is equal to the change in thermal energy (1st law). Since the temperature increased in the above diagram, the thermal energy must also have increased. Heat entered the system while zero work was done by the system, as a result it increased its temperature.
Isothermal: A process where the temperature remains constant is considered an isothermal process. The ideal gas law shows us that if the temperature is constand the volume increases, then the pressure must decrease. Because of the inverse relationship between pressure and volume when temperature (and number) is constant, isothermal transitions are always curvy lines in a PV-diagram.
Work = area under curve, for isothermal: $|W|=Nk_BT |\ln{\frac{V_f}{V_i}}|$
In an isothermal process the internal thermal energy remains constant, so from the 1st law, the heat must be equal to negative the work. If the system is expanding, like that shown above, the work is negative so the heat must be positive. Heat is entering into the system at the same rate work is taking that energy out of the system.
Adiabatic: A process where no heat is transfered into or out of the system is considered an adiabatic process. On a PV-diagram adiabatic curves are curvy like isotherms, but they are steeper.
Work = area under curve, w/ $Q = 0$, $W = \Delta E_{th}$
In an adiabatic process there is no explicit equation for the work in terms of the state variables. In these transitions, you can use the first law and the fact that the heat is zero to find the work equal to the change in internal thermal energy. For a monoatomic gas we know that $\Delta E_{th}=\frac{3}{2}Nk_B \Delta T$, an equation found in the kinetic theory of gases.
Now, take a look at the pre-lecture reading and videos below.
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