What are PV diagrams?

Consider a gas sealed in a container with a tightly fitting yet movable piston as seen below. We can do work on the gas by pressing the piston downward, and we can heat up the gas by placing the container over a flame or submerging it in a bath of boiling water. When we subject the gas to these thermodynamics processes, the pressure and volume of the gas can change.

A convenient way to visualize these changes in the pressure and volume is by using a Pressure Volume diagram or PV diagram for short. Each point on a PV diagram corresponds to a different state of the gas. The pressure is given on the vertical axis and the volume is given on the horizontal axis, as seen below.

Every point on a PV diagram represents a different state for the gas (one for every possible volume and pressure). As a gas goes through a thermodynamics process, the state of the gas will shift around in the PV diagram, tracing out a path as it moves (as shown in the diagram below).

Being able to decode the information shown in a PV diagram allows us to make statements about the change in internal energy $\Delta U$delta, U, heat transferred $Q$Q, and work done $W$W on a gas. In the sections below, we'll explain how to decipher the hidden information contained in a PV diagram.

Let's say our gas starts out in the state shown in the PV diagram below.

If we press the piston downward, the volume of the gas will decrease, so the state must shift to the left toward smaller volumes (as seen in the diagram below). Since the gas is being compressed we can also say for sure that positive work $W$W is being done on the gas.

Similarly, if we let the gas expand, pushing the piston upward, the volume of the gas will increase, so the state must shift to the right toward larger volumes (as seen in the diagram below). Since the gas is expanding we can also say for sure that negative work $W$W is being done on the gas.

So if we ever see a state shifting to the left on a PV diagram we can say for sure that the work done on the gas was positive. Similarly, if we ever see a state shifting to the right on a PV diagram we can say for sure that the work done on the gas was negative.

The work done during a thermodynamic process is equal to the area under the curve as seen in the diagram below.

The reason why work is equal to the area under the curve is that,

And since $P\Delta V$P, delta, V is just the $\text{height} \times \text{ width}$start text, h, e, i, g, h, t, end text, times, start text, space, w, i, d, t, h, end text of the rectangle shown above, the work is equal to the area. If we use pressure units of $\text{pascals}$start text, p, a, s, c, a, l, s, end text and volume units of $\text{m}^3$start text, m, end text, cubed then the energy we find will be in units of $\text{joules}$start text, j, o, u, l, e, s, end text.

We have to be really careful with signs though. If the path on a PV diagram is directed to the left, the volume is decreasing, and positive work is being done on the gas. If the path on a PV diagram is directed to the right (as in the diagram above), the volume is increasing, and negative work is being done on the gas since $W_\text{by gas}=-W_\text{on gas}$W, start subscript, start text, b, y, space, g, a, s, end text, end subscript, equals, minus, W, start subscript, start text, o, n, space, g, a, s, end text, end subscript.

It doesn't matter what shape the path takes, the area under the curve will still represent the work done. For any curved path we can imagine breaking the area into an infinite amount of infinitesimally thin rectangles.

The area of each rectangle would represent the work done during each infinitesimal step, and the sum of the areas would represent the total work done for the entire process.

It should be said that we are always going to assume these processes are taking place slowly enough that the entire gas can be at thermodynamic equilibrium at every moment (i.e. the same temperature throughout the gas). If this seems dubious to you, you're right to question it. However, even though basically no real world processes will exactly satisfy this requirement, our ability to model many thermodynamic processes are not fatally jeopardized by this lack of adherence to ideal circumstances.

Remember that internal energy and temperature are proportional $U \propto T$U, \propto, T. So if the temperature increases, the internal energy must also increase.

Now, if the gas we're considering is an ideal gas we also know that,

And if no gas is allowed to escape (so the number of molecules $N$N is constant) we can say that $PV \propto T$P, V, \propto, T. All of this means that,

So if the quantity of pressure times volume $(P \times V)$left parenthesis, P, times, V, right parenthesis increases, the temperature $T$T and internal energy $U$U must also increase (which makes $\Delta U$delta, U positive). This idea is represented in the diagram shown below.

This means that anytime the state in a PV diagram ends up further up and right than where it started, $\Delta U$delta, U is a positive number. Similarly, anytime the state in a PV diagram ends up further down and left than where it started, $\Delta U$delta, U is a negative number.

Now if the state in the PV diagram moves up and left (pressure increases and volume decreases), or down and right (pressure decreases and volume increases), it is a little ambiguous whether the quantity $(P \times V)$left parenthesis, P, times, V, right parenthesis actually increased or decreased (since one variable increased and the other variable decreased). To be sure, one would have to check the exact values of the initial and final $P$P and $V$V on the axes of the graph to tell if the quantity $(P \times V)$left parenthesis, P, times, V, right parenthesis actually increased or decreased.

It is also good to note that if the quantity $(P \times V)$left parenthesis, P, times, V, right parenthesis does not change, then the temperature $T$T and internal energy $U$U do not change either. For instance, if the pressure doubles, and the volume is cut in half, $(P \times V)$left parenthesis, P, times, V, right parenthesis remains the same value (since $2P \times \dfrac{V}{2}=PV$2, P, times, start fraction, V, divided by, 2, end fraction, equals, P, V). the temperature $T$T and internal energy $U$U will end the process with the same values they started with.

Given a PV diagram, we typically have to rely on the first law of thermodynamics $\Delta U=Q+W$delta, U, equals, Q, plus, W to determine the sign of the net heat that enters or exits a gas. If we solve this equation for the heat $Q$Q we get,

Now that we know this, we can use what we know about finding the sign of $\Delta U$delta, U and $W$W to find the sign of $Q$Q in many cases. So for instance, if the change in internal energy is positive and the work done is negative,

$Q=(+)-(-)=+\qquad$Q, equals, left parenthesis, plus, right parenthesis, minus, left parenthesis, minus, right parenthesis, equals, plus ...the net heat must be positive.

Which makes sense, since if the internal energy increased even though work was done by the gas, that implies that more heat must have entered the gas than energy lost due to the work done by the gas.

Or for example, if the internal energy decreases and the work is positive,

$Q=(-)-(+)=-\qquad$Q, equals, left parenthesis, minus, right parenthesis, minus, left parenthesis, plus, right parenthesis, equals, minus ...the net heat must be negative.

Which makes sense, since if the internal energy decreased even though work was done on the gas, that implies that more heat must have left the gas than energy gained by the gas from work being done on it.

An ideal gas in a sealed container is taken through the process shown in the PV diagram below.

An ideal gas in a sealed container is taken through the process shown in the PV diagram below. The initial volume of the gas is $V_i=0.25 \text{m}^3$V, start subscript, i, end subscript, equals, 0, point, 25, start text, m, end text, cubed and the final volume of the gas is $V_f=0.75 \text{m}^3$V, start subscript, f, end subscript, equals, 0, point, 75, start text, m, end text, cubed. The initial pressure of the gas is $P_i=70,000 \text{ Pa}$P, start subscript, i, end subscript, equals, 70, comma, 000, start text, space, P, a, end text and the final pressure of the gas is $P_f=160,000 \text{ Pa}$P, start subscript, f, end subscript, equals, 160, comma, 000, start text, space, P, a, end text.

Solution:

We can find the work done by determining the total area under the curve on a PV diagram. We have to make sure we use the total area, all the way down to the volume axis. For instance, we can imagine viewing the area under the curve in the example shown above as a triangle and a rectangle (as seen below).

Now we just find the sum of the areas of the triangle and rectangle. The height of the rectangle is the pressure $P_i$P, start subscript, i, end subscript and the width of the rectangle is the change in volume $\Delta V=V_f-V_i$delta, V, equals, V, start subscript, f, end subscript, minus, V, start subscript, i, end subscript. So,

$\blueD{\text{area 1}}=\text{height} \times \text{width} \quad$start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, start text, h, e, i, g, h, t, end text, times, start text, w, i, d, t, h, end text (area of the rectangle)

$\blueD{\text{area 1}}=P_i \times \Delta V \quad$start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, P, start subscript, i, end subscript, times, delta, V (height is $P_i$P, start subscript, i, end subscript and width is $\Delta V$delta, V)

$\blueD{\text{area 1}}=(70,000\text{ Pa}) \times (0.75\text{m}^3-0.25\text{m}^3)\quad$start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, left parenthesis, 70, comma, 000, start text, space, P, a, end text, right parenthesis, times, left parenthesis, 0, point, 75, start text, m, end text, cubed, minus, 0, point, 25, start text, m, end text, cubed, right parenthesis (plug in values)

$\blueD{\text{area 1}}=35,000 \text{ J} \quad$start color #11accd, start text, a, r, e, a, space, 1, end text, end color #11accd, equals, 35, comma, 000, start text, space, J, end text (calculate)

We can find the area of a triangle by using $A=\dfrac{1}{2}bh$A, equals, start fraction, 1, divided by, 2, end fraction, b, h.

$\greenD{\text{area 2}}=\dfrac{1}{2}bh \quad$start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, b, h (area of the triangle)

$\greenD{\text{area 2}}=\dfrac{1}{2}b(160,000\text{Pa}-70,000\text{ Pa}) \quad$start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, b, left parenthesis, 160, comma, 000, start text, P, a, end text, minus, 70, comma, 000, start text, space, P, a, end text, right parenthesis (the height of the triangle is the difference in pressures $P_f-P_i$P, start subscript, f, end subscript, minus, P, start subscript, i, end subscript)

$\greenD{\text{area 2}}=\dfrac{1}{2} (0.75\text{m}^3-0.25\text{m}^3)(160,000\text{Pa}-70,000\text{ Pa}) \quad$start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, 0, point, 75, start text, m, end text, cubed, minus, 0, point, 25, start text, m, end text, cubed, right parenthesis, left parenthesis, 160, comma, 000, start text, P, a, end text, minus, 70, comma, 000, start text, space, P, a, end text, right parenthesis (the base of the triangle is the difference in volumes $V_f-V_i$V, start subscript, f, end subscript, minus, V, start subscript, i, end subscript)

$\greenD{\text{area 2}}=22,500 \text{ J} \quad$start color #1fab54, start text, a, r, e, a, space, 2, end text, end color #1fab54, equals, 22, comma, 500, start text, space, J, end text (calculate)

So the total area under the curve is $35,000 \text{ J} + 22,500 \text{ J}=57,500 \text{ J}$35, comma, 000, start text, space, J, end text, plus, 22, comma, 500, start text, space, J, end text, equals, 57, comma, 500, start text, space, J, end text

This area represents the absolute value of the total work done during the process. To determine the sign of the work done on the gas we notice that the process moves the state to the right, causing the gas to expand. When gas expands the work done on the gas is negative. So,

$W_\text{on gas}=-57,500 \text{ J}\quad$W, start subscript, start text, o, n, space, g, a, s, end text, end subscript, equals, minus, 57, comma, 500, start text, space, J, end text (celebrate)