# Closed path term in Thermodynamics

• ChiralSuperfields
In summary: English language. It is not a semicircle.It seems to me that the person who drew the graph (as well as the final editor) should be fired.
ChiralSuperfields
Homework Statement
Relevant Equations
For part(a) of this problem,

The solution is, a. 160 J

In part (a), are they referring to the simi-circular path from R to S instead of the path from R back to R? I though the closed path would be the path from R to R, or the path from S to S, where both give W = 0 since ##W = P(V_f - V_i) = P(V_i - V_i) = 0##

Many thanks!

Callumnc1 said:
View attachment 324177
The solution is, a. 160 J

In part (a), are they referring to the simi-circular path from R to S instead of the path from R back to R? I though the closed path would be the path from R to R, or the path from S to S, where both give W = 0 since ##W = P(V_f - V_i) = P(V_i - V_i) = 0##

They are referring to the complete cycle. For example, start at R and go to S along the semicircle and then return to R along the straight section.

The formula ##W = P(V_f - V_i)## gives the work done by the gas if the pressure is constant from ##i## to ##f##. For the semicircular part, note that ##P## is not constant.

ChiralSuperfields and Lnewqban
Callumnc1 said:
In part (a), are they referring to the simi-circular path from R to S instead of the path from R back to R? I though the closed path would be the path from R to R, or the path from S to S, where both give W = 0 since ##W = P(V_f - V_i) = P(V_i - V_i) = 0##
The work done by a gas is not ##p\Delta V##. That is only the case when the pressure is constant. Here it is not. The general expression for the work done by the gas is $$W_{ab}=\int_{V_a}^{V_b}p~dV.$$Hint: The geometric interpretation of the integral is the "area under the curve."

PhDeezNutz, ChiralSuperfields and Lnewqban
TSny said:
They are referring to the complete cycle. For example, start at R and go to S along the semicircle and then return to R along the straight section.

The formula ##W = P(V_f - V_i)## gives the work done by the gas if the pressure is constant from ##i## to ##f##. For the semicircular part, note that ##P## is not constant.
kuruman said:
The work done by a gas is not ##p\Delta V##. That is only the case when the pressure is constant. Here it is not. The general expression for the work done by the gas is $$W_{ab}=\int_{V_a}^{V_b}p~dV.$$Hint: The geometric interpretation of the integral is the "area under the curve."
Thank you for your replies @TSny and @kuruman! I will try have another go at this problem given your hints and if I don't understand I will reply to this thread again.

Many thanks!

How can they talk about the curved path from R to S being a semi-circle? If you change the units of either P or V, the shape will change. It doesn't even look like a semicircle in the units shown in the figure. At the very least, they should call it half an ellipse.

SammyS and ChiralSuperfields
Chestermiller said:
How can they talk about the curved path from R to S being a semi-circle? If you change the units of either P or V, the shape will change. It doesn't even look like a semicircle in the units shown in the figure. At the very least, they should call it half an ellipse.
From @Callumnc1's other posts, it appears to be a non-calculus course, so one has to be able to calculate areas without integration

ChiralSuperfields
DrClaude said:
From @Callumnc1's other posts, it appears to be a non-calculus course, so one has to be able to calculate areas without integration
The area of an ellipse is ##\pi ab## where a is the semi-major axis and b is the semi-minor axis.

hutchphd and ChiralSuperfields
Chestermiller said:
The area of an ellipse is ##\pi ab## where a is the semi-major axis and b is the semi-minor axis.
If you look at the drawing, it certainly does not look like a circle because clearly the diameter is 2 units (L) long but the top of the arc is farther than 1 unit (atm) from the diameter. It seems to me that we should go along with the assertion that it is a semicircle instead of trying to read the value of the semi-minor axis from the drawing. The quoted answer is matches to 2 sig figs the semicircle assumption.

ChiralSuperfields
Chestermiller said:
At the very least, they should call it half an ellipse.

kuruman said:
It seems to me that we should go along with the assertion that it is a semicircle instead of trying to read the value of the semi-minor axis from the drawing.

It seems to me that the person who drew the graph (as well as the final editor) should be fired. This is nuts.

malawi_glenn, ChiralSuperfields and Chestermiller
Thank you for your replies @Chestermiller , @DrClaude , @kuruman and @hutchphd !

Yes I am very confused how to solve this problem. I am not sure what value for the radius of the simi-circle a am meant to use since ##r = 1~atm = 1~L## from the simi-circle, however, work should be in SI units so ##r = 101325~Pa = 2 \times 10^{-3}~m^3##

Many thanks!

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DrClaude said:
From @Callumnc1's other posts, it appears to be a non-calculus course, so one has to be able to calculate areas without integration
Thank you for your reply @DrClaude! I do know the integral definition of work done by a gas since we are using a calculus physics textbook for the course https://openstax.org/details/books/university-physics-volume-2

I am trying to solve this problem using both the area and integral definition.

For the integral definition, ##W_{RR} = W_{RS} + W_{SR}## since work is path dependent for thermodynamic processes, we must calculate the work for each path, however, since the pressure varies for the ##W_{RS}## path, then we cannot calculate the work since ##W_{RS} = nRTIn|\frac{V_f}{V_i}|## and ##n## and ##T## are not known.

Many thanks!

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Chestermiller said:
The area of an ellipse is ##\pi ab## where a is the semi-major axis and b is the semi-minor axis.
Thank you for your replies @Chestermiller! I tried that and got ##\pi(101325)(2 \times 10^{-3} = 640~J##. It is interested that you mention that the simi-circular assumption is clearly not valid if the units are changed (which I guess is most commonly done before calculating the area)

Many thanks!

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kuruman said:
If you look at the drawing, it certainly does not look like a circle because clearly the diameter is 2 units (L) long but the top of the arc is farther than 1 unit (atm) from the diameter. It seems to me that we should go along with the assertion that it is a semicircle instead of trying to read the value of the semi-minor axis from the drawing. The quoted answer is matches to 2 sig figs the semicircle assumption.
Thank you for your reply @kuruman! I am still trying to understand what value of ##r## they used for finding the area of the simi-circle. To find what ##r## they used: ##160 = \frac{\pi r^2}{2}## which means that ##10~J^{1/2}= r##. I do not understand how they got that from the diagram.

Also to make sure I am actually finding the correct area, we are meant to be finding the area in orange correct?

This is because I know from the integral definition of work that the work done by the gas from S to R ##W_{SR}## will be negative since it compresses so displacement of container is in opposite direction to force pushing due to gas pressure. However, I don't think there is anyway we know just by considering the area under the curve from the diagram that the work from S to R is negative. I think we just have to look at the integral definition of work.

Before I knew that the work done by gas was path dependent, I thought that we could just find the work by considering the endpoints on the PV diagram so I wrongly thought that ##W_{RR} = 0## since it has a finial volume is equal to the initial volume.

Many thanks!

Callumnc1 said:
Before I knew that the work done by gas was path dependent ##~\dots##
Where did you get that impression? The work done by the gas is most certainly not path-independent.
The change in internal energy is path-independent and depends on the end points in a pV diagram.
Callumnc1 said:
This is because I know from the integral definition of work that the work done by the gas from S to R ##W_{SR}## will be negative since it compresses so displacement of container is in opposite direction to force pushing due to gas pressure.
The work done by the gas from S to R is negative and the work done by the gs from R back to S is positive. The net work is the sum of the two works. Graphically ii is the area colored red. Do you see why that represents the net work?

ChiralSuperfields
Callumnc1 said:
Thank you for your replies @Chestermiller! I tried that and got ##\pi(101325)(2 \times 10^{-3} = 640~J##. It is interested that you mention that the simi-circular assumption is clearly not valid if the units are changed (which I guess is most commonly done before calculating the area)

Many thanks!
That’s not what I get.I get about 240 J.

ChiralSuperfields
kuruman said:
Where did you get that impression? The work done by the gas is most certainly not path-independent.
The change in internal energy is path-independent and depends on the end points in a pV diagram.

The work done by the gas from S to R is negative and the work done by the gs from R back to S is positive. The net work is the sum of the two works. Graphically ii is the area colored red. Do you see why that represents the net work?

Yeah, I think I failed to differentiate enough between internal energy and work. Sorry what is ii?

Many thanks!

Chestermiller said:
That’s not what I get.I get about 240 J.

Sorry how do you get 240J? Was there a mistake in my calculation?

Many thanks!

To do the problem "their way" requires us to find the area from the PV graph. The area of the unit semicircle according to their description will be pi/2 and the unit of area is L-Atm. Thus the semicircular path work is $$W= \frac {\pi} 2 lAtm(101.325\frac J {lAtm})$$

ChiralSuperfields
Callumnc1 said:

Sorry how do you get 240J? Was there a mistake in my calculation?

Many thanks!
##W=\frac{\pi (101325\times 1.5)(0.001)}{2}=239##

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Steve4Physics and ChiralSuperfields
hutchphd said:
To do the problem "their way" requires us to find the area from the PV graph. The area of the unit semicircle according to their description will be pi/2 and the unit of area is L-Atm. Thus the semicircular path work is $$W= \frac {\pi} 2 lAtm(101.325\frac J {lAtm})$$
Thank you for you reply @hutchphd ! I am not sure if I understand how expression. Why is it in terms of atm?

Many thanks!

Chestermiller said:
##W=\frac{\pi (101325\ x\ 1.5)(0.001)}{2}=239##
Thank you for your help @Chestermiller ! I see my mistake in that equation now!

Many thanks!

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kuruman said:
Where did you get that impression? The work done by the gas is most certainly not path-independent.
The change in internal energy is path-independent and depends on the end points in a pV diagram.

The work done by the gas from S to R is negative and the work done by the gs from R back to S is positive. The net work is the sum of the two works. Graphically ii is the area colored red. Do you see why that represents the net work?
It seems difficult to prove that from S to R is negative work since the pressure is on the y-axis and cannot be negative. Is the negative work done by the gas from S to R, ##W_{SR} = AREA~from~S~to~R~= (101325 Pa)(2 \times 10^{-3} m^3)## however, to make it negative, we must make the volume negative so ##(101325 Pa)(-2 \times 10^{-3} m^3)##. Normally negative area is shown below the graph thought like for negative work shown on a force vs position diagram. Is this just a convection for PV diagrams?

Many thanks!

That was their choice of label on the graph. It could have any unit of pressure.
This is just an abysmally badly conceived problem. If the graph had simply been drawn correctly to scale you could have been told to estimate the area. It is useful to remember that a circle is a degenerate ellipse and any uniform stretching of either x or y (x→αx or y→βy) turns a circle into an ellipse.

ChiralSuperfields
hutchphd said:
That was their choice of label on the graph. It could have any unit of pressure.
This is just an abysmally badly conceived problem. If the graph had simply been drawn correctly to scale you could have been told to estimate the area. It is useful to remember that a circle is a degenerate ellipse and any uniform stretching of either x or y (x→αx or y→βy) turns a circle into an ellipse.
Thank you for your reply @hutchphd! Do you please know whether my reasoning in post #22 is correct?

Many thanks!

I think you get the physics (I often get the sign wrong so I just fudge check it). The actual correct answer is that the sign of the integral depends upon the direction along the function (meaning whether dx is positive or negative). Going right to left is "backwards" and the result is negative area.

ChiralSuperfields
It is interesting to see the approaches to this problem by different people. @Chestermiller approach is to ignore the statement of the problem that asserts that the arc is a semi-circle and consider it half an ellipse. The shape could be declared an ellipse because the major axis is 2 units long while the semi-minor axis appears to be longer than one unit long.

However, if one measures the ratio of the horizontal axis to the height (I did with a plastic ruler against my computer screen), it is (to within experimental error) 2:1. So what does one believe? The tick marks or the statement of the problem and the proportions of the shape itself? I give the edge to the latter because it shows the intentions of the problem's author.

We have a semicircle of horizontal radius ##R_h= 1~\text{L}## and vertical radius ##R_v= 1~\text{atm}##. The area of this is the work $$W=\frac{1}{2}\pi R_h~R_v=\frac{1}{2}\pi~(1)\times(1) \text{(L}\cdot\text{atm)}.$$In MKS units and with the atmospheric pressure rounded off to 2 significant figures, $$1~\text{(L}\cdot\text{atm)}=(10^{-3}\text{m} ^{3}\cdot 10^5~\text{Pa}^3)=100~\text{J}.$$To 2 significant figures, which is the accuracy of the tick marks, ##W =100 \frac{\pi}{2}~\text{J}=160~\text{J}##. This answer matches the given answer.

It is clear to me that the author intended this solution. It is also clear to me that the author should have been more careful with the tick mark positioning to avoid the kind of confusion that has justifiably ensued.

Steve4Physics, hutchphd and ChiralSuperfields
hutchphd said:
I think you get the physics (I often get the sign wrong so I just fudge check it). The actual correct answer is that the sign of the integral depends upon the direction along the function (meaning whether dx is positive or negative). Going right to left is "backwards" and the result is negative area.
Thank you for your help @hutchphd! At first it seems counterintuitive that we can have negative area above the x-axis for a rectangle.

But, if area results from the change in x and y then we can write the area of a rectangle as ##A = \Delta x \Delta y##. If we consider a special case where we have a rectangle above the x-axis and we start at an x-value ##x_i## and end at an x-value ##x_f## and ##x_i > x_f## then ##A = (x_i - x_f)(\Delta y)## so therefore as ##\Delta y > 0## then ##A < 0##

I was always taught that area of a rectangle was always positive. This is because we were only concerned about the magnitude of the lengths, that is ##A = |\Delta x| |\Delta y|## so ##A >0##

Does anybody agree with the results of the proof?

Many thanks!

kuruman said:
It is interesting to see the approaches to this problem by different people. @Chestermiller approach is to ignore the statement of the problem that asserts that the arc is a semi-circle and consider it half an ellipse. The shape could be declared an ellipse because the major axis is 2 units long while the semi-minor axis appears to be longer than one unit long.

However, if one measures the ratio of the horizontal axis to the height (I did with a plastic ruler against my computer screen), it is (to within experimental error) 2:1. So what does one believe? The tick marks or the statement of the problem and the proportions of the shape itself? I give the edge to the latter because it shows the intentions of the problem's author.

We have a semicircle of horizontal radius ##R_h= 1~\text{L}## and vertical radius ##R_v= 1~\text{atm}##. The area of this is the work $$W=\frac{1}{2}\pi R_h~R_v=\frac{1}{2}\pi~(1)\times(1) \text{(L}\cdot\text{atm)}.$$In MKS units and with the atmospheric pressure rounded off to 2 significant figures, $$1~\text{(L}\cdot\text{atm)}=(10^{-3}\text{m} ^{3}\cdot 10^5~\text{Pa}^3)=100~\text{J}.$$To 2 significant figures, which is the accuracy of the tick marks, ##W =100 \frac{\pi}{2}~\text{J}=160~\text{J}##. This answer matches the given answer.

It is clear to me that the author intended this solution. It is also clear to me that the author should have been more careful with the tick mark positioning to avoid the kind of confusion that has justifiably ensued.
Thank you for your help @kuruman!

I agree this was a really interesting problem!

Many thanks!

kuruman said:
It is interesting to see the approaches to this problem by different people. @Chestermiller approach is to ignore the statement of the problem that asserts that the arc is a semi-circle and consider it half an ellipse. The shape could be declared an ellipse because the major axis is 2 units long while the semi-minor axis appears to be longer than one unit long.

However, if one measures the ratio of the horizontal axis to the height (I did with a plastic ruler against my computer screen), it is (to within experimental error) 2:1. So what does one believe? The tick marks or the statement of the problem and the proportions of the shape itself? I give the edge to the latter because it shows the intentions of the problem's author.

We have a semicircle of horizontal radius ##R_h= 1~\text{L}## and vertical radius ##R_v= 1~\text{atm}##. The area of this is the work $$W=\frac{1}{2}\pi R_h~R_v=\frac{1}{2}\pi~(1)\times(1) \text{(L}\cdot\text{atm)}.$$In MKS units and with the atmospheric pressure rounded off to 2 significant figures, $$1~\text{(L}\cdot\text{atm)}=(10^{-3}\text{m} ^{3}\cdot 10^5~\text{Pa}^3)=100~\text{J}.$$To 2 significant figures, which is the accuracy of the tick marks, ##W =100 \frac{\pi}{2}~\text{J}=160~\text{J}##. This answer matches the given answer.

It is clear to me that the author intended this solution. It is also clear to me that the author should have been more careful with the tick mark positioning to avoid the kind of confusion that has justifiably ensued.

Strangely enough you can actually also get the correct answer from ##W = \frac{1}{2}\pi(1 \times 10^{-3} m^3)(101325 Pa) = 160##. It really interesting how we don't get the correct answer if we do ##W = \frac{1}{2}\pi(1 \times 10^{-3} m^3)^2## or ##W = \frac{1}{2}\pi(101325 Pa)^2## from ##W = \frac{\pi}{2}r^2##, when according to the question statement (ignoring the diagram) ##r = 101325~Pa = 1 \times 10^{-3} m^3##

Maybe something to do with units?

Many thanks!

Callumnc1 said:

Strangely enough you can actually also get the correct answer from ##W = \frac{1}{2}\pi(1 \times 10^{-3} m^3)(101325 Pa) = 160##. It really interesting how we don't get the correct answer if we do ##W = \frac{1}{2}\pi(1 \times 10^{-3} m^3)^2## or ##W = \frac{1}{2}\pi(101325 Pa)^2## from ##W = \frac{\pi}{2}r^2##, when according to the question statement (ignoring the diagram) ##r = 101325~Pa = 1 \times 10^{-3} m^3##

Maybe something to do with units?

Many thanks!
Never mind it is because if we convert the units beforehand then it is no longer a simi-circle (imagine a graph of Pressure (Pa) vs volume (m^3) for the values, which means that we have to use the ellipse formula. Is this please correct @kuruman?

Many thanks!

Callumnc1 said:
Never mind it is because if we convert the units beforehand then it is no longer a simi-circle (imagine a graph of Pressure (Pa) vs volume (m^3) for the values, which means that we have to use the ellipse formula. Is this please correct @kuruman?

Many thanks!
You don't need to convert units to anything as long as it is clear what units are appropriate for each axis. And yes, if you rescale the axes by different factors, the shape will not retain its previous proportions. A circle may become an ellipse but it is also possible for an ellipse to become a circle if the axes are scaled just right.

ChiralSuperfields
kuruman said:
You don't need to convert units to anything as long as it is clear what units are appropriate for each axis. And yes, if you rescale the axes by different factors, the shape will not retain its previous proportions. A circle may become an ellipse but it is also possible for an ellipse to become a circle if the axes are scaled just right.

That is a very interesting way to think about substituting in values!

Dose anybody please know how to solve this problem using integration instead of graphically calculating the area?

Many thanks!

It also looks the author did it again for part(c) and (d). This time they don't state the simi-circular assumption for (c), so I did an elliptical shape from the figure to get 240J which dose not match the solutions. The correct answer according to the solutions for (c) is 160J.

Many thanks!

I’m coming in rather late but I hope this helps.

Callumnc1 said:
It also looks the author did it again for part(c) and (d). This time they don't state the simi-circular assumption for (c), so I did an elliptical shape from the figure to get 240J which dose not match the solutions. The correct answer according to the solutions for (c) is 160J.
I see no part d)!

Can I pull together a few points...

The ‘horizontal’ radius of the semicircle is 1.0L.

As @kuruman showed in post #26, we get the official answer (160J) if we assume the pressure scale is incorrectly drawn such that the ‘vertical’ radius is in fact 1.0 atm.

As @Chestermiller showed in Post #19, if we use 1.5 atm as the ‘vertical’ length, then we get 240J.

A general point (if not already apparent):
The area inside a closed loop on a PV diagram represents the work done by (or on) the gas. And this equals the amount of heat entering (or leaving).
Going clockwise round the loop, the area represents the work done by the gas in the cycle and this equals the amount of heat entering;
Going anticlockwise round the loop, the area represents the work done on the gas in the cycle and this equals the amount of heat leaving.

If you wanted to solve this by integration, it would be simply an exercise to find half the area of a circle (or ellipse) by integration. You would start by finding the equation of the curve.

ChiralSuperfields and DrClaude

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