# Is My Calculation for Bicycle Pump Pressure Incorrect?

• Hak
Hak said:
Is the following graph (first stroke) correct? Or is the arrow pointing in the opposite direction?
Opposite direction

TSny said:
Opposite direction
So it is an expansion, not a compression... Right?

TSny said:
Compression generally means a decrease in volume of the system.

The isothermal work done on an ideal gas is given by ##W = - \int PdV = - nRT\log\frac{V_f}{V_i} = nRT\log\frac{V_i}{V_f}## and this formula holds for compression (##V_f < V_i##) and for expansion (##V_f > V_i##). For compression, the formula yields a positive work done on the gas. For expansion, the same formula yields a negative work done on the gas.I'm a little confused here. Our system consists of the air in the pump and in the container. Work is done on this system of air by the piston pushing on the air in the pump. So I'm not sure what you mean by work "carried out by the system on the air".
I mean "carried out by the system of air to the piston". Sorry.

TSny said:
Opposite direction
I just can't understand this and I am very much in distress and confusion. If it were directed in the opposite way, the initial volume would be less than the final volume, so it would be an expansion, not a compression... No?

Hak said:
So it is an expansion, not a compression... Right?
Oh, I think I might see the confusion now.

If you are just about to push down on the handle to transfer air into the container, then the air that is initially inside the pump with volume ##V_P## ends up inside the container with volume ##V_C##. So the volume of that air expands. But, we are not taking our system to be the air initially in the pump. The system is the air inside the pump plus the air inside the container. This system has an initial volume equal to ##V_P + V_C## and a final volume ##V_C##. So, this system undergoes compression and the pressure of the system increases. Positive work is done on this system by the piston.

I just noticed that your volumes ##V_C## and ##V_C + V_P## should be interchanged on the horizontal axis of your graph.

TSny said:
I just noticed that your volumes ##V_C## and ##V_C + V_P## should be interchanged on the horizontal axis of your graph.
Why?

TSny said:
Oh, I think I might see the confusion now.

If you are just about to push down on the handle to transfer air into the container, then the air that is initially inside the pump with volume ##V_P## ends up inside the container with volume ##V_C##. So the volume of that air expands. But, we are not taking our system to be the air initially in the pump. The system is the air inside the pump plus the air inside the container. This system has an initial volume equal to ##V_P + V_C## and a final volume ##V_C##. So, this system undergoes compression and the pressure of the system increases. Positive work is done on this system by the piston.
Thanks, I got it.

TSny said:
I just noticed that your volumes ##V_C## and ##V_C + V_P## should be interchanged on the horizontal axis of your graph.
If the initial volume is ##V_P +V_C## and the initial pressure is ##p_{atm}##, why must the two volumes be interchanged on the horizontal axis?

TSny said:
I just noticed that your volumes ##V_C## and ##V_C + V_P## should be interchanged on the horizontal axis of your graph.
I understand now. I swapped the values, the volumes and pressures I entered as lower, are actually higher than the latter. I am an idiot.

Edit. Actually, I did not understand how to make the graph. Maybe the latter is just wrong. I can't get the corresponding values on the axes of the Clapeyron diagram. How should this drawing be done?

Hak said:
I understand now. I swapped the values, the volumes and pressures I entered as lower, are actually higher than the latter.
Yes.
Hak said:
I am an idiot.
No. It was just an oversight.

Hak said:
Edit. Actually, I did not understand how to make the graph. Maybe the latter is just wrong. I can't get the corresponding values on the axes of the Clapeyron diagram. How should this drawing be done?
In your P-V diagram, you just need to switch the positions of ##V_C## and ##V_C+V_P## on the horizontal axis and switch the two pressure values on the vertical axis. Then, you can see that the direction of the process on the diagram is from lower right to upper left. The pressure increases as the volume decreases.

Hak said:
I understand now. I swapped the values, the volumes and pressures I entered as lower, are actually higher than the latter. I am an idiot.

Edit. Actually, I did not understand how to make the graph. Maybe the latter is just wrong. I can't get the corresponding values on the axes of the Clapeyron diagram. How should this drawing be done?
Is the following graph correct?

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Hak said:
Is the following graph correct?
Yes, that's correct.

TSny said:
In your P-V diagram, you just need to switch the positions of ##V_C## and ##V_C+V_P## on the horizontal axis and switch the two pressure values on the vertical axis. Then, you can see that the direction of the process on the diagram is from lower left to upper right. The pressure increases as the volume decreases.
I understand now. I had already fixed it, check if it is right, but it should be. Thank you very much.

TSny said:
Yes, that's correct.
Ok. So, in light of that, heat is absorbed by the system of the air and is positive, right? Thank you so much.

Hak said:
Ok. So, in light of that, heat is absorbed by the system of the air and is positive, right? Thank you so much.
In addition, I made these diagrams. Don't mind the Italian writing (unfortunately, carelessly, I forgot to translate from my native language to English), they are in ascending order from the first to the fourth stroke of the small pump, and finally there is that of the large pump. The part in yellow is the work done by each stroke. How do they look to you? Could anything be added?

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Hak said:
Ok. So, in light of that, heat is absorbed by the system of the air and is positive, right? Thank you so much.

Hak said:
Ok. So, in light of that, heat is absorbed by the system of the air and is positive, right?
No. Treating the air as an ideal gas, what can you say about the change in internal energy of a fixed amount of air as it is compressed isothermally? Use the first law to decide if the heat transfer to the air is positive or negative.

TSny said:
No. Treating the air as an ideal gas, what can you say about the change in internal energy of a fixed amount of air as it is compressed isothermally? Use the first law to decide if the heat transfer to the air is positive or negative.
I get ##\Delta U =0##, i.e. ##|Q| = |W|##. How can I tell if it is positive or negative? Normally, in a compression, the work done and the heat exchanged are both negative, but since in this case 'compression' takes on a different meaning... I had thought that the heat was absorbed by the air because of the work done by the piston. Do you have any advice?

Hak said:
I get ##\Delta U =0##, i.e. ##|Q| = |W|##.
OK. In post #24 you wrote the first law as
Hak said:
$$\displaystyle \Delta U = Q + W$$

Does ##Q## represent the heat that is gained by the system during the process or does it represent the heat that that is lost by the system?

Does ##W## represent the work done on the system during the process or the work done by the system?

For the compression of air in our problem, is ##W## positive or negative? So, is ##Q## positive or negative?

TSny said:
OK. In post #24 you wrote the first law asDoes ##Q## represent the heat that is gained by the system during the process or does it represent the heat that that is lost by the system?

Does ##W## represent the work done on the system during the process or the work done by the system?

For the compression of air in our problem, is ##W## positive or negative? So, is ##Q## positive or negative?
I made a mistake writing internal energy that way, the plus should be replaced by a minus. At first thought, I would say that ##Q## is the heat lost from the system and ##W## the work done on the system, in general. In our particular case, for air compression, ##W## is positive because it is the work done by the piston on the air. But how to deduce the sign of ##Q## from here? I don't understand. It could be positive as well as negative, couldn't it? How to deduce it? Thank you.

Hak said:
I made a mistake writing internal energy that way, the plus should be replaced by a minus.
Whether or not you write the first law as ##\Delta U = Q+W## or as ##\Delta U = Q-W## depends on whether you define ##W## as the work done on the system or as the work done by the system. It is OK to write the first law as ##\Delta U = Q + W##, but you need to be clear on the meaning of ##W## when writing the first law this way.

Once you have this cleared up, you can decide if ##W## is positive or negative for the air that is compressed. Then you should be able to deduce from ##\Delta U = Q + W## the sign of ##Q##. From the sign of ##Q## you should be able to conclude whether the air gained heat or lost heat.

However, the author of the problem says that the problem is not as intuitive as it might seem at first glance (four strokes of the small pump equal one stroke of a pump four times as large), not least because the work done by the first stroke of the small pump is different from a quarter of the work done by the large pump. Interesting, no? I'd say he's right... This problem is not as trivial as it seems...

TSny said:
Whether or not you write the first law as ##\Delta U = Q+W## or as ##\Delta U = Q-W## depends on whether you define ##W## as the work done on the system or as the work done by the system. It is OK to write the first law as ##\Delta U = Q + W##, but you need to be clear on the meaning of ##W## when writing the first law this way.

Once you have this cleared up, you can decide if ##W## is positive or negative for the air that is compressed. Then you should be able to deduce from ##\Delta U = Q + W## the sign of ##Q##. From the sign of ##Q## you should be able to conclude whether the air gained heat or lost heat.
In this case, the work is assumed to be positive because it is done by the piston on the air, so we have that ##W = Q##. For ##W## to be positive, ##Q## must also be positive, so I conclude that air gain heat. Where do I go wrong?

Hak said:
In this case, the work is assumed to be positive because it is done by the piston on the air, so we have that ##W = Q##. For ##W## to be positive, ##Q## must also be positive, so I conclude that air gain heat. Where do I go wrong?
What are the steps that led you from ##\Delta U = Q+W## to ##Q =W##?

TSny said:
What are the steps that led you from ##\Delta U = Q+W## to ##Q =W##?
I considered ##\Delta U = Q - W##, not ##\Delta U = Q+W##.

Hak said:
I considered ##\Delta U = Q - W##, not ##\Delta U = Q+W##.
OK. You can write the first law as ##\Delta U = Q - W##. But then, what is the meaning of ##W##?

Does positive ##W## mean that a positive amount of work was done on the system by the environment, or does positive ##W## mean that a positive amount of work was done on the environment by the system?

To help decide this, consider a process for which ##Q = 0## so that your way of writing the first law reduces to ##\Delta U = -W##. This says that the internal energy of the system decreases when ##W## is positive.

TSny said:
OK. You can write the first law as ##\Delta U = Q - W##. But then, what is the meaning of ##W##?

Does positive ##W## mean that a positive amount of work was done on the system by the environment, or does positive ##W## mean that a positive amount of work was done on the environment by the system?

To help decide this, consider a process for which ##Q = 0## so that your way of writing the first law reduces to ##\Delta U = -W##. This says that the internal energy of the system decreases when ##W## is positive.
I assumed the work done by the system on the environment.

Take a look at this link. Be sure to click on the additional information immediately under the equation of the first law.

TSny said:
Take a look at this link. Be sure to click on the additional information immediately under the equation of the first law.
OK, maybe I got it. With the expression ##\Delta U = Q - W##, ##W## is positive work done by the gas (system) on the environment (piston), whereas we have assumed positive work done on the gas (system of air) by the environment (piston). Therefore, we use ##\Delta U = Q + W##. We therefore have ##Q = - W##. With positive ##W##, ##Q## must be negative. Therefore, heat is transferred from the gas to the environment, not gained. Right? Thank you.

Hak said:
OK, maybe I got it. With the expression ##\Delta U = Q - W##, ##W## is positive work done by the gas (system) on the environment (piston), whereas we have assumed positive work done on the gas (system of air) by the environment (piston). Therefore, we use ##\Delta U = Q + W##. We therefore have ##Q = - W##. With positive ##W##, ##Q## must be negative. Therefore, heat is transferred from the gas to the environment, not gained. Right? Thank you.
Yes. Very nice.

TSny said:
Yes. Very nice.
Thank you very much, I learnt a lot.
Are the graphs at post #86 correct?

TSny said:
Take a look at this link. Be sure to click on the additional information immediately under the equation of the first law.
However, the source of my confusion was that work is considered positive when the system does work on its surroundings, negative when the opposite happens. Does this situation fall into these two categories? It doesn't seem so to me. What answers can you give me?

Last edited:
Hak said:
Thank you very much, I learnt a lot.
Are the graphs at post #86 correct?
Yes, they look correct.

TSny said:
Yes, they look correct.
Thanks.

Hak said:
However, the source of my confusion was that work is considered positive when the system does work on its surroundings, negative when the opposite happens. Does this situation fall into these two categories? It doesn't seem so to me. What answers can you give me?
It is a matter of convention as to whether the symbol ##W## in the first law is positive when the system does work on the surroundings or is positive when the surroundings does work on the system.

For the first convention where ##W## represents the work done by the system, the first law is $$\Delta U = Q - W.$$ For the second convention where ##W## represents the work done on the system, the first law is $$\Delta U = Q+W.$$ ##W## will differ in sign for the two conventions. You are free to choose either convention for this problem. If you use the first convention and calculate ##W## for, say, the first stroke of the small pump, then you would get a negative value for ##W##. This means that the air in the pump did negative work on the piston. Therefore, the piston did positive work on the air in the pump. (This follow's from Newton's third law of motion.)

If you use the second convention, then you would find that ##W## has a positive value. Therefore, the air did positive work on the air in the pump. This agrees with the first convention.

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