Is My Calculation for Bicycle Pump Pressure Incorrect?

[Hak]

Yes, that looks good. But, note that your expressions for $W_{3a}$ and $W_{3b}$ will be negative quantities since the arguments of the logarithms are less than one. This has to do with Chestermiller's expressions that represent the work done by the air instead of the work done on the air. I believe the problem is asking for work done on the air by the person operating the pumps.

Sorry to bother, rereading this statement of yours gave me a doubt. Why would the work be done by the system on the environment (and not by the environment on the system)? The problem talks about "isothermal compression," and in a compression the work is always done by the system on the environment, no? Where am I wrong? Thank you for any response.

 

[TSny]

Sorry to bother, rereading this statement of yours gave me a doubt. Why would the work be done by the system on the environment (and not by the environment on the system)? The problem talks about "isothermal compression," and in a compression the work is always done by the system on the environment, no? Where am I wrong? Thank you for any response.

The environment does positive work on the air in the pump while, at the same time, the air in the pump does negative work on the environment. This is the same as when you compress a spring. You do positive work on the spring while the spring does negative work on you. Your force on the spring is equal and opposite to the force of the spring on you (Newton's 3rd law).

 

[Hak]

The environment does positive work on the air in the pump while, at the same time, the air in the pump does negative work on the environment. This is the same as when you compress a spring. You do positive work on the spring while the spring does negative work on you. Your force on the spring is equal and opposite to the force of the spring on you (Newton's 3rd law).

Of course, I understand, it is right what you say. I, however, meant that, perhaps, by "isothermal compression" the text might mean the negative work done by the air on the system rather than the positive work done by the system on the air (although, in fact, it is essentially the same, but you had said that you thought the text meant the second option). Let me know. Thank you very much.

 

[TSny]

Of course, I understand, it is right what you say. I, however, meant that, perhaps, by "isothermal compression" the text might mean the negative work done by the air on the system rather than the positive work done by the system on the air (although, in fact, it is essentially the same, but you had said that you thought the text meant the second option). Let me know. Thank you very much.

I took the system to be the air that was initially in the container plus the air in the atmosphere that would eventually be pumped into the container. I believe $W_1$ and $W_2$ in the part (2) of the problem refer to the work done on this system by the piston in the pump (or, equivalently by the person pushing the piston). This work is positive as the air in the pump is compressed.

The formula $dW = PdV$ gives the work done by the air inside the pump on the piston. This work is negative (note $dV$ of the system is negative for a compression). For an isothermal compression, integration of this formula leads to $W = nRT\log(V_f/V_i)$ which you can see is also negative since $V_f$ is less than $V_i$.

If you want the work done on the air in the pump, then you would use $dW = -PdV$ which yields $W = -nRT\log(V_f/V_i) = nRT\log(V_i/V_f)$. This work is positive. It's the work that I think is asked for in the problem.

 

[Hak]

I took the system to be the air that was initially in the container plus the air in the atmosphere that would eventually be pumped into the container. I believe $W_1$ and $W_2$ in the part (2) of the problem refer to the work done on this system by the piston in the pump (or, equivalently by the person pushing the piston). This work is positive as the air in the pump is compressed.

The formula $dW = PdV$ gives the work done by the air inside the pump on the piston. This work is negative (note $dV$ of the system is negative for a compression). For an isothermal compression, integration of this formula leads to $W = nRT\log(V_f/V_i)$ which you can see is also negative since $V_f$ is less than $V_i$.

If you want the work done on the air in the pump, then you would use $dW = -PdV$ which yields $W = -nRT\log(V_f/V_i) = nRT\log(V_i/V_f)$. This work is positive. It's the work that I think is asked for in the problem.

Okay, I think I understand. I didn't understand, however, whether the work you define as positive leads to an isothermal expansion (compression of air), with initial volume less than final volume, or whether the initial and final volume values are always the same as given by @Chestermiller (i.e., initial volume greater than final volume), but occupy a different place within the natural logarithm argument. I think there is a definitional disconnect between how a "compression" is vulgarly understood (work carried out by the external environment on the gas system) and how you understand it more deeply (compression of air, then work carried out by the gas on the air). Could you clarify the above doubt? Thank you very much.

 

[TSny]

I believe $W_1$ and $W_2$ in the part (2) of the problem refer to the work done on this system by the piston in the pump (or, equivalently by the person pushing the piston).

On second thought, I don't believe that the work done by the pump's piston on the air in the pump is equal to the work done by the person pushing the piston. This is because external atmospheric pressure will help to push the piston inward. So, the force that the person pushes on the handle of the pump is less than the force that the air inside the pump pushes outward on the piston. The person does less work on the pump handle than the work done by the piston on the air inside the pump. But this doesn't affect the result that the work done by the person for the small pump is the same as the work done by the person for the large pump.

 

[TSny]

Okay, I think I understand. I didn't understand, however, whether the work you define as positive leads to an isothermal expansion (compression of air), with initial volume less than final volume, or whether the initial and final volume values are always the same as given by @Chestermiller (i.e., initial volume greater than final volume), but occupy a different place within the natural logarithm argument. I think there is a definitional disconnect between how a "compression" is vulgarly understood (work carried out by the external environment on the gas system) and how you understand it more deeply (compression of air, then work carried out by the gas on the air). Could you clarify the above doubt? Thank you very much.

I'm sorry, but your question is not clear to me.

 

[Hak]

I'm sorry, but your question is not clear to me.

Are the initial and final volumes the same as those given by @Chestermiller? If so, is what you mean by "compression" what is generally called "isothermal expansion"? Or does isothermal compression always occur, and you have to choose a posteriori whether the text means positive work because it is carried out by the system on the air, or negative work because it is carried out by the air on the system?

 

[Hak]

Are the initial and final volumes the same as those given by @Chestermiller? If so, is what you mean by "compression" what is generally called "isothermal expansion"? Or does isothermal compression always occur, and you have to choose a posteriori whether the text means positive work because it is carried out by the system on the air, or negative work because it is carried out by the air on the system?

Is the following graph (first stroke) correct? Or is the arrow pointing in the opposite direction?

 

[TSny]

Are the initial and final volumes the same as those given by @Chestermiller? If so, is what you mean by "compression" what is generally called "isothermal expansion"?

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.

Or does isothermal compression always occur, and you have to choose a posteriori whether the text means positive work because it is carried out by the system on the air, or negative work because it is carried out by the air on the system?

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".

 

[TSny]

Is the following graph (first stroke) correct? Or is the arrow pointing in the opposite direction?

Opposite direction

 

[Hak]

Opposite direction

So it is an expansion, not a compression... Right?

 

[Hak]

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.

 

[Hak]

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?

 

[TSny]

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.

 

[TSny]

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

 

[Hak]

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

Why?

 

[Hak]

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.

 

[Hak]

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?

 

[Hak]

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?

 

[TSny]

I understand now. I swapped the values, the volumes and pressures I entered as lower, are actually higher than the latter.

Yes.

I am an idiot.

No. It was just an oversight.

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]

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?

 

[TSny]

Is the following graph correct?

Yes, that's correct.

 

[Hak]

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.

 

[Hak]

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]

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?

 

[TSny]

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

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.

 

[Hak]

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?

 

[TSny]

I get $\Delta U =0$, i.e. $|Q| = |W|$.

OK. In post #24 you wrote the first law as

\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?

 

[Hak]

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.