Is My Calculation for Bicycle Pump Pressure Incorrect?

[Hak]

Yes, your original formula happens to agree with the correct formula to first order in the small quantity $V_p/V_C$.
From the binomial expansion we have $(1+x)^n \approx 1 + nx$ for small $x$.

Thank you. I cannot, however, understand why. The most general possible formula should be @Chestermiller's, not mine.

 

[Hak]

I've also found the attached document. What do you think? Can it be useful to me in any way for some application of a similar physical situation? Do you recommend reading it carefully or is it a waste of time? Thanks.

 

[TSny]

Thank you. I cannot, however, understand why. The most general possible formula should be @Chestermiller's, not mine.

Let $x = V_p/V_C$. The correct formula for the final pressure when there are "n transfers" of the small pump is $$P = P_{atm}\left(1+\frac{nV_p}{V_C}\right) = P_{atm}(1+nx).$$
Your formula in post #1 is $$P= P_{atm}\left(\frac{V_C + V_p}{V_C}\right)^n = P_{atm}(1+x)^n. $$
Are you asking why these two formulas give approximately the same result when $x$ is much less than 1?

 

[Hak]

Let $x = V_p/V_C$. The correct formula for the final pressure when there are "n transfers" of the small pump is $$P = P_{atm}\left(1+\frac{nV_p}{V_C}\right) = P_{atm}(1+nx).$$
Your formula in post #1 is $$P= P_{atm}\left(\frac{V_C + V_p}{V_C}\right)^n = P_{atm}(1+x)^n. $$
Are you asking why these two formulas give approximately the same result when $x$ is much less than 1?

Thank you very much, but unfortunately that was not my question. What I meant was that, if it meant anything, the most general possible formula for $p$ would be the one I set out in post #1, i.e. the one with the power term. Doing the necessary approximations for small $x$, applying the binomial formula, you get @Chestermiller's formula (which is actually the correct one). So, if that were the case, the most correct expression possible would be mine, and that of @Chestermiller only a special approximation for small $x$. On the other hand, if the most correct possible formula is @Chestermiller's (which is indeed true), how to start from it to obtain my equation? One would be starting from a particular equation to obtain a general one, which is contradictory. This physical assumption (@Chestermiller's equation is an approximation of my equation for small $x$) would therefore not be correct. This is what I wanted to know, and why. Thank you.

 

[TSny]

What I meant was that, if it meant anything, the most general possible formula for $p$ would be the one I set out in post #1, i.e. the one with the power term. Doing the necessary approximations for small $x$, applying the binomial formula, you get @Chestermiller's formula (which is actually the correct one). So, if that were the case, the most correct expression possible would be mine, and that of @Chestermiller only a special approximation for small $x$.

I don't understand your argument here or what you mean when you say that your formula is "the most general possible formula".

The correct formula for $P$ is $P = P_A(1+nx)$. There are an infinite number of incorrect formulas that would approximate the correct formula for small $x$. For example, $P = P_A\left(1+\frac{1}{2} nx \right)^2$, $P = P_Ae^{nx}$, and $P = P_A\left[1 + n\tan(x)\right]$. They are certainly incorrect formulas for the final pressure. But they do give approximately the right answer for small $x$. How do you decide which of these incorrect formulas is "the most general"? Even if you did have a criterion for deciding "most general", what does being "the most general expression" have to do with being "the most correct expression"?

On the other hand, if the most correct possible formula is @Chestermiller's (which is indeed true), how to start from it to obtain my equation?

I don't understand the motivation for trying to obtain an incorrect formula starting from the correct formula.

 

[Hak]

I don't understand your argument here or what you mean when you say that your formula is "the most general possible formula".

The correct formula for $P$ is $P = P_A(1+nx)$. There are an infinite number of incorrect formulas that would approximate the correct formula for small $x$. For example, $P = P_A\left(1+\frac{1}{2} nx \right)^2$, $P = P_Ae^{nx}$, and $P = P_A\left[1 + n\tan(x)\right]$. They are certainly incorrect formulas for the final pressure. But they do give approximately the right answer for small $x$. How do you decide which of these incorrect formulas is "the most general"? Even if you did have a criterion for deciding "most general", what does being "the most general expression" have to do with being "the most correct expression"?I don't understand the motivation for trying to obtain an incorrect formula starting from the correct formula.

Thank you. I wanted to say that if my expression was correct, it would be as general as possible, since from it, via the binomial formula, we get @Chestermiller's formula. If my expression is wrong (which it unquestionably is), that makes no sense. When I had asked "why, via the binomial formula, do I get @Chestermiller's correct one from my result?", you understood my question in a purely mathematical sense, telling me what binomial formula had to be used for this to make sense. Instead, I wanted to know if my solution could somehow be correct, since there was this particularity. I now had the answer to my doubt: my solution is incorrect, full stop. Binomial calculus and Taylor series have no relevance here from a physical point of view. Thank you for clarifying my doubt.
You and @Chestermiller have been very helpful.

Also, I wanted to add that the author of the exercise said that he had not yet found an intuitive solution for point 2.) I think the intuitive method may be your brilliant solution, @TSny, or am I wrong? I will refer it to him. As you can see, everyone was convinced to do calculations, no one but you would ever think there was a faster solution. Thanks again.

 

[TSny]

Also, I wanted to add that the author of the exercise said that he had not yet found an intuitive solution for point 2.) I think the intuitive method may be your brilliant solution, @TSny, or am I wrong? I will refer it to him. As you can see, everyone was convinced to do calculations, no one but you would ever think there was a faster solution. Thanks again.

I appreciate the compliments on my solution. But, I did not see that solution at first! Initially, I went through essentially the same calculations as outlined by @Chestermiller. When I saw that $W_1$ and $W_2$ came out the same, I stepped back to see if there was a different way. It took me a while.

Here's another way to see the answer. The volume of the large pump is four times the volume of the small pump. So, we can think of the initial air in the large pump as consisting of four "parcels" of air each of which has an initial volume equal to that of the small pump.

First step: Push the knob of the large pump until the first parcel of air is transferred to the container. The three parcels of air remaining in the large pump are now at the new pressure, $P_1$, of the container. So, each remaining parcel has been compressed from atmospheric pressure to $P_1$. The total work done so far is the work done to transfer the first parcel plus the work done to compress the other 3 parcels. Note that the work done on just the first parcel equals the work that is done by the small pump during its first stroke. So, the total work done by the large pump at this point is greater than the work done by the first stroke of the small pump.

Second step: Push the knob of the large pump farther until the second parcel of air has been transferred. The pressure of the system increases from $P_1$ to $P_2$. The total work done in this step is the work done on parcel number 2 as it was transferred to the container plus the work done on the two remaining parcels to compress them so that their pressure increases to $P_2$.

Note that the total work that was done on parcel number 2 is the work that was done on it in the first step to raise its pressure from atmospheric pressure to $P_1$ plus the work that was done on it as it was transferred to the container in the second step. Thus, this total work on parcel 2 matches the work done by the small pump during its second stroke.

Third step: Parcel number 3 will be transferred and the total work done on parcel 3 will be the work done on it during steps 1 and 2 to compress it from atmospheric pressure to $P_2$ plus the work done in step 3 to transfer it to the container. So, the total work done on parcel 3 matches the work done by stroke 3 of the small pump.

Fourth step: (Left for you).

 

[Hak]

I appreciate the compliments on my solution. But, I did not see that solution at first! Initially, I went through essentially the same calculations as outlined by @Chestermiller. When I saw that $W_1$ and $W_2$ came out the same, I stepped back to see if there was a different way. It took me a while.

Here's another way to see the answer. The volume of the large pump is four times the volume of the small pump. So, we can think of the initial air in the large pump as consisting of four "parcels" of air each of which has an initial volume equal to that of the small pump.

First step: Push the knob of the large pump until the first parcel of air is transferred to the container. The three parcels of air remaining in the large pump are now at the new pressure, $P_1$, of the container. So, each remaining parcel has been compressed from atmospheric pressure to $P_1$. The total work done so far is the work done to transfer the first parcel plus the work done to compress the other 3 parcels. Note that the work done on just the first parcel equals the work that is done by the small pump during its first stroke. So, the total work done by the large pump at this point is greater than the work done by the first stroke of the small pump.

Second step: Push the knob of the large pump farther until the second parcel of air has been transferred. The pressure of the system increases from $P_1$ to $P_2$. The total work done in this step is the work done on parcel number 2 as it was transferred to the container plus the work done on the two remaining parcels to compress them so that their pressure increases to $P_2$.

Note that the total work that was done on parcel number 2 is the work that was done on it in the first step to raise its pressure from atmospheric pressure to $P_1$ plus the work that was done on it as it was transferred to the container in the second step. Thus, this total work on parcel 2 matches the work done by the small pump during its second stroke.

Third step: Parcel number 3 will be transferred and the total work done on parcel 3 will be the work done on it during steps 1 and 2 to compress it from atmospheric pressure to $P_2$ plus the work done in step 3 to transfer it to the container. So, the total work done on parcel 3 matches the work done by stroke 3 of the small pump.

Fourth step: (Left for you).

Thank you. For the fourth step, I would follow the same path as the third step. I would say that parcel number 4 will be transferred and the total work done on parcel 4 will be the work done on it during steps 1, 2, and 3 to compress it from atmospheric pressure to $\displaystyle P_3$ plus the work done in step 4 to transfer it to the container. So, the total work done on parcel 4 matches the work done by stroke 4 of the small pump.

Therefore, we can see that the total work done by one stroke of the large pump is equal to the sum of the work done by each parcel of air, which is equal to the sum of the work done by each stroke of the small pump. This would confirm our previous result that $W_{four} = W_{one}.$

If I have made mistakes, please point them out to me. I would appreciate any subsequent response from you that better formulates the answer to Fourth step. Please do not hesitate to let me know. Thanks.

 

[TSny]

Thank you. For the fourth step, I would follow the same path as the third step. I would say that parcel number 4 will be transferred and the total work done on parcel 4 will be the work done on it during steps 1, 2, and 3 to compress it from atmospheric pressure to $\displaystyle P_3$ plus the work done in step 4 to transfer it to the container. So, the total work done on parcel 4 matches the work done by stroke 4 of the small pump.

Therefore, we can see that the total work done by one stroke of the large pump is equal to the sum of the work done by each parcel of air, which is equal to the sum of the work done by each stroke of the small pump. This would confirm our previous result that $W_{four} = W_{one}.$

Yes, that looks good.

 

[Hak]

Yes, that looks good.

Thank you so much!

 

[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.

 

[TSny]

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.

 

[Hak]

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...

 

[Hak]

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?

 

[TSny]

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 $Q =W$?

 

[Hak]

What are the steps that led you from $\Delta U = Q+W$ to $Q =W$?

I considered $\Delta U = Q - W$, not $\Delta U = Q+W$.

 

[TSny]

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.

 

[Hak]

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.

 

[TSny]

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

 

[Hak]

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.

 

[TSny]

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.