Is My Calculation for Bicycle Pump Pressure Incorrect?

[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.

 

[Hak]

Yes. Very nice.

Thank you very much, I learnt a lot.
Are the graphs at post #86 correct?

 

[Hak]

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?

 

[TSny]

Thank you very much, I learnt a lot.
Are the graphs at post #86 correct?

Yes, they look correct.

 

[Hak]

Yes, they look correct.

Thanks.

 

[TSny]

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.

 

[Hak]

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.

Thank you so much. I understand now.

 

[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).

Sorry, but an additional doubt came to me. You say (rightly) that the total work of the large pump at step 1 is greater than that carried out by the first stroke of the small pump, while all others are equal. How would that sanction that the work of four strokes of the small pump is equal to that of one stroke of the large pump, if it is greater at point 1? A quarter of the total work of the large pump is not equal to the work carried out by the first stroke of the small pump. Thank you very much, sorry again for the trouble.

 

[TSny]

Sorry, but an additional doubt came to me. You say (rightly) that the total work of the large pump at step 1 is greater than that carried out by the first stroke of the small pump,

Yes

while all others are equal.

I don't believe this is correct. Check it.

 

[Hak]

Yes

I don't believe this is correct. Check it.

I checked, it should say in your post that they are all the same from the second to the fourth stroke (post #57). Maybe I misunderstood it.... What advice can you give me? Thank you.

 

[TSny]

I checked, it should say in your post that they are all the same from the second to the fourth stroke (post #57). Maybe I misunderstood it.... What advice can you give me? Thank you.

Summarizing from post #57:

In step 1 of the large pump, parcel 1 is transferred to the container. During this step work is done on all four parcels of air. (Even though parcels 2, 3, and 4 are not transferred to the container in this step, work is done on these parcels to compress their volume.)

In step 2 of the large pump, parcel 2 is transferred to the container and work is done on parcels 2, 3, and 4.

In step 3 of the large pump, work is done on parcels 3 and 4 as parcel 3 is transferred.

In step 4 of the large pump, work is done only on parcel 4.

The net work done on parcel 2, for example, is the work done on parcel 2 in step 1 plus the work done on parcel 2 in step 2. This net work done on parcel 2 turns out to match the total work done by the small pump during stroke 2 of the small pump.

But, nowhere in post #57 is it claimed that the total work done in step 2 of the large pump equals the total work done by the small pump during stroke 2 of the small pump.

 

[Hak]

Summarizing from post #57:

In step 1 of the large pump, parcel 1 is transferred to the container. During this step work is done on all four parcels of air. (Even though parcels 2, 3, and 4 are not transferred to the container in this step, work is done on these parcels to compress their volume.)

In step 2 of the large pump, parcel 2 is transferred to the container and work is done on parcels 2, 3, and 4.

In step 3 of the large pump, work is done on parcels 3 and 4 as parcel 3 is transferred.

In step 4 of the large pump, work is done only on parcel 4.

The net work done on parcel 2, for example, is the work done on parcel 2 in step 1 plus the work done on parcel 2 in step 2. This net work done on parcel 2 turns out to match the total work done by the small pump during stroke 2 of the small pump.

But, nowhere in post #57 is it claimed that the total work done in step 2 of the large pump equals the total work done by the small pump during stroke 2 of the small pump.

Thank you very much, I had misunderstood a large part of the content of the message, as I had anticipated. It is all clear now, and of mirror-like clarity to say the least.

 

[TSny]

Glad I could help.

 

[Hak]

Glad I could help.

Thanks. Here is what the person who proposed the problem told me:

"Nice solution with entropy. It is a bit strange, though, to identify as a 'system' the air that will eventually be inside the wheel. Some of this air is initially outside and it is not really identifiable a priori which particles will end up inside, and it is mixed on a microscopic level with other particles that are not part of the system. However, I don't think this invalidates the validity of the solution."

What do you think? What solutions or adjustments can we make?

Thanks again.

 

[TSny]

"Nice solution with entropy. It is a bit strange, though, to identify as a 'system' the air that will eventually be inside the wheel. Some of this air is initially outside and it is not really identifiable a priori which particles will end up inside, and it is mixed on a microscopic level with other particles that are not part of the system. However, I don't think this invalidates the validity of the solution."

I don't see a problem with identifying the system. Suppose we set up the system as shown below

We have 4 small pumps attached to the container. The container and the four pumps are initially filled with air at atmospheric pressure. Our thermodynamic system is the air in the container and the air in the four pumps. This is a well-defined system. If the handle of each pump is pushed down, one after the other, the work done on the system is clearly the same as the work $W_1$ that is asked for in the problem statement. [The pumps have a valve as explained in the problem statement that prevents any air in the container from flowing into a pump when the pressure in the container is greater than the pressure in the pump.]

Now start over but connect the pump handles together to make one handle so that all four small pumps will be compressed simultaneously when the handle is pushed down.

The four pumps now act as a single, larger pump of volume $4 V_P$. The work done for a single stroke of this larger pump is $W_2$. It should be clear that we are dealing with the same thermodynamic system as before in both pictures. [Edited for clarity.]

 

[Hak]

It should be clear that we are dealing with the same thermodynamic system as before.

Thank you very much for your reply. However, it is not clear to me: why is the last system you mentioned also as well defined as the first? Sorry if I still don't quite understand.

However, I still don't understand what the author's digression about the microscopic level, the presence of other particles, etc. has to do with it... What relevance does it have on the system and the discussion? That is, why are the objections raised by the author irrelevant? If they do have it, in light of your explanation (complete, as always), how would the author's objection I quoted above be resolved?

Thank you very much.

 

[TSny]

Thank you very much for your reply. However, it is not clear to me: why is the last system you mentioned also as well defined as the first? Sorry if I still don't quite understand.

However, I still don't understand what the author's digression about the microscopic level, the presence of other particles, etc. has to do with it... What relevance does it have on the system and the discussion? That is, why are the objections raised by the author irrelevant? If they do have it, in light of your explanation (complete, as always), how would the author's objection I quoted above be resolved?

I think that the author was concerned about taking the initial system to include the particular molecules of air in the atmosphere that will eventually get pumped into the container by strokes 2, 3, and 4 of the small pump. These molecules would be initially dispersed in the atmosphere and would not have a well-defined volume. So, the thermodynamic state of this system is problematic.

However, our goal is to compare $W_1$ with $W_2$. $W_1$ and $W_2$ in the scenario of post #114 are equal to $W_1$ and $W_2$ of the original statement of the problem. In post #114 I think the initial state of the system is well-defined.

 

[Hak]

I think that the author was concerned about taking the initial system to include the particular molecules of air in the atmosphere that will eventually get pumped into the container by strokes 2, 3, and 4 of the small pump. These molecules would be initially dispersed in the atmosphere and would not have a well-defined volume. So, the thermodynamic state of this system is problematic.

However, our goal is to compare $W_1$ with $W_2$. $W_1$ and $W_2$ in the scenario of post #114 are equal to $W_1$ and $W_2$ of the original statement of the problem. In post #114 I think the initial state of the system is well-defined.

Thank you.