Hak
- 709
- 56
Thank you very much, I learnt a lot.TSny said:Yes. Very nice.
Are the graphs at post #86 correct?
Thank you very much, I learnt a lot.TSny said:Yes. Very nice.
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 said:Take a look at this link. Be sure to click on the additional information immediately under the equation of the first law.
Yes, they look correct.Hak said:Thank you very much, I learnt a lot.
Are the graphs at post #86 correct?
Thanks.TSny said:Yes, they look correct.
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.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?
TSny said: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.
Thank you so much. I understand now.TSny said: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.
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 said: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).
YesHak said: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,
I don't believe this is correct. Check it.Hak said:while all others are equal.
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 said:Yes
I don't believe this is correct. Check it.
Summarizing from post #57:Hak said: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.
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 said: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.
Thanks. Here is what the person who proposed the problem told me:TSny said:Glad I could help.
Hak said:"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."
TSny said:It should be clear that we are dealing with the same thermodynamic system as before.
Hak said: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.TSny said: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.