Adiabatic expansion work far exceeds isobaric of same volume, why?

In summary,Using the adiabatic process formula, I've calculated the change in volume for a volume of gas with an initial pressure of 10 psig expanding to 0 psig. The initial volume is 100 cubic inches and the expanded volume is 144.9. This is a difference of 44.9. The total work done ends up being about 96.2 joules.
  • #1
MysticDream
80
3
Using the adiabatic process formula, I've calculated the change in volume for a volume of gas with an initial pressure of 10 psig expanding to 0 psig. The initial volume is 100 cubic inches and the expanded volume is 144.9. This is a difference of 44.9. The total work done ends up being about 96.2 joules.

Now, if I use a simple pressure-area-volume formula to calculate work at the constant pressure of 10 psig and use that same displacement volume of 44.9, the total work done is about 50.7 joules. This is almost half the work done with the first formula and it's not even adiabatic expansion, but constant pressure (isobaric).

How is this possible?
 
Science news on Phys.org
  • #2
If you do not show your work, the most we can say is you made a mistake.
 
  • Like
Likes PhDeezNutz and Chestermiller
  • #3
Fair enough.

To calculate pressure change for increase in volume during adiabatic process:
(initial pressure * (initial volume^(7/5)))/(final volume^(7/5))
=
(24.7*(100^(7/5)))/(144.9^(7/5))
=
14.7 psi

To calculate work done after that process:
((initial volume * initial pressure)-(final volume * final pressure))/(7/5-1)
=
((100*0.0000163871*24.7*6894.75728)-(144.9*0.0000163871*14.7*6894.75728))/(7/5-1) ; I converted units here from cubic inches to cubic meters; and from psi to pascal
=
96.2 joules (I've verified this with two online calculators)

To calculate work (ft lb) done for isobaric expansion with the same volume change (against atmospheric pressure):
pressure * area * distance
=
(10*(44.9/12))
=
37.4 ft lbs

Convert ft lbs to joules:
37.4/0.7376
=
50.7 joules
 
  • #4
MysticDream said:
Fair enough.

To calculate pressure change for increase in volume during adiabatic process:
(initial pressure * (initial volume^(7/5)))/(final volume^(7/5))
=
(24.7*(100^(7/5)))/(144.9^(7/5))
=
14.7 psi

To calculate work done after that process:
((initial volume * initial pressure)-(final volume * final pressure))/(7/5-1)
=
((100*0.0000163871*24.7*6894.75728)-(144.9*0.0000163871*14.7*6894.75728))/(7/5-1) ; I converted units here from cubic inches to cubic meters; and from psi to pascal
=
96.2 joules (I've verified this with two online calculators)

To calculate work (ft lb) done for isobaric expansion with the same volume change (against atmospheric pressure):
pressure * area * distance
=
(10*(44.9/12))
=
37.4 ft lbs

Convert ft lbs to joules:
37.4/0.7376
=
50.7 joules
You used the gauge pressure rather than the absolute pressure in calculating the isobaric work.
 
  • Like
Likes Bystander
  • #5
Yes, but what matters is the pressure difference, does it not? A pressure of 10 psig working against atmospheric pressure is the same as 24.7 psia working against 14.7 psia.
 
  • #6
It didn‘t matter in the first calculation. I also do not see any exponents in it.
 
Last edited:
  • #7
For future reference, it is generally preferable if you give the problem statement and perform the calculation symbolically before plugging in the numbers. PF uses latex to help format equations. https://www.physicsforums.com/help/latexhelp/
 
  • #8
MysticDream said:
Yes, but what matters is the pressure difference, does it not? A pressure of 10 psig working against atmospheric pressure is the same as 24.7 psia working against 14.7 psia.
No. The PV work is always ##\int{P_{ext}dV}##, where ##P_{ext}## Is the absolute external pressure. In the first calculation you did, you considered an adiabatic reversible expansion, starting with an external pressure 0f 10 psi gauge and lowering the external pressure gradually to 0 psi gauge. The equation you used to get to work was specifically for this case.

For the adiabatic irreversible isobaric expansion you were considering in the second case, I believe you intended to start out at 10 psi gauge again, but in this case to suddenly drop the external pressure to zero psi gauge and to hold that external pressure constant until the gas had expanded by the same volume as in the first case. So ##\Delta V=44.9\ in^3=0.000736\ m^3## and ##P_{ext}=14.7\ psi=101.325\ kPa##, and the work is W = 74.6 J.
 
  • #9
Frabjous said:
For future reference, it is generally preferable if you give the problem statement and perform the calculation symbolically before plugging in the numbers. PF uses latex to help format equations. https://www.physicsforums.com/help/latexhelp/

Thanks for the tip. Yes, the exponents above were after the "^" which is how you write it in excel.
 
  • #10
Chestermiller said:
No. The PV work is always ##\int{P_{ext}dV}##, where ##P_{ext}## Is the absolute external pressure. In the first calculation you did, you considered an adiabatic reversible expansion, starting with an external pressure 0f 10 psi gauge and lowering the external pressure gradually to 0 psi gauge. The equation you used to get to work was specifically for this case.

For the adiabatic irreversible isobaric expansion you were considering in the second case, I believe you intended to start out at 10 psi gauge again, but in this case to suddenly drop the external pressure to zero psi gauge and to hold that external pressure constant until the gas had expanded by the same volume as in the first case. So ##\Delta V=44.9\ in^3=0.000736\ m^3## and ##P_{ext}=14.7\ psi=101.325\ kPa##, and the work is W = 74.6 J.
Sorry for the misunderstanding, but that is not the first calculation I was intending to do. I was expanding a 10 psig volume of gas adiabatically against an external pressure of 0 psig. In both cases the external pressure was 0 psig.

This could be visualized as a cylinder with a piston in it. Both cylinders and pistons are of the same exact size, will travel the same volume while doing the work, and have the same starting pressure and external pressure. The only difference is that one cylinder will expand adiabatically to 0 psig, and the other will maintain pressure.

Using the formulas I'm familiar with, I get the opposite of what I would expect. The work done with the adiabatic expansion ends up being more than the work done with constant pressure. The work done during adiabatic expansion drops off in pressure as it expands and cools whereas the work done during constant pressure does not drop off but both examples are the exact same volume of displacement. Logically, the constant pressure example should be more work done than the adiabatic expansion example since it doesn't lose pressure during the work.
 
  • #11
Starting volume = 100 cubic in
Final volume = 144.9 cubic in
Starting pressure = 10 psig (or 24.7 psi)
External pressure = 0 psig (or 14.7 psi)
Adiabatic expansion formula to calculate final pressure:
##24.7*(100^{7/5})/(144.9^{7/5})=14.7## psi

Adiabatic expansion formula to calculate work done:
##((100*0.0000163871*24.7*6894.76)-(144.9*0.0000163871*14.7*6894.76))/(7/5-1)##
=96.2 joules
; I converted units here from cubic inches to cubic meters; and from psi to pascal. Again, these solutions were verified with two online calculators.

Constant pressure example:
Starting volume = 100 cubic in (same as above)
Final volume = 144.9 cubic in (same as above)
Starting pressure = 10 psig (same as above)
External pressure = 0 psig (same as above)
To calculate work done using pressure*volume formula:
##10*44.9 = 449## in lbs
Convert in lbs to joules:
##449*0.112984=50.7## joules
 
Last edited:
  • #12
MysticDream said:
Sorry for the misunderstanding, but that is not the first calculation I was intending to do. I was expanding a 10 psig volume of gas adiabatically against an external pressure of 0 psig. In both cases the external pressure was 0 psig.

This could be visualized as a cylinder with a piston in it. Both cylinders and pistons are of the same exact size, will travel the same volume while doing the work, and have the same starting pressure and external pressure. The only difference is that one cylinder will expand adiabatically to 0 psig, and the other will maintain pressure.

Using the formulas I'm familiar with, I get the opposite of what I would expect. The work done with the adiabatic expansion ends up being more than the work done with constant pressure. The work done during adiabatic expansion drops off in pressure as it expands and cools whereas the work done during constant pressure does not drop off but both examples are the exact same volume of displacement. Logically, the constant pressure example should be more work done than the adiabatic expansion example since it doesn't lose pressure during the work.
The final volumes are not the same if the final pressures are the same.
 
  • #13
MysticDream said:
Starting volume = 100 cubic in
Final volume = 144.9 cubic in
Starting pressure = 10 psig (or 24.7 psi)
External pressure = 0 psig (or 14.7 psi)
Adiabatic expansion formula to calculate final pressure:
##24.7*(100^{7/5})/(144.9^{7/5})=14.7## psi

Adiabatic expansion formula to calculate work done:
##((100*0.0000163871*24.7*6894.76)-(144.9*0.0000163871*14.7*6894.76))/(7/5-1)##
=96.2 joules
; I converted units here from cubic inches to cubic meters; and from psi to pascal. Again, these solutions were verified with two online calculators.

Constant pressure example:
Starting volume = 100 cubic in (same as above)
Final volume = 144.9 cubic in (same as above)
Starting pressure = 10 psig (same as above)
External pressure = 0 psig (same as above)
To calculate work done using pressure*volume formula:
##10*44.9 = 449## in lbs
You should be using the absolute pressure, not the gauge pressure, and the constantl pressure should be 0 psig (14.7 psi), not 10 psig. Plus, the final volume will not be 144.9cubic inches, it will be greater than this..
MysticDream said:
Convert in lbs to joules:
##449*0.112984=50.7## joules
Constant pressure example for adiabatic irreversible case:
$$\Delta U=nC_v(T_f-T_i)=-P_{f}(V_f-V_i)=-P_f\left(\frac{nRT_f}{P_f}-\frac{nRT_i}{P_i}\right)$$or
$$C_v(T_f-T_i)=R\left(T_f-\frac{P_i}{P_f}T_i\right)$$Solving for the final temperature gives:$$\frac{T_f}{T_i}=\frac{C_v+R\frac{P_f}{P_I}}{C_v+R}=\frac{1+(\gamma-1)\frac{P_f}{P_i}}{\gamma}=\frac{1+0.4\frac{14.7}{24.7}}{1.4}=0.8843$$So, from the ideal gas law, $$\frac{V_f}{V_i}=\frac{T_f}{T_i}\frac{P_i}{P_f}=0.8843\frac{24.7}{14.7}=1.4859$$So the final volume is 148.6 cubic in and the volume change is 48.6 cubic in, not 44.9 cubic in. So the adiabatic irreversible work is $$W=(14.7)(48.6)(1.35882)/12=80.9\ J$$
 
  • #14
The final pressures are not the same. The final volumes are the same. That's the whole purpose of the comparison. One expands to 0 psig adiabatically and the other does not expand but pushes on the piston at constant pressure to the same volume.

I don't understand why you'd need to use an adiabatic formula for constant pressure. A constant pressure formula is simply volume * pressure, is it not? If my volume is in cubic inches and my pressure is in psig, then I simply multiply the two to get work in in-lbf. I can then convert to joules which is the work units I want to end up with.

Perhaps my formula to calculate adiabatic expansion is wrong, but I did verify it with two online calculators, one of them being here:
https://www.omnicalculator.com/physics/thermodynamic-processes

I do appreciate any help
 
  • #15
MysticDream said:
The final pressures are not the same. The final volumes are the same. That's the whole purpose of the comparison. One expands to 0 psig adiabatically and the other does not expand but pushes on the piston at constant pressure to the same volume.

I don't understand why you'd need to use an adiabatic formula for constant pressure. A constant pressure formula is simply volume * pressure, is it not? If my volume is in cubic inches and my pressure is in psig, then I simply multiply the two to get work in in-lbf. I can then convert to joules which is the work units I want to end up with.

Perhaps my formula to calculate adiabatic expansion is wrong, but I did verify it with two online calculators, one of them being here:
https://www.omnicalculator.com/physics/thermodynamic-processes

I do appreciate any help
Your original calculation was for an adiabatic reversible expansion. This is where you control the motion of the piston so that the gas expands gradually and does maximum work. This calculation was done correctly.

Your second calculation was for an adiabatic irreversible expansion. This is where you drop the external pressure suddenly and allow the gas to expand against the piston in an uncontrolled manner. There are two possible end states for this expansion, one in which you allow the gas to expand until it equilibrates with the external surroundings, and the other where you stop the expansion when the volume change is exactly the same as for the reversible case. In the first scenario, the final pressures willl be the same, but the final volumes will be different, and iii the second scenario, the final volumes will be the same, but the final pressures will not. It is not possible to expand the gas adiabatically and irreversibly and to reach the final end state as for adiabatic reversible expansion. Something how got to give.

If you disagree with what I have been saying, you are incorrect. As one of the resident experts on thermodynamics in Physics Forums, I can assure you that I am the one who is correct. I hate to pull rank like this, but if you continue to suggest that you are correct and I am not (without any effort to search for why you are mistaken), I will have to close this thread.
 
  • Like
Likes weirdoguy
  • #16
Yeah, you'd be correct in your own interpretation of what I described but you've misunderstood what I described.

The second calculation was for an isobaric process which is as simple as P * V = W. The change in volume was kept the same as the first calculation for the adiabatic process. These are different amounts of gas. Every source I've read agrees with this including wiki: https://en.wikipedia.org/wiki/Isobaric_process.

If you want to close the thread, that's up to you.
 
  • #17
Chestermiller said:
(without any effort to search for why you are mistaken)
That's quite an assumption. Thanks anyway, though.
 
  • #18
MysticDream said:
Fair enough.

To calculate pressure change for increase in volume during adiabatic process:
(initial pressure * (initial volume^(7/5)))/(final volume^(7/5))
=
(24.7*(100^(7/5)))/(144.9^(7/5))
=
14.7 psi

To calculate work done after that process:
((initial volume * initial pressure)-(final volume * final pressure))/(7/5-1)
=
((100*0.0000163871*24.7*6894.75728)-(144.9*0.0000163871*14.7*6894.75728))/(7/5-1) ; I converted units here from cubic inches to cubic meters; and from psi to pascal
=
96.2 joules (I've verified this with two online calculators)

To calculate work (ft lb) done for isobaric process with the same volume change (against atmospheric pressure):
pressure * area * distance
=
(10*(44.9/12))
=
37.4 ft lbs

Convert ft lbs to joules:
37.4/0.7376
=
50.7 joules
 
  • #19
Chestermiller said:
For the adiabatic irreversible isobaric expansion you were considering in the second case, I believe you intended to start out at 10 psi gauge again, but in this case to suddenly drop the external pressure to zero psi gauge and to hold that external pressure constant until the gas had expanded by the same volume as in the first case.
I think you believe wrong. The OP never mentioned the isobaric case was adiabatic.

His second case is that the exterior pressure is constant at 10 psig and the volume increases.

What the OP doesn't realize is that, in the first case, energy is released from the fluid to produce work. In the second case, energy must be added to the fluid while producing work.

?u=https%3A%2F%2Ftse1.mm.bing.net%2Fth%3Fid%3DOIP.jpg

From the graph above, the work (i.e. the area under the curves) between the isobaric and adiabatic cases is very different.

What you should also notice from the above graph is that the final temperatures are also much different. The isothermal case gives you an indication of the direction of constant temperature. The adiabatic case must drop in temperature but the isobaric case must increase - a lot - in temperature. This means that energy must be added to the system somehow.

This is essentially the difference between the Otto cycle and the Diesel cycle: The former adds all the energy (combustion) before it expands and the latter adds the energy while it expands. But for the same amount of energy addition, the initial conditions for the expansion are much much different.
 
  • #20
jack action said:
I think you believe wrong. The OP never mentioned the isobaric case was adiabatic.

His second case is that the exterior pressure is constant at 10 psig and the volume increases.

What the OP doesn't realize is that, in the first case, energy is released from the fluid to produce work. In the second case, energy must be added to the fluid while producing work.


From the graph above, the work (i.e. the area under the curves) between the isobaric and adiabatic cases is very different.

What you should also notice from the above graph is that the final temperatures are also much different. The isothermal case gives you an indication of the direction of constant temperature. The adiabatic case must drop in temperature but the isobaric case must increase - a lot - in temperature. This means that energy must be added to the system somehow.

This is essentially the difference between the Otto cycle and the Diesel cycle: The former adds all the energy (combustion) before it expands and the latter adds the energy while it expands. But for the same amount of energy addition, the initial conditions for the expansion are much much different.
Thanks, and yes I understand what you are describing. Heat energy would have to be added during an isobaric process.

This whole problem is about the work calculation.

The problem seems to be with my formulas or my misunderstanding of the results. I'm trying to calculate the work done for both processes with the same displacement volume, but the work done for the adiabatic process is more according to these formulas. The work done for the isobaric process should be more.
 
  • #21
Mystic Dream said "
To calculate work (ft lb) done for isobaric process with the same volume change (against atmospheric pressure):
pressure * area * distance
=
(10*(44.9/12))
=
37.4 ft lbs

Convert ft lbs to joules:
37.4/0.7376
=
50.7 joules"

This calculation is not done correctly. To get the work done by a gas, you need to use absolute pressure, not gauge pressure. 10 psig =24.7 psi. So, for isobaric expansion at 10 psig with a volume change of 44.9 in^3, the work is 50.7 x 24.7 / 10 = 125.2 Joules
 
  • #22
MysticDream said:
Yes, but what matters is the pressure difference, does it not? A pressure of 10 psig working against atmospheric pressure is the same as 24.7 psia working against 14.7 psia.
Where do you get your 14.7 psia in the isobaric case? There is not even a pressure difference in the isobaric case.

In both cases, the area under the curve is bound at the bottom by ##p = 0##.

Redoing your previous calculations:
MysticDream said:
Constant pressure example:
Starting volume = 100 cubic in (same as above)
Final volume = 144.9 cubic in (same as above)
Starting pressure = 10 psig (same as above)
External pressure = 0 psig (same as above)
To calculate work done using pressure*volume formula:
##10*44.9 = 449## in lbs
Convert in lbs to joules:
##449*0.112984=50.7## joules
Constant pressure example:
Starting volume = 100 cubic in (same as above)
Final volume = 144.9 cubic in (same as above)
Starting pressure = 10 psig (same as above)
External pressure = 0 psig (same as above)
To calculate work done using pressure*volume formula:
##24.7*44.9 = 449## in lbs
Convert in lbs to joules:
##449*0.112984=125.3## joules
 
  • #23
Well, in both cases I had the volume expanding against atmospheric pressure. For the isobaric case, I just took the pressure difference, which ends up being 10 psi and multiplied that by the displacement volume in cubic inches which gives me work in in-lbf (or divide by 12 to get ft-lbf).

In the adiabatic case I just used a starting pressure of 10 psig (24.7 psia) and expanded that against 0 psig (14.7 psia). I understand I had to use absolute pressure in this formula and this may be where the misunderstanding is. Perhaps I was not getting as much work out of the adiabatic process as I thought and I'm misinterpreting the results. I want to know what the work would be expanding this displacement volume against atmospheric pressure.

Again, both cases are the same displacement volume.
 
  • #24
MysticDream said:
Well, in both cases I had the volume expanding against atmospheric pressure. For the isobaric case, I just took the pressure difference, which ends up being 10 psi and multiplied that by the displacement volume in cubic inches which gives me work in in-lbf (or divide by 12 to get ft-lbf).
MysticDream said:
That is the net work done on the piston, not the work done by the gas on the piston.
 
  • #25
MysticDream said:
Well, in both cases I had the volume expanding against atmospheric pressure. For the isobaric case, I just took the pressure difference, which ends up being 10 psi and multiplied that by the displacement volume in cubic inches which gives me work in in-lbf (or divide by 12 to get ft-lbf).

In the adiabatic case I just used a starting pressure of 10 psig (24.7 psia) and expanded that against 0 psig (14.7 psia). I understand I had to use absolute pressure in this formula and this may be where the misunderstanding is. Perhaps I was not getting as much work out of the adiabatic process as I thought and I'm misinterpreting the results. I want to know what the work would be expanding this displacement volume against atmospheric pressure.

Again, both cases are the same displacement volume.
Now what you are saying is that you are considering the constant [atmospheric] pressure on the other side of the piston. You basically want to exclude the work you have done on the atmosphere (which is compressed at the same time your fluid expands).

It is also an isobaric process, with the same volume change, but this one does work on the fluid because it pushes in the opposite direction on the piston. So you are correct in considering the pressure differential in the second case as both are isobaric processes. But in the adiabatic case, you have only considered the expansion on the inside of the cylinder from ##p## to ##p_{atm}##, but did not consider the isobaric compression at ##p_{atm}## on the other side.

So modifying your calculations again from that point of view:
Starting volume = 100 cubic in
Final volume = 144.9 cubic in
Starting pressure = 10 psig (or 24.7 psi)
External pressure = 0 psig (or 14.7 psi)
Adiabatic expansion formula to calculate final pressure:
##24.7*(100^{7/5})/(144.9^{7/5})=14.7## psi

Adiabatic expansion formula to calculate work done:
##((100*0.0000163871*24.7*6894.76)-(144.9*0.0000163871*14.7*6894.76))/(7/5-1)##
##- (44.9*0.0000163871*14.7*6894.76)##
= 21.5 joules
; I converted units here from cubic inches to cubic meters; and from psi to pascal. Again, these solutions were verified with two online calculators.

Constant pressure example:
Starting volume = 100 cubic in (same as above)
Final volume = 144.9 cubic in (same as above)
Starting pressure = 10 psig (same as above)
External pressure = 0 psig (same as above)
To calculate work done using pressure*volume formula:
##10*44.9 = 449## in lbs
Convert in lbs to joules:
##449*0.112984=50.7## joules
This is like having a complete cycle where the piston goes down the cylinder with the expansion and then goes up with the isobaric process with ##p=p_{atm}## in both cases.

One way or the other, the net isobaric work is greater than the adiabatic one.
 
  • #26
jack action said:
Now what you are saying is that you are considering the constant [atmospheric] pressure on the other side of the piston. You basically want to exclude the work you have done on the atmosphere (which is compressed at the same time your fluid expands).

It is also an isobaric process, with the same volume change, but this one does work on the fluid because it pushes in the opposite direction on the piston. So you are correct in considering the pressure differential in the second case as both are isobaric processes. But in the adiabatic case, you have only considered the expansion on the inside of the cylinder from ##p## to ##p_{atm}##, but did not consider the isobaric compression at ##p_{atm}## on the other side.

So modifying your calculations again from that point of view:

This is like having a complete cycle where the piston goes down the cylinder with the expansion and then goes up with the isobaric process with ##p=p_{atm}## in both cases.

One way or the other, the net isobaric work is greater than the adiabatic one.
I don’t even understand this. Let’s see your analysis for either of the irreversible cases I specified
 
  • #27
Chestermiller said:
I don’t even understand this. Let’s see your analysis for either of the irreversible cases I specified
That is because it goes above thermodynamics. It considers the pressure differential across the piston and the net work on the piston.

From what I understand from @MysticDream , "it's the pressure difference that matters". I think he's referring to the pressure difference across the piston. That is the only way I can explain why he's bringing the atmospheric pressure into the isobaric process.

How else can we explain such a contradictory statement:
MysticDream said:
For the isobaric case, I just took the pressure difference, which ends up being 10 psi
Where's the pressure difference in an isobaric process?
 
  • #28
jack action said:
That is because it goes above thermodynamics. It considers the pressure differential across the piston and the net work on the piston.

From what I understand from @MysticDream , "it's the pressure difference that matters". I think he's referring to the pressure difference across the piston. That is the only way I can explain why he's bringing the atmospheric pressure into the isobaric process.

How else can we explain such a contradictory statement:

Where's the pressure difference in an isobaric process?
It is not the pressure difference across the piston that determines work done by the gas inside the cylinder on its surroundings. It is the absolute force per unit area multiplied by the volume charge , represented by Pext dV.
 
Last edited:
  • Like
Likes jack action
  • #29
Aha, very interesting. It seems I've really been after the net work done on the piston in both cases and did not understand what "work" implied when using the adiabatic formula. I suppose I have to subtract the work done on the atmosphere which stays at a constant pressure to obtain the "net work" done on the piston. The terminology gets confusing when you say the external atmosphere gets "compressed", as it doesn't change pressure.

This is what I really need to understand, the difference between net work and the work done by the gas, because I thought they were one in the same.

Chestermiller says "It is not the pressure difference across the piston that determines work done by the gas inside the cylinder on its surroundings. It is the absolute force per unit area multiplied by the volume charge , represented by Pext dV."

To me work = force * distance, so why wouldn't work be the pressure difference? The external pressure is a force opposing the internal pressure force.

If the net work and gas work are different, what happens to the work done by the gas that is not used? Is it just wasted and this is how you calculate efficiency?

Thank you both for your insight.
 
  • #30
MysticDream said:
Aha, very interesting. It seems I've really been after the net work done on the piston in both cases and did not understand what "work" implied when using the adiabatic formula. I suppose I have to subtract the work done on the atmosphere which stays at a constant pressure to obtain the "net work" done on the piston. The terminology gets confusing when you say the external atmosphere gets "compressed", as it doesn't change pressure.

This is what I really need to understand, the difference between net work and the work done by the gas, because I thought they were one in the same.

Chestermiller says "It is not the pressure difference across the piston that determines work done by the gas inside the cylinder on its surroundings. It is the absolute force per unit area multiplied by the volume charge , represented by Pext dV."

To me work = force * distance, so why wouldn't work be the pressure difference? The external pressure is a force opposing the internal pressure force.

If the net work and gas work are different, what happens to the work done by the gas that is not used? Is it just wasted and this is how you calculate efficiency?

Thank you both for your insight.
If the piston is massless and frictionless, then by Newton's 2nd law, the force per unit area exerted by the external atmosphere on the outside face of the piston ##P_{ext}## must be equal in magnitude to the force per unit area exerted by the gas on the inside face of the piston. That means that the work done by the internal gas on the inside face of the piston must be $$dW=P_{ext}dV$$
 
  • Like
Likes MysticDream
  • #31
Chestermiller said:
If the piston is massless and frictionless, then by Newton's 2nd law, the force per unit area exerted by the external atmosphere on the outside face of the piston ##P_{ext}## must be equal in magnitude to the force per unit area exerted by the gas on the inside face of the piston. That means that the work done by the internal gas on the inside face of the piston must be $$dW=P_{ext}dV$$
Could you give me the formula for the net work done on the piston in the adiabatic case?

If I'm not mistaken, you seem to be in agreement with jack action. I'd just like to clarify.
 
  • #32
MysticDream said:
Could you give me the formula for the net work done on the piston in the adiabatic case?

If I'm not mistaken, you seem to be in agreement with jack action. I'd just like to clarify.
In the adiabatic reversible case, the gas within the cylinder is only slightly removed from thermodynamic equilibrium along the entire process path, so the force per unit area on the inside face of the piston is described in this situation by the ideal gas law: $$dW=P_{ext}dV=\frac{nRT}{V}dV$$Note that this means that the external pressure ##P_{ext}## must be forced to change as a function of volume in the adiabatic reversible case.
 
  • #33
Chestermiller said:
In the adiabatic reversible case, the gas within the cylinder is only slightly removed from thermodynamic equilibrium along the entire process path, so the force per unit area on the inside face of the piston is described in this situation by the ideal gas law: $$dW=P_{ext}dV=\frac{nRT}{V}dV$$Note that this means that the external pressure ##P_{ext}## must be forced to change as a function of volume in the adiabatic reversible case.
Thanks.

So the net work done on the piston only (not the atmosphere) for the adiabatic process, would be:
$$W=(P_{1}*V_{1}-P_{2}*V_{2})/(7/5-1)-P_{ext}dV$$
 
Last edited:
  • #34
MysticDream said:
Thanks.

So the net work done on the piston only (not the atmosphere) for the adiabatic process, would be:
If it is a massless frictionless piston, in this adiabatic reversible version of the process, the net work done on the piston is zero. $$Work\ done\ by\ gas\ in\ cylinder\ on\ piston=\frac{(P_1V_1-P_2V_2)}{\gamma-1}$$
$$Work\ done\ by\ externally\ controlled\ pressure\ on\ piston = -\int{P_{ext}(V)dV}=-\frac{(P_1V_1-P_2V_2)}{\gamma-1}$$For a massless frictionless piston, the net work done on the piston is always zero, throughout the process.
 
  • #35
Chestermiller said:
If it is a massless frictionless piston, in this adiabatic reversible version of the process, the net work done on the piston is zero. $$Work\ done\ by\ gas\ in\ cylinder\ on\ piston=\frac{(P_1V_1-P_2V_2)}{\gamma-1}$$
$$Work\ done\ by\ externally\ controlled\ pressure\ on\ piston = -\int{P_{ext}(V)dV}=-\frac{(P_1V_1-P_2V_2)}{\gamma-1}$$For a massless frictionless piston, the net work done on the piston is always zero, throughout the process.

Ah yes, but if the piston had any mass at all, that formula would hold true, correct?
 

Similar threads

  • Thermodynamics
Replies
8
Views
657
Replies
22
Views
2K
Replies
56
Views
3K
Replies
1
Views
734
Replies
3
Views
1K
Replies
12
Views
1K
Replies
5
Views
2K
Replies
4
Views
992
  • Thermodynamics
Replies
11
Views
5K
Replies
3
Views
1K
Back
Top