Adiabatic Gas expansion

  • Thread starter hyddro
  • Start date
  • #1
74
2

Homework Statement


Hi, I did a lab experiment where I took a 5L vessel made of some material that isolates gas inside and thus behaves like an isolated system (adiabatic). I then pumped gas from ~90kPa to up to 150 kPa... recorded the temperature, then let the gas 'expand' by opening the piston and recorded the change in Temperature (usually of about 0.4 °C) with an about ~20-30 kPa change in internal pressure.

I have the following data.

T1 P1 T2 P2
Trial 1 27.1 142.5 26.7 95.7
Trial 2 28.1 143.5 27.7 83.9
trial 3 29.2 147.16 28.7 108.00


I need to calculate the amount of work done by this gas.


Homework Equations



ΔU = q + w

adiabatic

ΔU = w

w = pΔV

The Attempt at a Solution



I dont know how to begin... I think i approximated the volume to be constant (Despite some gas being released.

But that doesnt make sense because then no expansion, means no work....

How then do i calculate the work done adiabatically? The temperature dropped, and so the pressure... any help?
 

Answers and Replies

  • #2
74
2
I have come up with something.

Since w = p (V2 - V1) i am thinking....

w = p ( nRT2/P2 - nRT1/P1) ... assuming n = approximately constant...

would this work?

thanks
 
  • #4
74
2
Sorry. The vessel had a stopper that you could open and let the gas expand... That's what i meant..
 
  • #5
Matterwave
Science Advisor
Gold Member
3,966
327
If the expansion is free then no work is being done because the gas is not pushing on a piston or anything like that. If the expansion is adiabatic as well, the ideal gas law tells you that the temperature of the gas will remain constant since U=Q+W and Q=0 due to adiabaticity and W=0 due to free expansion, and an ideal gas's temperature is dependent only on U (U=3/2NkT). This is the adiabatic expansion of a free gas
 
  • #6
74
2
isnt the gas pushing against the atmosphere?
 
  • #7
Matterwave
Science Advisor
Gold Member
3,966
327
I thought that the gas expanded into a vacuum. You are just letting the gas out? In that case, it will quickly just come to equilibrium with the atmosphere...

Both processes are non-reversible, and non-quasistatic.
 
  • #8
21,488
4,864
Some questions:

Are those absolute pressures or gage pressures?

Is the gas allowed to expand freely against the atmosphere, or does it pass through some kind of constraint like a valve?

Why isn't the final pressure equal to atmospheric pressure?

Chet
 
  • #9
74
2
Absolute pressure.

The gas is allowed to expand against the atmosphere, no valve involved.

Because the stopper was opened for a split second and thus some of the compressed air pushed against atmosphere (left the vessel) while still some remained slightly compressed.



I think I didnt explain it correctly so let me do it once more.

The vessel had a wide opening which a rubber stopper (like a flask stopper but bigger and with holes) was placed in order to seal the vessel. A tube that pumps air was connected to the stopper (through one of the holes) into the vessel. Also, a temperature probe was placed through the stopper (through another hole) into the vessel.. Finally, a tube connected to the pressure reader was also connected to the stopper into the vessel. The sealed vessel then is filled with AIR until a certain arbitrary (usually close to 150 kPa) pressure was reached. My friend was in charge of holding the stopper in place and not allow it to pop out due to the high pressure (risky move hehehe). Then after the temperature increased and stabilized at some point, he let the stopper go for a split second (allowing the compressed air to expand against the atmosphere), and quickly closed it again. Not all the air left the vessel which is why the last reading is still higher than atmosphere.

Attaching an image.

hope this helps.

EDIT: I just noticed.. I dont have n !!! the number of moles of 'air' ... so i cant use that formula.... ARG!! i dont know what to do :(

EDIT2: What about using dw = Cv (dT)... we assume constant volume?
 

Attachments

  • IMG_20140904_171510.jpg
    IMG_20140904_171510.jpg
    34.3 KB · Views: 395
Last edited:
  • #10
21,488
4,864
So when your friend opened the stopper for a split second, air was escaping around the stopper. This is similar to a valve, although it was more open than most valves. If the air were released gradually, the expansion would have been reversible, but if it were released to the atmosphere, you would have had irreversible expansion against a constant pressure. Your situation was somewhere in between. Because the pressure change was relatively small, you will probably get the same answer for the temperature change doing it either way. So do it both ways and see what you get.

Also, you can get the number if moles in the container at the higher pressure by using the ideal gas law.

Chet
 
  • #11
21,488
4,864
Incidentally, your closed system is going to be the air that still remains I'm the flask after the stopper is closed. This air has undergone an expansion. To get the temperature change for this air, you don't need to know the number of moles.

Chet
 
  • #12
74
2
well, i know the temperature change because the probe was measuring the temperature inside of the vessel at all times... what is the difference between a reversible and irreversible expansion... sorry but i took p.chem long ago and i forgot already some stuff...

EDIT: Also, in a reversible expansion... the external pressure is equal to the internal pressure... why would I approach this experiment as being reversible when clearly the external pressure is ~100 kPa and the internal was 50kPA higher? (just read this on a book)
 
Last edited:
  • #13
74
2
Also, we did not intend to reach any specific final or initial pressure reading (due to the difficulty of the experiment itself, holding the stopper of the flask was very difficult to do and thus we needed to be quick). Due to this, our readings vary a lot and thus the degree of expansion is different and not necessarily the same... do I still calculate error? I mean, when we perform an experiment in which an error is calculated, we usually have fixed values we expect to obtain (for instance, burning 5 g of X chemical releases 5 cal of heat <- we expect to obtain this number using the same initial conditions in many different trials). In this case, we started from totally different pressures and ended at some different final pressure after releasing the stopper. It is not that we tried to start at say at 150kPA and by letting go the stopper we expected to reach 100kPa... like i said the difficulty of the experiment itself made it difficult to reach any specific target and thus for each trial I calculated each work done by the expansion without comparing to other trials since clearly they were performed at different initials and final pressures. do i still have to calculate the error? thank you

By the way we have 20 trials, I only included the first 3.
 
  • #14
21,488
4,864
well, i know the temperature change because the probe was measuring the temperature inside of the vessel at all times... what is the difference between a reversible and irreversible expansion... sorry but i took p.chem long ago and i forgot already some stuff...

EDIT: Also, in a reversible expansion... the external pressure is equal to the internal pressure... why would I approach this experiment as being reversible when clearly the external pressure is ~100 kPa and the internal was 50kPA higher? (just read this on a book)
You are using the term "external pressure" incorrectly. It is not the pressure external to the chamber. It is the pressure acting on the interface between your closed system and its surroundings. In this case, your closed system is the gas that eventually remains in the vessel after part of the original gas has escaped through the stopper. The external pressure on this closed system is the pressure of the gas that it is pushing ahead of itself (which constitutes its surroundings) out through the clearance between the cork and the wall. There is a pressure drop across the cork, and the pressure on the gas that will eventually remain in the chamber is higher than the pressure of the outside atmosphere throughout the entire expansion. In my judgement, this expansion will be close to reversible.
 
  • #15
21,488
4,864
Also, we did not intend to reach any specific final or initial pressure reading (due to the difficulty of the experiment itself, holding the stopper of the flask was very difficult to do and thus we needed to be quick). Due to this, our readings vary a lot and thus the degree of expansion is different and not necessarily the same... do I still calculate error? I mean, when we perform an experiment in which an error is calculated, we usually have fixed values we expect to obtain (for instance, burning 5 g of X chemical releases 5 cal of heat <- we expect to obtain this number using the same initial conditions in many different trials). In this case, we started from totally different pressures and ended at some different final pressure after releasing the stopper. It is not that we tried to start at say at 150kPA and by letting go the stopper we expected to reach 100kPa... like i said the difficulty of the experiment itself made it difficult to reach any specific target and thus for each trial I calculated each work done by the expansion without comparing to other trials since clearly they were performed at different initials and final pressures. do i still have to calculate the error? thank you

By the way we have 20 trials, I only included the first 3.
My understanding of what you are trying to do is to compare the observed temperature drops with theoretical estimates of the temperature drops that you make based on thermodynamic principles. Do you remember the equation for determining the temperature change for a gas undergoing a reversible adiabatic expansion from one pressure to an lower pressure? Look that up, and apply it to all 20 cases, and see how the results compare. Maybe it will match up pretty well with the observed temperature drops. If not, you can start to think about ways to improve the theoretical model.

Chet
 
  • #16
74
2
You are using the term "external pressure" incorrectly. It is not the pressure external to the chamber. It is the pressure acting on the interface between your closed system and its surroundings. In this case, your closed system is the gas that eventually remains in the vessel after part of the original gas has escaped through the stopper. The external pressure on this closed system is the pressure of the gas that it is pushing ahead of itself (which constitutes its surroundings) out through the clearance between the cork and the wall. There is a pressure drop across the cork, and the pressure on the gas that will eventually remain in the chamber is higher than the pressure of the outside atmosphere throughout the entire expansion. In my judgement, this expansion will be close to reversible.

I am having difficulties trying to understand the external pressure... You are saying that the pressure of the gas 'escaping' is now the external pressure of the gas that remains inside?

How then do I calculate the work done by this expansion?

w = nRT ln(v2/v1) ? but T isnt constant.. im very confused. thanks for your help though


Also, to answer your previous question...isnt it V2/V1 = (T2/T1)^gama ?
 
  • #17
21,488
4,864
I am having difficulties trying to understand the external pressure... You are saying that the pressure of the gas 'escaping' is now the external pressure of the gas that remains inside?

Yes. I'm saying that the pressure of the gas 'escaping' is the external pressure on the gas that will be remaining inside after the expansion is complete. Think of there being an imaginary boundary that, throughout the expansion, surrounds the gas that will eventually remain in the chamber after the expansion is complete.
How then do I calculate the work done by this expansion?

w = nRT ln(v2/v1) ? but T isnt constant.. im very confused. thanks for your help though
This is the equation for the reversible work in an isothermal process. Your process is not isothermal. It's adiabatic. For an adiabatic reversible expansion, you have:
[tex]dU = nC_vdT=PdV=nRT\frac{dV}{V}[/tex]
where n is the number of moles of gas that will be remaining in the cylinder after the expansion is complete. Note that n cancels out between the left hand side and the right hand side of the equation.

Also, to answer your previous question...isnt it V2/V1 = (T2/T1)^gama ?
No. Integrate the above equation, and tell us what you get.

Chet
 
  • #18
74
2
Thank you. That is indeed very difficult to visualize... i can see the difference between this and piston that just allows the gas to expand rather than expand and escape.

integrating gives me

ΔU = nCv*ΔT = nRTln(v2/v1)

is this correct?

then what? do i use the left side of the equation ? or the right side?
 
  • #19
21,488
4,864
Thank you. That is indeed very difficult to visualize... i can see the difference between this and piston that just allows the gas to expand rather than expand and escape.

integrating gives me

ΔU = nCv*ΔT = nRTln(v2/v1)

is this correct?

then what? do i use the left side of the equation ? or the right side?
Neither. You integrated incorrectly.

Try dividing both sides by nT first. Then, from the ideal gas law, ΔlnV=ΔlnT-ΔlnP, so combine this with the result of your integration (i.e., after you integrate correctly this time).

Also, I am again asking, "are those absolute pressures or gage pressures that you measured?"

Chet
 
  • #20
74
2
Neither. You integrated incorrectly.

Try dividing both sides by nT first. Then, from the ideal gas law, ΔlnV=ΔlnT-ΔlnP, so combine this with the result of your integration (i.e., after you integrate correctly this time).

Also, I am again asking, "are those absolute pressures or gage pressures that you measured?"

Chet

ΔU = Cv * ΔlnT = R * Δ(lnV) = R (ΔlnT - ΔlnP)

do I keep going?

Cv ΔlnT = R*ΔlnT - R*ΔlnP -> ΔlnT ( Cv - R) = - R*ΔlnP ?


still lost... sorry :(

(could you share a link where I can read about this topic, maybe i can save you some time if you could just direct me to the right topic)
 
  • #21
21,488
4,864
In post #17, I got the sign of the right hand side wrong. It should be a minus sign. Please try again with this change. You are now very close.

Incidentally, I'm guessing that those pressures you measured are gage pressures. If so, to get absolute pressures, you need to add the atmospheric pressure. The reversible thermo relationships are based on absolute pressure.

Chet
 
Last edited:
  • #22
74
2
so then,

R ΔlnP = ΔlnT (Cv + R) ?
 
  • #23
21,488
4,864
Yes. Now, do you happen to remember the relationship between Cp, Cv, and R (for an ideal gas), and do you remember the definition of the specific heat ratio γ? If so, please express your result in terms of γ.

Chet
 
  • #24
74
2
ok, so I got it down to

ΔlnT/ΔlnP = 1 - 1/γ

what do i do next? I am still puzzled as to how this is gonna help me find the work done by the expanding gas... :/

thanks though
 
  • #25
21,488
4,864
At this point, if your objective is to compare the predicted and observed temperature changes, there is no longer a need to calculate the total work. The energy balance, involving the differential changes in internal energy and differential work, was already taken into account earlier in deriving the equation from your previous post. From that equation, can you show that
[tex]\frac{T_2}{T_1}=\left(\frac{P_2}{P_1}\right)^{\frac{γ-1}{γ}},[/tex]
where the P's are absolute pressures?

Chet
 

Related Threads on Adiabatic Gas expansion

  • Last Post
2
Replies
29
Views
7K
Replies
1
Views
7K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
9K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
5
Views
18K
  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
3
Views
2K
Top