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Thermal question: Reversible cycle

  • #1
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Homework Statement




Hi all

I'm a little confused about work in gases. I thought I knew it now I'm not so sure. Basically I'm answering a question about work done on the gas in a reversible cycle where a change in temperature (T1 -T2)at constant volume (V1) is followed by a change in volume at constant temperature (V1-V2), and back again (T2V2-T1V2 then T1V2-T1V1).

I've gone through the maths numerous times and the only way I can think to make my answer match the given one is two have the cycle begin with cooling (T2<T1).

However, there is nothing in the problem set up that indicates it should be in that direction. Therefore starting the cycle with heating (T1>T2) seems equally valid, however the resulting answer is the opposite sign of the given one.

Am I missing some fact that would tell me why it should be cooling. There's no mention of the process being adiabatic as well as reversible so I can't immediately assume Joule cooling.

Homework Equations



For a reversible process Work done on the gas:

[tex] dW = - P.dV[/tex]

The Attempt at a Solution



For net work done if (T1<T2) [tex] W = (T_{1}-T_{2})ln(\frac{V_{2}}{V_{1}})[/tex]

Given answer is [tex] W = (T_{2}-T_{1})ln(\frac{V_{2}}{V_{1}})[/tex]

Which implies T2<T2 as I read it.
 

Answers and Replies

  • #2
Andrew Mason
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I'm a little confused about work in gases. I thought I knew it now I'm not so sure. Basically I'm answering a question about work done on the gas in a reversible cycle where a change in temperature (T1 -T2)at constant volume (V1) is followed by a change in volume at constant temperature (V1-V2), and back again (T2V2-T1V2 then T1V2-T1V1).
This cannot be a reversible process. The entropy change of the system + surroundings will always be greater than 0.
For net work done if (T1<T2) [tex] W = (T_{1}-T_{2})ln(\frac{V_{2}}{V_{1}})[/tex]

Given answer is [tex] W = (T_{2}-T_{1})ln(\frac{V_{2}}{V_{1}})[/tex]

Which implies T2<T1 as I read it.
Plot the cycle on a PV diagram. P = nRT/V. When T is constant, P is inversely proportional to V. When V is constant, P is proportional to T.

The work done by the gas is the area inside the cycle path. That area is the area under the higher P = nRT2/V curve LESS the area under the lower curve P = nRT1/V.

AM
 
  • #3
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This cannot be a reversible process. The entropy change of the system + surroundings will always be greater than 0.
Plot the cycle on a PV diagram. P = nRT/V. When T is constant, P is inversely proportional to V. When V is constant, P is proportional to T.

The work done by the gas is the area inside the cycle path. That area is the area under the higher P = nRT2/V curve LESS the area under the lower curve P = nRT1/V.
AM
The problem sheet this comes from is limited to discussion of the first law. Entropy has not been introduced in the text yet so I'm not sure I should be taking it into account.

I solved the problem by integrating each constant temperature curve, which is equivalent to taking the area under the curve, and subtracting.

I could see how that could give the right answer except that I'm considering the work done on the gas. Which is [tex] dW = -P.dV[/tex] and its resolving that negative which seems to swap my Temperature values in the solution. However, I can't see any physical reason to ignore that negative.
 
  • #4
LeonhardEuler
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The problem sheet this comes from is limited to discussion of the first law. Entropy has not been introduced in the text yet so I'm not sure I should be taking it into account.

I solved the problem by integrating each constant temperature curve, which is equivalent to taking the area under the curve, and subtracting.

I could see how that could give the right answer except that I'm considering the work done on the gas. Which is [tex] dW = -P.dV[/tex] and its resolving that negative which seems to swap my Temperature values in the solution. However, I can't see any physical reason to ignore that negative.
The only mistake I see that you've made is losing some factors from the ideal gas law (PV=nRT). If the net direction of work is from the system to the surroundings, then the work done on the gas is negative. There's nothing wrong with that. Also, the work done on the gas will be positive if T2>T1 and V1>V2.

As far as this being a reversible process, there is nothing about this process that would prevent it from being as reversible as, say, a Carnot cycle. That is, an idealization that can be approached. For this to be reversible, the heating and cooling would have to be done over an infinitesimal temperature gradient, which would obviously take forever, but could be approached in reality, for instance by putting this gas in thermal contact with another gas undergoing expansion or compression reversibly, and letting the temperature equilibrate by going very slowly. The entropy change can approach 0.
 
  • #5
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The only mistake I see that you've made is losing some factors from the ideal gas law (PV=nRT). If the net direction of work is from the system to the surroundings, then the work done on the gas is negative. There's nothing wrong with that. Also, the work done on the gas will be positive if T2>T1 and V1>V2.
What if I say V1<V2, then the work is only positive if T1<T2, correct?

Again there is no indication of if V1 is smaller or larger than V2. However the previous question worked best if V2>V1. Also, I expected that if the temperature increased from T1-T2 in the first step, the next step would be expansion of the gas.

However, the given answer seems to imply cooling from T1-T2 (where T1>T2) followed by expansion. Which makes less sense to me.
 
  • #6
LeonhardEuler
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The question doesn't imply that because it never said the work done on the gas was positive.
 
  • #7
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The question doesn't imply that because it never said the work done on the gas was positive.
Maybe if I show my work in more detail my error might stand out more.

So, taking each step of the four step cycle separately, we can see that the increase/decrease of temperature (steps 1 and 3) at constant volume does not do and work because dV=0.

That leaves the two processes of expansion (step 2) at T2 and compression (step 4) at T1. Taking T2<T1 (i.e. in step 1 the gas increases in temp from T1V1 to T2V1, then expands in step 2 to T2V2.)

For expansion at T2.

[tex] W_{23} =-\int^{2}_{1}P.dV =-\int^{V_{2}}_{V_{1}}\frac{RT_{2}}{V}.dV = -RT_{2}ln\left(\frac{V_{2}}{V_{1}}\right) [/tex]

For compression at T1.

[tex] W_{41} =-\int^{4}_{1}P.dV =-\int^{V_{1}}_{V_{2}}\frac{RT_{1}}{V}.dV =\int^{V_{2}}_{V_{1}}\frac{RT_{1}}{V}.dV = RT_{1}ln\left(\frac{V_{2}}{V_{1}}\right)[/tex]

Total work done on gas.

[tex] W = W_{23}+W_{41}= R(T_{1}-T_{2})ln\left(\frac{V_{2}}{V_{1}}\right)[/tex]


The given answer in the book is:

[tex] W = R(T_{2}-T_{1})ln\left(\frac{V_{2}}{V_{1}}\right)[/tex]

Which I can only replicate if I either say the work done on the gas is positive or if I've botched my calculation somewhere and am missing it.
 
  • #8
LeonhardEuler
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...Taking T2<T1 (i.e. in step 1 the gas increases in temp from T1V1 to T2V1, then expands in step 2 to T2V2.)...
This assumption has no bearing on anything else you calculate. If you assumed T2>T1 you would have gotten the same answer.
Either the answer given is wrong, or the exact wording of the question differs from your interpretation of it. Could you give the exact wording?
 
  • #9
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Okay.

The first question had a similar cycle involving changes from P1-P2 at constant volume and isobaric changes from V1-V2. It wanted the work done on the gas. Which I found with little difficulty (my one hang up was to realise that for the gas expansion to expand the P1>P2).

The second part simply states.

The same as the previous problem with the four stages of the cycle (i)at constant volume from T1V1 to T2V2. (ii)isothermally to T2V2. (iii) At constant volume from T2V2 to T1V2. (iv) Isothermally back to T1V1.
 
  • #10
LeonhardEuler
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Ok, it looks like the answer they gave is wrong, but to be sure, how was the first question worded?
 
  • #11
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One mole of perfect gas performs a quasistatic cycle which consists of the following four succesive stages: i)From the state P1V1 at constant pressure to the state P1V2 ii) at constant volume to the state P2V2 iii) At constant pressure to to P1V2 iv) at constant volume back to the initial state P1V1. Find the work done on the gas in the cycle and trhe heat absorbed by the gas in the cycle.
 
  • #12
LeonhardEuler
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Then I think the book is wrong. Your answer makes sense when you think about it. Assuming T2>T1, V2>V1, then you are expanding at a higher temperature than you are contracting at. Because of the ideal gas law, this means you are expanding at a higher pressure. Since work is done by the system when expanding and on the system when contracting, and because the work is proportional to the pressure, it must be that net work is done by the system in this case.
 
  • #13
Andrew Mason
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One mole of perfect gas performs a quasistatic cycle which consists of the following four succesive stages: i)From the state P1V1 at constant pressure to the state P1V2 ii) at constant volume to the state P2V2 iii) At constant pressure to to P1V2 iv) at constant volume back to the initial state P1V1. Find the work done on the gas in the cycle and trhe heat absorbed by the gas in the cycle.
The cycle can go in either direction. The question does not give you the direction.

If it is run in the forward direction so that on a pv diagram the gas expands at the higher temperature/pressure and is compressed at the lower temperature/ pressure, it is a heat engine. In that case, the net work done ON the gas is negative (ie. the work done by the gas over a full cycle is positive). If it is run in the reverse direction (compressed at higher T and expands at lower T) as a refrigerator, the net work done ON the gas is positive.

AM
 
  • #14
Andrew Mason
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For this to be reversible, the heating and cooling would have to be done over an infinitesimal temperature gradient, which would obviously take forever, but could be approached in reality, for instance by putting this gas in thermal contact with another gas undergoing expansion or compression reversibly, and letting the temperature equilibrate by going very slowly. The entropy change can approach 0.
I am not sure how the entropy of the system + surroundings can be 0 in the expansion or compression parts. The temperature of the reservoir with which the system is in thermal contact would have to increase/decrease at the same rate as the system. How does it do that without allowing heat flow into or out of the surroundings from another reservoir of some kind?

AM
 
  • #15
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So my answer is right but possibly the wrong direction the author solved it for?

If T2<T1 then the gas is expanded at low pressure then I should get the author's answer with net positive work. If T2>T1 then my answer is correct. Since the question doesn't give the direction either is correct.

However, intuitively if one reduced the pressure (and thus the Temp) of a gas in a vessel (e.g. by reducing the force on a piston) one would expect the gas to expand at the new low temp.

Would that line of though be sufficient to make the author's answer more correct than mine?
 
  • #16
LeonhardEuler
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I am not sure how the entropy of the system + surroundings can be 0 in the expansion or compression parts. The temperature of the reservoir with which the system is in thermal contact would have to increase/decrease at the same rate as the system. How does it do that without allowing heat flow into or out of the surroundings from another reservoir of some kind?

AM
There is heat flow, but because of the definition of entropy:
[tex]dS=\frac{dQ_{rev}}{T}[/tex]
and the fact that during a reversible process the system is always in equilibrium, so that the temperature of both system and reservoir are uniform, the total entropy change for heat transfer during a reversible process is
[tex]dS_{tot} =\frac{dQ_{ab}}{T}+\frac{dQ_{ba}}{T}=\frac{dQ_{ab}}{T}-\frac{dQ_{ab}}{T} = 0[/tex]
 
  • #17
LeonhardEuler
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So my answer is right but possibly the wrong direction the author solved it for?

If T2<T1 then the gas is expanded at low pressure then I should get the author's answer with net positive work. If T2>T1 then my answer is correct. Since the question doesn't give the direction either is correct.

However, intuitively if one reduced the pressure (and thus the Temp) of a gas in a vessel (e.g. by reducing the force on a piston) one would expect the gas to expand at the new low temp.

Would that line of though be sufficient to make the author's answer more correct than mine?
No, the answer does not depend on whether T1>T2 or not. The formula works for all cases and gives a positive number when work on the system is positive and a negative one in situations where it is negative. I was just giving that as a plausibility check by taking one situation the formula covers and reasoning through what should be expected.
 
  • #18
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So I'm correct and the book is probably wrong?

That did occur to me, after going through the calculation for the 10th time, but I dismissed it. Maybe I'm too insecure.
 
  • #19
LeonhardEuler
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I believe so.
 
  • #20
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You have no idea how much better that makes me feel.

It's been 3 years since I last did a problem sheet and I'm so out of practise it's quite disheartening. So, it's good to know that I actually got that one right after banging my head against a wall trying to get the given answer.
 
  • #21
Andrew Mason
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There is heat flow, but because of the definition of entropy:
[tex]dS=\frac{dQ_{rev}}{T}[/tex]
and the fact that during a reversible process the system is always in equilibrium, so that the temperature of both system and reservoir are uniform, the total entropy change for heat transfer during a reversible process is
[tex]dS_{tot} =\frac{dQ_{ab}}{T}+\frac{dQ_{ba}}{T}=\frac{dQ_{ab}}{T}-\frac{dQ_{ab}}{T} = 0[/tex]
We both agree that a process is reversible only if the system is in equilibrium at all times with all the surroundings. One cannot look at only some of the surroundings.

If you have a finite reservoir as the source of reversible heat flow into a system whose temperature is constantly increasing, that reservoir's temperature must be constantly increasing at the same rate. That means that the finite reservoir must have a source of positive heat flow with which it is in thermal equilibrium at all times. That just means a bigger reservoir... And so on. So it requires an arbitrarily large reservoir whose temperature is constantly increasing but without an external source of heat flow. Can you provide an example of such a reservoir?

AM
 
  • #22
LeonhardEuler
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An example would be if you put your system inside an adiabatic chamber containing a gas. You compress the chamber reversibly, increasing it's temperature and allowing the system surrounded by this gas to constantly equilibrate to the new temperature.
 
  • #23
Andrew Mason
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An example would be if you put your system inside an adiabatic chamber containing a gas. You compress the chamber reversibly, increasing it's temperature and allowing the system surrounded by this gas to constantly equilibrate to the new temperature.
And how do you get the chamber back to its original temperature, reversibly, for the beginning of the next cycle?

AM
 
  • #24
LeonhardEuler
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And how do you get the chamber back to its original temperature, reversibly, for the beginning of the next cycle?

AM
You just do a reversible adiabatic expansion.
 
  • #25
Andrew Mason
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You just do a reversible adiabatic expansion.
First of all, expansions and compressions of the surrounding gas cannot be adiabatic because heat has to flow out of and into the surrounding gas / into and out of the system. Second, if the surrounding gas could undergo a reversible expansion, the only way it could expand back to its original temperature is by the gas doing at least the same amount of work that was done on it in the compression phase. Where does this energy come from? It obviously cannot come from the system because the system cannot convert 100% of the heat flow from the surrounding gas into work.

I think you will find, if you pursue this, that the only (theoretically) possible reversible thermodynamic cycle is one in which all heat flows occur isothermally.

AM
 

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