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I Confusion between adiabatic and energy

  1. Dec 22, 2017 #1
    upload_2017-12-22_20-46-28.png Source: https://en.wikipedia.org/wiki/Irreversible_process

    I think I can see it mathematically but it is not very intuitive to me as it seems like there is creation of energy?

    For instance,
    1) Adiabatic expansion
    1st law of Thermodynamics to describe conservation of energy
    (-dU) = dQ - W
    where dU-change in internal energy, dQ-energy in/out the system, W-work done against the environment
    (-dU) as internal energy of system decreases(temperature decreases) due to expansion
    Since is adiabatic, dQ = 0, thus dU = W.

    2nd law of Thermodynamics to describe direction of natural process
    (-dU) = TdS - PdV1
    TdS = (-dU) + PdV1

    when the system is irreversible, dS > 0. For dS > 0, -dU + PdV1 > 0, thus PdV1 > dU.
    Hence PdV1 > W and thus, TdS > Q.
    Therefore, increase in entropy for adiabatic expansion.

    2) Subsequently, when we compress it adiabatically
    Again, 1st law of Thermodynamics to describe conservation of energy
    dU = dQ - (-W)
    -W as work on system due to compression
    Since is adiabatic, dQ = 0, thus dU = W.

    2nd law of Thermodynamics to describe direction of natural process
    dU = TdS - (-PdV2)
    TdS = dU - PdV2

    when the system is irreversible, dS > 0. For dS > 0, dU - PdV2 > 0, thus PdV2 < dU.
    Hence PdV2 < W and thus, TdS > Q.
    Since PdV2 < W and pressure change is the same, dV2 < dV1.

    Therefore, we cannot get back to the previous volume at the same temperature.

    Adiabatic means that there is no energy/matter transfer between the boundaries right? The only thing that is allowed is work against environment through expansion or compression.

    From my above, why won't the system be such that PdV = W especially since it is adiabatic? Entropy cannot decrease but it can remain the same which means the process can be reversible. The only reason I can think why PdV > W is that the system is not perfect as maybe there are friction? However won't that not be adiabatic? Where did that excess PdV - W comes from?
    Last edited: Dec 22, 2017
  2. jcsd
  3. Dec 22, 2017 #2


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    If only volume work is taken into account, this is really not so obvious. It is easier to see if you include another means of doing work on the system, i.e. a stirring wheel. The weel is operated using mechanical energy, but friction leads to an increase of internal energy of the system (i.e. work is converted into heat inside the system, hence there is no flow of heat over the boundaries and the process is still adiabatic). But even volume work can be irreversible due to longitudinal viscosity. In non-ideal gasses, the distribution of the gas molecules depends on the pressure. If the pressure is changed, the gas molecules need some time to adapt to the new situation. Hence the pressure P(t) of a non-ideal gas in a rapid process will not be equal to the sequence of equilibrium pressures when the process is run slowly. An extreme example where this is obvious is a gas which is compressed above the pressure where it would condense under ordinary conditions. If the gas has no time to condense, its pressure will be much higher than that of a gas which condenses into a liquid. Hence you will do considerable more work to compress the gas rapidly than slowly.
  4. Dec 22, 2017 #3
    You can't use the ideal gas law to calculate the amount of work for an irreversible expansion or compression using dW=PdV (even if there is no friction) because, in such an expansion or compression, the pressure of the gas in the cylinder is not uniform spatially, so which pressure do you use in the integral? The fact of the matter is that, in an irreversible expansion, the gas is not close to equilibrium, so the ideal gas law (or other equation of state) does not give the correct relationship. In addition to the pressure not being uniform within the cylinder, the pressure at the piston face (where the work is actually done) depends not only on the current volume, but on the rate of change of volume (which is unknown).

    There is enough informationin pictures you have provided to precisely calculate the conditions at the end of each step, and the exact changes in internal energy and entropy. If you'd like, I can help you determine these. Any interest?
  5. Dec 23, 2017 #4
    Yes that will be great! I give it a shot but i doubt it is correct. Also for non-ideal work done, it is obtained through experimental data right?

    1) Adiabatic expansion
    (-dU) = dQ - dW
    $$c_{v1}(T_2-T_1) = -W$$
    where cv1 is the specific heat capacity at V1
    Since is adiabatic, dQ = 0, thus U = W.

    2nd law of Thermodynamics
    (-dU) = TdS - PdV1
    Work done will be that of adiabatic expansion, thus the formula:
    $$ΔW_1 = \frac{P_1*V_1-P_2*V_2}{n-1} $$
    where n is the specific heat ratio for the gas shown in the diagram.
    $$Q_1 = (c_{v1}(T_2-T_1)) + \frac{P_1*V_1-P_2*V_2}{n-1}$$
    $$ΔS_1 = \frac{1}{T_2-T_1}[(c_{v1}(T_2-T_1)) + \frac{P_1*V_1-P_2*V_2}{n-1}]$$

    2) Subsequently, when we compress it adiabatically
    dU = dQ - (-W)
    $$c_{v2}(T_1-T_2) = W$$
    where cv2 is the specific heat capacity at V2
    Since is adiabatic, dQ = 0, thus dU = W.

    2nd law of Thermodynamics
    dU = TdS - (-PdV2)
    Work done will be that of adiabatic expansion, thus the formula:
    $$ΔW_2 = \frac{P_3*V_3-P_2*V_2}{n-1} $$
    $$Q_2 = (c_{v2}(T_1-T_2)) + \frac{P_1*V_3-P_2*V_2}{n-1}$$
    $$ΔS_2 = \frac{1}{T_1-T_2}[(c_{v2}(T_1-T_2)) + \frac{P_1*V_3-P_2*V_2}{n-1}]$$
    as P1 = P3
  6. Dec 23, 2017 #5
    Virtually all of this is incorrect. The reason is that the equations you have used are valid exclusively for an adiabatic reversible process, and we are dealing here with an adiabatic irreversible process.

    To simplify things, what we are going to do in our analysis is to assume that the piston-cylinder system in the figures is situated inside a vacuum chamber (external pressure = 0) and that the piston in the figures is both massless and frictionless. These simplifying assumptions do not in any way change the key factor that we are interested in considering in our analysis, which is the irreversibility of the expansion and subsequent compression.

    Based on these assumptions, if you do a force balance on the piston in the initial state of the system, how is the initial pressure ##p_1## (in State 1) related to the mass ##m_1##, the acceleration of gravity g, and the cross sectional area of the cylinder A? How is the pressure ##p_2## in State 2 related to the mass ##m_2##, the acceleration of gravity g, and the cross sectional area of the cylinder A? In terms of the mass ##m_2##, the acceleration of gravity g, and the change in elevation of the piston ##\Delta h##, what is the work W done by the gas on its surroundings (the piston) in the transition from State 1 to State 2?

    We'll continue after you answer these questions.
  7. Dec 23, 2017 #6
    $$p_1 = \frac{m_1g}{A}$$
    $$p_2 = \frac{m_2g}{A}$$
    Work done by gas on its surroundings, $$W=\frac{m_2g\Delta h}{A}$$
  8. Dec 24, 2017 #7
    This last equation is incorrect. The correct expression for the work is ##m_2g\Delta h##, the change in potential energy of m2. So, for the work done by the gas on its surroundings in the transition between States 1 and 2, we have: $$W_{1,2}=m_2g\Delta h=\frac{m_2g}{A}(A\Delta h)$$But, geometrically, $$A\Delta h=(V_2-V_1)$$Therefore, $$W_{1,2}=p_2(V_2-V_1)$$Note that this equation differs from what you obtained by assuming that the ideal gas law is satisfied in all intermediate states occurring during the transition from State 1 to State 2. In an irreversible expansion or compression of an ideal gas, the ideal gas law is not satisfied during the process, since the system is not in thermodynamic equilibrium. It can only be applied to the two end states, where the system is in thermodynamic equilibrium. Therefore, since our two ends states are thermodynamic equilibrium states, we can write: $$p_1V_1=nRT_1\tag{1}$$ and $$p_2V_2=nRT_2\tag{2}$$

    Applying the first law of thermodynamics to the transition between states 1 and 2, we have: $$nC_v(T_2-T_1)=-p_2(V_2-V_1)\tag{3}$$ If we next use Eqns. 1 and 2 to eliminate V1 and V2 from this equation, what do we obtain?
  9. Dec 24, 2017 #8
    Oops sorry! I confuse the pressure with the force.

    I obtain this,

    Just to confirm, the Cv will be the specific heat capacity for the volume at V1 right?
  10. Dec 24, 2017 #9
    Excellent!! Now, see if you can re-express this equation as:
    $$\frac{T_2}{T_1}=1-\frac{(\gamma-1)}{\gamma}\left(1-\frac{p_2}{p_1}\right)$$where ##\gamma=C_p/C_v=(C_v+R)/C_v##. Next, use the ideal gas law to determine the volume ratio ##V_2/V_1## for the change from State 1 to State 2, exclusively as a function of the pressure ratio ##p_2/p_1##

    Now, do you know the equation for the change in entropy of an ideal gas in terms of the temperature ratio and pressure ratio between the initial and final states?

    The internal energy of an ideal gas is a function only of temperature, and Cv is thus given by $$C_v=\frac{1}{n}\frac{dU}{dT}$$independent of the gas volume. So Cv is the same for all volumes of the gas. It is a function only of temperature.
  11. Dec 24, 2017 #10
    The formula for change in entropy is
    $$\Delta S = C_p ln\frac{T_2}{T_1} - Rln\frac{P_2}{P_1}$$
    Now all I do is sub in the above temperature and pressure ratio to get the change in entropy.

    I think the really confusing part is doing the irreversible and reversible process differently. And how differently am I suppose to do It I have no idea. It will be easier if values are provided because I can see if the process is reversible by checking that the change in entropy is zero. But how do I go about doing with variables instead? Like what steps are used for reversible and what steps are used for irreversible process? Thanks!
  12. Dec 25, 2017 #11
    This equation gives the entropy change per mole of ideal gas. You need to multiply by the number of moles to get the actual entropy change.
    So you are able to identify whether a process is reversible or irreversible. What characteristics of reversible vs irreversible are different?

    I think the real problem is knowing how to analyze an irreversible process. Here is a Physics Forums Insights article I wrote which focuses on this: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

    If you have any specific questions, please feel free to ask.

  13. Dec 25, 2017 #12
    Are you familiar with the concepts of ideal springs and dampers (aka dashpots) in mechanical systems (a damper is essentially the same as a shock absorber in a car, where the force to extend the damper is proportional to the relative velocity of the two ends of the damper). If so, I think I can give you a better feel for reversible vs irreversible.
  14. Dec 27, 2017 #13
    Thanks a lot!! The article really help a lot!

    However I am still a bit fuzzy about reversible vs irreversible.
    1) For step 3 when I devise a reversible process to take the system from its initial to its final thermodynamic equilibrium state, shouldn't the individual entropy change be equal to zero.
    For instance for example 2, when I find the change in internal energy,
    Then, $$ds_{rev}=nC_v\frac{dT}{T}$$
    $$\Delta s_{rev}=nC_vln(\frac{T_f}{T_i})>0$$
    And now if I were to reverse all the process to get back to the original temperature, my final entropy is
    $$\Delta s_{rev}=nC_vln(\frac{T_f}{T_i})+nC_vln(\frac{T_i}{T_f})>0$$
    The Ti/Tf and Tf/Ti are always positive. Won't the entropy always be increasing? If it is always increasing then shouldn't reversible process be impossible?

    2) As for a vacuum condition, when I let the gas in it expand, I know there shouldn't be work done on the environment. However won't there be work done by the gas because it is spreading further apart? Will the ideal gas law work for it too because its pressure will decrease as its volume increase.
  15. Dec 27, 2017 #14
    I am not familiar with damper but i am familiar with ideal spring.

    Is this a good example to use?

    Attached Files:

  16. Dec 27, 2017 #15
    In these equations for the change in entropy, you left out the contribution of the pressure term. For an adiabatic reversible process, the pressure term exactly cancels out the temperature term, so that the entropy change is zero. For an adiabatic irreversible process, the terms do not cancel.

    In Example 2 (and in our present problem), we are dealing with an irreversible adiabatic process. To get the change in entropy, we need to follow a reversible path between the same two end states. But it is impossible to devise and adiabatic reversible path to take us between the same two end states as our irreversible adiabatic process.. Any reversible path between these states will involve some heat transfer. So for this irreversible process, even though the actual process is adiabatic, there is still an entropy increase. And then when we return our system to its original pressure by restoring the original weight to the piston, the final state 3 will be different from the initial state 1, in that both its temperature and its volume will be different.

    So, in short, it is impossible to devise an adiabatic reversible process between the same two end states as an adiabatic irreversible process.
    Just because no work is done on the air in the room does not mean that the gas does not do any work. It is exerting a force on the piston to raise the weight, and the increase in potential energy of the weight is just equal to the amount of work that the gas does. The "environment" includes both the surrounding vacuum in the chamber "plus" the weight on top of the piston. In analyzing a thermodynamic system, you must be very precise in identifying exactly what comprises your "system." In this case, the system is the gas.

    Also, as I said in an earlier post, for an irreversible process, the ideal gas law gives the wrong answer for the work. For an irreversible process, the pressure exerted by the gas on the piston depends not just on the volume but also on the time rate of change of the volume. A very crude approximation to this would be: $$P=\frac{nRT}{V}-\frac{k}{V}\frac{dV}{dt}$$where k is a constant that depends on the viscosity of the gas. In an irreversible expansion or compression, the gas volume is changing very rapidly.
  17. Dec 30, 2017 #16
    Thanks! But as for isothermal reversible expansion of gas, I can't seem to find the term to get my change in entropy zero.
    $$\Delta S = nRln(\frac{V_2}{V_1}) $$
    My temperature term will be zero because temperature remains constant. What did I miss out?
  18. Dec 30, 2017 #17
    In an isothermal reversible expansion, the change in entropy is not zero. Heat is transferred reversibly to the gas from the surroundings.
  19. Dec 30, 2017 #18
    I am sorry now I am really confused. Am I right to say that isothermal expansion is always irreversible then?
    And what does it mean by heat is transferred reversibly to the gas? Is it that the transfer of heat to the system will not change the entropy of the system?
  20. Dec 30, 2017 #19
    What makes you think that if the entropy of a system increases, the process is irreversible? This is not correct.

    There are two ways that the entropy of a system can change. One is by exchange of heat between the system and its surroundings during the process. The other is by generation of entropy within the system itself as a result of irreversibilities occurring within the system. In an irreversible process, the second of these mechanisms is always present. In a reversible process, only the first of these mechanisms is present. So, in isothermal reversible expansion, the entropy of the system increases as a result of heat transferred from the surroundings to the system, but not as a result of irreversibilites occurring within the system.
  21. Dec 30, 2017 #20
    Okay I think I get it. As for heat transfer between system and surroundings, is it always reversible?
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