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Finding Pressure/Temperature w/o Knowing Work

  1. Jan 26, 2015 #1
    1. The problem statement, all variables and given/known data
    An insulated vertical cylindrical vessel of cross-sectional area A= .12m^2 contains 20 mol of He gas at To= 300K and Po= .5 bar. It is capped by a movable piston of mass m = 150kg supported from the ceiling by a wire. The top of the vessel is open to the atmosphere (Patm = 1bar, Tatm = 300K). The wire is cut and the weight falls and compresses the gas.

    a) Obtain the pressure and temperature of gas a short time after the wire has been cut, when a negligible amount of heat transfer has taken place.

    b)After a long time, the He gas thermally equilibrates with the enviornment by heat transfer through the piston. How much heat is exchanged with the environment during this process?

    Helium can be considered to be an ideal gas at all conditions of interest, with Cv= 3*R/2

    2. Relevant equations
    Not sure.....There are so many I don't know which to use. I'm guessing:

    delta U =n*Cv*delta(T) <------from my thermo textbook

    Work= integral for force times displacement

    also Pf= Ppiston + Patmosphere

    3. The attempt at a solution for part a):

    F= 9.8 m/s^2 *(150 kg) = 1,470 N
    P=F/A----> 1,470 N/0.12 m2 =12,250 Pa

    I know the above is a calculation I will have to make, but I'm not sure what else to do from here. This is neither a constant pressure or constant volume process, so I don't know why I'm given Cv.

    Confused...Please help. :/
     
  2. jcsd
  3. Jan 27, 2015 #2

    Andrew Mason

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    What kind of compression is involved? (eg. isothermal, adiabatic, isobaric, isochoric). What is the relationship between P and V in such a process?

    AM
     
  4. Jan 27, 2015 #3
    This adiabatic compression, I believe, so delta U= W.... Work is the integral of PdV or Fdx.... However, in this case, we know neither the change in volume or displacement. Also, I'm confused on how to incorporate the atmospheric pressure in this.
     
  5. Jan 27, 2015 #4

    Andrew Mason

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    You have to apply the adiabatic condition. How are P and V related in an adiabatic compression (hint: it involves [itex]\gamma = C_p/C_v[/itex]

    The pressure in the gas inside the cylinder is the result of the cylinder weight + atmospheric pressure.

    AM
     
  6. Jan 27, 2015 #5
    Ok. I considered using Cv earlier, But then I realized that this is neither a constant pressure or constant voume process, because Helium gas is being compressed. ...Would I use the ideal gas law, PV= nRT to solve for the initial volume and then Cp= Cv +R to find Cp? After that, I could use PV^(gamma)= constant?
     
  7. Jan 27, 2015 #6
    This is a rather complicated problem, and, the reason it is complicated is that the compression is irreversible, so that you can't apply the ideal gas law to the gas within the cylinder during the course of the deformation (i.e., the deformation is not quasistatic). However, that does not mean that the problem can't be solved.

    Let me help you get started. Even though we don't know how the pressure of the gas within the cylinder is varying with time, we can still call PI(t) the pressure of the gas at the interface with the piston at time t. We will not know how PI(t) is varying with time, but we will not let that stop us from solving the problem. Our focus will be on determining the integral of PI(t) dV over the time it takes for the system to equilibrate adiabatically. This will be the total work that the gas does on the surroundings. Let A be the area of the piston and let x represent the location of the piston. Please write down a Newton's second law force balance equation for the motion of the piston. This will be our starting point. (During the deformation, the piston will be oscillating up and down, but this oscillation will damp out as time progresses. We won't have to consider the oscillation in detail.)

    Chet
     
  8. Jan 27, 2015 #7

    Andrew Mason

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    Chet is technically correct that [itex]PV^\gamma=K[/itex] is for reversible adiabatic processes. But it is a very good approximation for this kind of non-reversible adiabatic process. This is thermodynamics, which deals with a changes between states of equilibrium rather than an analysis of a dynamic process. There is no easy way to analyse the process itself. So I would say that the author must have wanted you to apply the adiabatic condition.

    AM
     
  9. Jan 27, 2015 #8
    Hi Andrew. It is not extremely difficult to do the problem for the irreversible case, as we'll see when the OP solves it that way (The difficult part is figuring out a simple way to do it. I will lead the OP through this, which only involves a couple of steps). But, you make a very interesting point. I think we should do the problem both ways and see how the results compare. Would you be willing to provide the results for the calculated work assuming reversible?

    Chet
     
  10. Jan 27, 2015 #9
    Thanks Chet and AM for all your help.

    Chet, I have written the force balance for the piston below:

    The forces on the piston are the mass (since it fell due to gravity) and the force of the atmosphere (air molecues above the piston). These forces should be balanced by the pressure of helium gas acting on the piston.

    So:

    Fgravity= 1,470 N

    Fgravity + Patm/0.12 m2 = PHe gas/0.12 m2
     
  11. Jan 27, 2015 #10

    Andrew Mason

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    Be careful: P = F/A so F = PA

    P is in N/m^2 = Pascals.

    AM
     
  12. Jan 28, 2015 #11
    Oh, no.....How silly of me! :) Good catch, AM!

    Revised:

    Fgravity= 1,470 N

    Fgravity + Patm*0.12m 2 = PHe gas*0.12 m2
     
  13. Jan 28, 2015 #12

    Andrew Mason

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    So work it out. What is the pressure of gas in cylinder required to balance piston and atmosphere? What is the initial P?

    AM
     
  14. Jan 28, 2015 #13
    Not so fast. You left out a term in the force balance. During the irreversible deformation of the gas, the piston is accelerating and decelerating. The force balance on the piston should read:
    $$P_I(t)A-Mg-P_{atm}A=M\frac{dv}{dt}$$
    where M is the mass of the piston and v is the velocity of the piston. If we take this equation, multiply by vdt, and integrate between 0 and t, we obtain:
    $$\int_{V_0}^{V(t)}{P_IdV}=\left(\frac{Mg}{A}+P_{atm}\right)(V(t)-V_0)+\frac{1}{2}Mv^2(t)$$
    The left hand side of this equation represents the work done by the gas on the piston up to time t, W(t), and the last term on the right hand side represents the kinetic energy of the oscillating piston at time t.

    As time progresses, viscous damping in the gas will cause the mechanical oscillation of the piston to decay to zero, and at final equilibrium, the piston will no longer be moving. So, at final steady state, we have:
    $$W_F=\left(\frac{Mg}{A}+P_{atm}\right)(V_F-V_0)$$
    where the subscript F refers to the final equilibrium state.

    The above analysis yield's the same result as your analysis, but it gives proper consideration to the kinetic energy of the piston.

    Now, to solve for the final steady state of the system, you need to determine the final equilibrium volume and temperature. That can be done by setting the change in internal energy of the gas equal to minus the amount of work done by the gas. Of course, in the initial and final equilibrium states of the system, it valid to apply the ideal gas law.

    Once you find the final volume of the gas, we can use it to compare the irreversible work done by the gas with the corresponding quantity the Andrew calculates for the reversible case. We can also determine the change in entropy of the gas, which, for the reversible case, is equal to zero.

    Chet
     
  15. Jan 28, 2015 #14
    I don't understand how we can tell if this a reversible or irreversible process to begin with. Moreover, why are we using time in the force-balance equation. I was under the impression that thermodynamics does not attempt to make claims about what is happening over time, but only what is happening during equilibrium states.
     
  16. Jan 28, 2015 #15
    The key characteristic of a reversible process is that it passes through a continuous sequence of equilibrium states. In order for this to happen, the deformation must be carried out quasistatically. In our process, allowing the piston to drop rapidly after the rope is cut prevents the deformation from occurring quasistatically. The rate of viscous dissipation of mechanical energy to heat in a system is proportional to the square of the rate of deformation. So the rapid deformation caused by the rapidly falling piston causes substantial viscous dissipation, and prevents the gas deformation from being reversible. In order for the process to be quasistatic and reversible, we would have to lower the piston gradually so that the system is close to thermodynamic equilibrium at every stage of the deformation, and the rate of viscous dissipation is insignificant.

    In any event, we are going to solve the problem for both the irreversible case (rapid drop of the piston) and the reversible case (gradual lowering of the piston) so you can see the difference in the results for yourself.

    There is nothing wrong with using time to characterize the path of the heat added and the work done. Do you feel that we incorrectly determined the work done by the gas on the piston? I personally like to use time as a parameter in the analysis, especially for irreversible processes, because it makes it easier for the student to understand what is happening in the process. In our problem, we still focus on the initial and final equilibrium states of the system in getting a solution, but, we examine the time dependent motion of the piston to help us determine the total amount of work done by the gas on the piston between the initial and final equilibrium states. Otherwise, we wouldn't have understood that a piston oscillation occurs, and that it is damped out by viscous dissipation in the gas. So the time-dependent analysis helped us interpret physically what is happening between the initial and final equilibrium states.

    When we still were allowed to have blogs on Physics Forums, I had a blog on the first and second laws of thermo (that was very well received) that showed the development of the relationships based on time dependence of the heat flow and the work done. I think you will find it interesting and informative. I saved a copy on my computer, and am inserting it below. It only takes a short amount of time to read. I hope you find it of value, and would be happy to answer questions. Meanwhile, please continue with your analysis of our irreversible process.

    FIRST LAW OF THERMODYNAMICS

    Suppose that we have a closed system that at initial time ti is in an initial equilibrium state, with internal energy Ui, and at a later time tf, it is in a new equilibrium state with internal energy Uf. The transition from the initial equilibrium state to the final equilibrium state is brought about by imposing a time-dependent heat flow across the interface between the system and the surroundings, and a time-dependent rate of doing work at the interface between the system and the surroundings. Let [itex]\dot{q}(t)[/itex] represent the rate of heat addition across the interface between the system and the surroundings at time t, and let [itex]\dot{w}(t)[/itex] represent the rate at which the system does work on the surroundings at the interface at time t. According to the first law (basically conservation of energy),
    [tex]\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W[/tex]
    where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

    The time variation of [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] between the initial and final states uniquely characterizes the so-called process path. There are an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

    If a process path is irreversible, then the temperature and pressure within the system are typically inhomogeneous (i.e., non-uniform, varying with spatial position), and one cannot define a unique pressure or temperature for the system (except at the initial and the final equilibrium state). However, the pressure and temperature at the interface can be measured and controlled using the surroundings to impose the temperature and pressure boundary conditions that we desire. Thus, TI(t) and PI(t) can be used to impose the process path that we desire. Alternately, and even more fundamentally, we can directly control, by well established methods, the rate of heat flow and the rate of doing work at the interface [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex]).

    Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:
    [tex]\dot{w}(t)=P_I(t)\dot{V}(t)[/tex]
    where [itex]\dot{V}(t)[/itex] is the rate of change of system volume at time t. However, if the process path is reversible, the pressure P within the system is uniform, and

    [itex]P_I(t)=P(t)[/itex] (reversible process path)

    Therefore, [itex]\dot{w}(t)=P(t)\dot{V}(t)[/itex] (reversible process path)

    Another feature of reversible process paths is that they are carried out very slowly, so that [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] are both very close to zero over then entire process path. However, the amount of time between the initial equilibrium state and the final equilibrium state (tf-ti) becomes exceedingly large. In this way, Q-W remains constant and finite.

    SECOND LAW OF THERMODYNAMICS

    In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate [itex]\dot{q}(t)[/itex] and the rate of doing work [itex]\dot{w}(t)[/itex] as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:
    [tex]Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}[/tex]
    [tex]W=\int_{t_i}^{t_f}{\dot{w}(t)dt}[/tex]
    In the present section, we will be introducing a third integral of this type (involving the heat transfer rate [itex]\dot{q}(t)[/itex]) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

    The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800's.

    Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:
    [tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}[/tex]
    where TI(t) is the temperature at the interface with the surroundings at time t. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing. He found that, for any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) was not arbitrary; instead, there was a unique upper bound (maximum) to the value of the integral. Clausius also found that this result was consistent with all the "word definitions" of the Second Law.

    Clearly, if there was an upper bound for this integral, this upper bound had to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

    But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there existed a well-defined subset, each member of which gave the same maximum value for the integral. This subset consisted of what we call today the reversible process paths. So, to determine the change in entropy between two equilibrium states, one must first conceive of a reversible path between the states and then evaluate the integral. Any other process path will give a value for the integral lower than the entropy change.

    So, mathematically, we can now state the Second Law as follows:

    [tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq \Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}[/tex]
    where [itex]\dot{q}_{rev}(t)[/itex] is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is the system temperature at time t (which, for a reversible path, is equal to the temperature at the interface with the surroundings). This constitutes a precise mathematical statement of the Second Law of Thermodynamics.
     
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