Terry Bing said:Consider the following problem:
Gaseous helium (assumed ideal) filled in a horizontal cylindrical vessel is separated from its surroundings by a massless piston. Both piston and cylinder are thermally insulating. The ambient pressure is suddenly tripled without changing the ambient temperature. How many times of its initial value will the volume of the helium become, when the piston finally stops?
Initial pressure and volume be P1 and V1. Now the ambient pressure is suddenly made P2 = 3 P1. The gas compresses under constant pressure P2, so the process must be isobaric. However the walls are also thermally insulating, so the process must also be Adiabatic. I find this confusing.
If I use the expression for adiabatic process i.e [tex] PV^\gamma =constant [/tex], then I don't get the answer at the back.
However, I can use 1st law to write[tex] W=-P_2\Delta V=3/2 nR\Delta T=3/2\Delta (PV)=3/2(P_2V_2-P_1V_1) [/tex] to get the correct answer.
I think the second way is fine. But why doesn't the first method work?
If the piston is not MASSLESS which is the case in real world , is it solution still relevant ? Bcoz seems like we can't do anything practical with it .Terry Bing said:Consider the following problem:
Gaseous helium (assumed ideal) filled in a horizontal cylindrical vessel is separated from its surroundings by a massless piston. Both piston and cylinder are thermally insulating. The ambient pressure is suddenly tripled without changing the ambient temperature. How many times of its initial value will the volume of the helium become, when the piston finally stops?
Initial pressure and volume be P1 and V1. Now the ambient pressure is suddenly made P2 = 3 P1. The gas compresses under constant pressure P2, so the process must be isobaric. However the walls are also thermally insulating, so the process must also be Adiabatic. I find this confusing.
If I use the expression for adiabatic process i.e [tex] PV^\gamma =constant [/tex], then I don't get the answer at the back.
However, I can use 1st law to write[tex] W=-P_2\Delta V=3/2 nR\Delta T=3/2\Delta (PV)=3/2(P_2V_2-P_1V_1) [/tex] to get the correct answer.
I think the second way is fine. But why doesn't the first method work?
It's just a limiting case (approximation) for a piston with small mass. Are you also saying anything we do with geometric figures (such as circles, rectangles, spheres) is not relevant to the real world bcuz there is no such thing as a perfect geometric figure in the real world?chanchal pandey said:If the piston is not MASSLESS which is the case in real world , is it solution still relevant ? Bcoz seems like we can't do anything practical with it .
well we can still perform the experiment whether the geometry of the given figure (circles , rectangles , spheres ) is perfect or not . But the issue whether the piston is massless or not is more important I think .Chestermiller said:It's just a limiting case (approximation) for a piston with small mass. Are you also saying anything we do with geometric figures (such as circles, rectangles, spheres) is not relevant to the real world bcuz there is no such thing as a perfect geometric figure in the real world?
I guess we have a difference of opinion. I don't think it's any more important. If you want to design equipment and carry out experiments without using the laws of geometry (because it's too approximate), that's OK with me. I'm willing to give up massless pistons in everything I do if you're willing to never use geometry again to design equipment and calculate volumes, area, and lengths. Do we have a deal?chanchal pandey said:well we can still perform the experiment whether the geometry of the given figure (circles , rectangles , spheres ) is perfect or not . But the issue whether the piston is massless or not is more important I think .
Here's a better example. In real world, there is no such thing as an ideal gas, so is it solution still relevant ? Bcoz seems like we can't do anything practical with it .chanchal pandey said:If the piston is not MASSLESS which is the case in real world , is it solution still relevant ? Bcoz seems like we can't do anything practical with it .
Well I was about to bring Ideal gases into our conversation. But u did first. I have problem with ideal gas too . Whole Thermodynamics chapter(still in High school) is based on the assumption that for thermodynamic equations to work , the process must be Reversible ( means at every stage equilibrium state is present in the process) and reversible processes happen very very slow . And in real world most of the processes are irreversible (non-equilibrium at every stage of the process).Chestermiller said:Here's a better example. In real world, there is no such thing as an ideal gas, so is it solution still relevant ? Bcoz seems like we can't do anything practical with it .
chanchal pandey said:Well I was about to bring Ideal gases into our conversation. But u did first. I have problem with ideal gas too . Whole Thermodynamics chapter(still in High school) is based on the assumption that for thermodynamic equations to work , the process must be Reversible ( means at every stage equilibrium state is present in the process) and reversible processes happen very very slow . And in real world most of the processes are irreversible (non-equilibrium at every stage of the process).
Example change in entropy = heat / change in temperature . This equation is only valid for REVERSIBLE processes and if I want to find out change in entropy in my room , I cannot apply this equation..
I agree with virtually all of this, except that, irrespective of whether the process is reversible or irreversible, the PV work is ##W=\int{P_{ext}dV}##. The problem is that, for irreversible processes, only in special cases can we know in advance (in the absence of detailed gas dynamics calculations on the system) the relationship between ##P_{ext}## and V; one of these cases is if we control ##P_{ext}## so that it is constant.Philip Koeck said:We've had both opinions in this thread now: You can only calculate work for reversible processes. You can also calculate work for a sudden, irreversible process using W = Pext ΔV.
If we go back to the original post, how would you replace the problem by a reversible process and calculate work. The problem I see is that you need the work to find out what the P, V and T of the final state actually are. As long as you don't know the final state, how do you choose the correct reversible process?
Who says you can't find the change in entropy of your room? As @Philip Koeck correctly points out, all you need to know is the initial and final states, irrespective of whether the actual process from the initial state to the final state was irreversible. Didn't they teach you how to get the entropy change for an irreversible process?chanchal pandey said:Well I was about to bring Ideal gases into our conversation. But u did first. I have problem with ideal gas too . Whole Thermodynamics chapter(still in High school) is based on the assumption that for thermodynamic equations to work , the process must be Reversible ( means at every stage equilibrium state is present in the process) and reversible processes happen very very slow . And in real world most of the processes are irreversible (non-equilibrium at every stage of the process).
Example change in entropy = heat / change in temperature . This equation is only valid for REVERSIBLE processes and if I want to find out change in entropy in my room , I cannot apply this equation..
If we go back to the original post, how would you replace the problem by a reversible process and calculate work. The problem I see is that you need the work to find out what the V and T of the final state actually are. As long as you don't know the final state, how do you choose the correct reversible process?Let'sthink said:We construct or imagine a reversible process which takes the system from the given equilibrium state through a series of equilibrium states to the given final state of the system. For such processes we can calculate, or write expressions for work done or heat exchanged and then use laws of thermodynamics to get the changes between state variables of the system's final and initial equilibrium states.
Chester Miller has given answer to that: We forget thermodynamics and calculate integration of Pext which is given as 3P and calculate work done by environment on the system but in this case lasowe assume this is done quasi statically or with zero mass piston. Using first law we know this must increase internal energy of the system.Knowing that this is ideal gas and final state is equilibrium state with volume V2 and using Cp we can find the change in temperature.Philip Koeck said:If we go back to the original post, how would you replace the problem by a reversible process and calculate work. The problem I see is that you need the work to find out what the V and T of the final state actually are. As long as you don't know the final state, how do you choose the correct reversible process?
Do you calculate W for a reversible process then?Let'sthink said:Chester Miller has given answer to that: We forget thermodynamics and calculate integration of Pext which is given as 3P and calculate work done by environment on the system but in this case lasowe assume this is done quasi statically or with zero mass piston. Using first law we know this must increase internal energy of the system.Knowing that this is ideal gas and final state is equilibrium state with volume V2 and using Cp we can find the change in temperature.
No I am not calculating integral pdx, I am just using mechanics consideration. And even this mechanical work is to be assumed to be done quasi statically. We use work energy theorem of Mechanics to conclude that the energy of the system on which work has been done must increase its energy when the final equilibrium state is reached. To conclude the change in internal energy we will have to consider another reversible process of heating quasi statically to achieve the final state with the help of a reservoir of infinite thermal capacity passing through series of equlibrium states of increasing temperature.Philip Koeck said:My impression is that, if you use W = Pext ΔV in this example, you are doing the calculation for an irreversible process.
I have no understanding of what you are trying to say here, but I can tell you as one of Physics Forums experts on thermodynamics that I agree totally with what Philip Kosck and and Terry Bing have said. If what you are saying disagrees with them, then I too disagree with you.Let'sthink said:No I am not calculating integral pdx, I am just using mechanics consideration. And even this mechanical work is to be assumed to be done quasi statically. We use work energy theorem of Mechanics to conclude that the energy of the system on which work has been done must increase its energy when the final equilibrium state is reached. To conclude the change in internal energy we will have to consider another reversible process of heating quasi statically to achieve the final state with the help of a reservoir of infinite thermal capacity passing through series of equlibrium states of increasing temperature.
It is too much conditional Sir. After all I had used your.earlier post, to write what I wanted, where in you said that we can assume that external pressure can be assumed to be held constant at 3P, then we can calculate the external work done. My submission is that this kind of quasi statical integration is also present in the definition of potential energy in mechanics. Rather the work energy theorem it self talks about reversible work. I think it is not true that these considerations just pop up in thermodynamics alone but they are present in mechanics too. Having known that force varies as 1/r^2 our assertion that potential energy will be proportional to 1/r, does involve the idea of reversible work done.Chestermiller said:I have no understanding of what you are trying to say here, but I can tell you as one of Physics Forums experts on thermodynamics that I agree totally with what Philip Kosck and and Terry Bing have said. If what you are saying disagrees with them, then I too disagree with you.
The compression process that occurs as a result of increasing the external pressure on the gas from P to 3P and then holding the pressure constant is neither quasi static nor reversible. It is a highly irreversible process, resulting in significant dissipation of mechanical energy to internal energy as a result of viscous friction. What does the work energy theorem say about processes in which friction is involved to dissipate mechanical energy?Let'sthink said:It is too much conditional Sir. After all I had used your.earlier post, to write what I wanted, where in you said that we can assume that external pressure can be assumed to be held constant at 3P, then we can calculate the external work done.
My submission is that this kind of quasi statical integration is also present in the definition of potential energy in mechanics.
For that work energy theorem says nothing, as work energy theorem is originally conceptualized for a point particle change in its KE and PE. In the problem posed by OP it was not written that P was made 3P in a gradual manner. It was written P was suddenly made 3P it is mentioned in the problem that the process is isobaric and finally it reaches an equilibrium state. Whether it is stated or not we know the process is irreversible and adiabatic. It is only externally isobaric. P as a parameter of the gas is just not defined during the process. So we need to use the information that it is ideal gas and guess the final state and then connect these by reversible processes to evaluate the final equilibrium state values using the tenets of thermodynamics.Chestermiller said:The compression process that occurs as a result of increasing the external pressure on the gas from P to 3P and then holding the pressure constant is neither quasi static nor reversible. It is a highly irreversible process, resulting in significant dissipation of mechanical energy to internal energy as a result of viscous friction. What does the work energy theorem say about processes in which friction is involved to dissipate mechanical energy?
We don't need to guess anything. In this problem, during the deformation, 3P is the pressure of the gas at the piston face, and, from this we have enough information to precisely establish the final equilibrium state. We can precisely calculate the work done by the gas on the piston for this irreversible process, and we know that there is no heat transfer, so, from the first law of thermodynamics, we know the relationship between the work and the change in internal energy. We don't need to connect the initial and final states by a reversible process to evaluate the final equilibrium state; we already know the final thermodynamic equilibrium state for the irreversible process by applying the first law in conjunction with the ideal gas law.Let'sthink said:For that work energy theorem says nothing, as work energy theorem is originally conceptualized for a point particle change in its KE and PE. In the problem posed by OP it was not written that P was made 3P in a gradual manner. It was written P was suddenly made 3P it is mentioned in the problem that the process is isobaric and finally it reaches an equilibrium state. Whether it is stated or not we know the process is irreversible and adiabatic. It is only externally isobaric. P as a parameter of the gas is just not defined during the process. So we need to use the information that it is ideal gas and guess the final state and then connect these by reversible processes to evaluate the final equilibrium state values using the tenets of thermodynamics.
Yes, I stand by this. This is work done by the gas on the piston face (which constitutes its surroundings), and it is also minus the work done by the piston face on the gas. If you want to call it external work, that's your prerogative. But it is really the work done by the gas on its surroundings.Do you not now stand for your statement on this thread that external work done will be Pext*(V2 - V1)
I perfectly agree with this Sir. But if some one wants to know the work done in the irreversible process then we can connect by a reversible process and tell the work done in general especially in this case as the process is adiabatic.[/QUOTE]Chestermiller said:We don't need to guess anything. In this problem, during the deformation, 3P is the pressure of the gas at the piston face, and, from this we have enough information to precisely establish the final equilibrium state. We can precisely calculate the work done by the gas on the piston for this irreversible process, and we know that there is no heat transfer, so, from the first law of thermodynamics, we know the relationship between the work and the change in internal energy. We don't need to connect the initial and final states by a reversible process to evaluate the final equilibrium state; we already know the final thermodynamic equilibrium state for the irreversible process by applying the first law in conjunction with the ideal gas law.
No Way. For this pair of initial and final states, there is no adiabatic reversible process that gives the same amount of work as for the present adiabatic irreversible path. The work for the present irreversible adiabatic process is ##3P(V_2-V_1)##. I challenge you to specify an adiabatic reversible path that even goes between these same two end states (temperatures, pressures, and volumes). Every reversible path will have to involve some transfer of heat.Let'sthink said:I perfectly agree with this Sir. But if some one wants to know the work done in the irreversible process then we can connect by a reversible process and tell the work done in general especially in this case as the process is adiabatic.
Let'sthink said:I agree again Sir, But Cp(T2-T1), is it not a reversible heat transfer? Or you say it is irreversible heat transfer that will take place in a real process? It is reversible heat transfer through reservoirs of infinite thermal capacity at temperatures increasing steadily from T1 to T2 in infinite infinitesimal steps.Then using the fact that internal energy is a state function this heat transfer must be equal to the work done in the actual irreversible process. What I wish to state Sir is whenever we use integration to calculate either work done by the system or heat supplied to the system we make use of the concept of a reversible process only.
I might have a way of solving the original problem via the entropy, but I'm not sure it's right. Anyway, the idea is this: The entropy of the surroundings is constant. If we imagine a reversible process between the same initial and final state, the entropy of the system is also constant (This might be be the problem.), since the total entropy has to be constant. Now we can use the Sackur-Tetrode equation to write down the entropy of the system for initial and final state and equate them. Then we also have the ideal gas law to eliminate the temperature in initial and final state. The result I get is V2 p23 = V1 p13. Does that make sense?Chestermiller said:Who says you can't find the change in entropy of your room? As @Philip Koeck correctly points out, all you need to know is the initial and final states, irrespective of whether the actual process from the initial state to the final state was irreversible. Didn't they teach you how to get the entropy change for an irreversible process?
Generally, this equation does not describe heat transfer at all. It describes the change in enthalpy per mole ##\Delta h## of an ideal gas between two thermodynamic equilibrium states. For an irreversible process, it applies only to the two end states. For a reversible process, since all the intermediate states are also thermodynamic equilibrium states, it applies to all intermediate states as wellLet'sthink said:I agree again Sir, But Cp(T2-T1), is it not a reversible heat transfer?
Irreversible heat transfer will take place in a real process.Or you say it is irreversible heat transfer that will take place in a real process?
In the process being discussed in this thread, the system is adiabatic, so there is no heat transfer taking place at all (that is the definition of adiabatic). And there are no reservoirs, and the zero heat transfer is not equal to the finite amount of work done.It is reversible heat transfer through reservoirs of infinite thermal capacity at temperatures increasing steadily from T1 to T2 in infinite infinitesimal steps.Then using the fact that internal energy is a state function this heat transfer must be equal to the work done in the actual irreversible process. What I wish to state Sir is whenever we use integration to calculate either work done by the system or heat supplied to the system we make use of the concept of a reversible process only.
This is an adiabatic irreversible process, so entropy is generated within the system, and the change in entropy of the system is positive. Therefore, any reversible process between the same two end states must involve heat transfer to the system. It is therefore impossible to devise an adiabatic reversible path between the two end states of an adiabatic irreversible process.Philip Koeck said:I might have a way of solving the original problem via the entropy, but I'm not sure it's right. Anyway, the idea is this: The entropy of the surroundings is constant. If we imagine a reversible process between the same initial and final state, the entropy of the system is also constant (This might be be the problem.), since the total entropy has to be constant. Now we can use the Sackur-Tetrode equation to write down the entropy of the system for initial and final state and equate them. Then we also have the ideal gas law to eliminate the temperature in initial and final state. The result I get is V2 p23 = V1 p13. Does that make sense?