Two states of a gas

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  • #1
davidwinth
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Summary:: Irreversible process required to go from state 1 to state 2

If we start with an ideal gas at state 1 and undergo an isentropic compression, then follow this with an isothermal expansion, we could end up in a state, call it state 2, that has higher temperature, higher pressure, and larger volume than state 1. Is there a one step process to go from state 1 to 2 directly? I understand that it wouldn't be plottable on P-V because there would be huge irreversibilities. Still, I wonder what could happen to a gas that would go from s1 to s2 in one process rather than two.

What would such a pricess look like?
 

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  • #2
anuttarasammyak
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We have to appoint only two of P,V,T to define the state thanks to the state equation. Say State1(T_1,V_1) and State2(T_2,V_2). For an example, change volume from V_1 to V_2 by compression/expansion in any way. Then remove/add heat by contact with heat sink/source so that temperature becomes T_2.
State2 has entropy S_2 with no regard to the above mentioned process.

[EDIT]
Still, I wonder what could happen to a gas that would go from s1 to s2 in one process rather than two.
In one process for an example expand/compress to V_2 keep in contact with thermal source of temperature T_2.
 
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  • #3
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Are you familiar with the term "polytropic process" in which ##PV^n=const##?
 
  • #4
davidwinth
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Are you familiar with the term "polytropic process" in which ##PV^n=const##?
Yes. Could that be used? More specifically, what would that look like? Polytropic processes are theoretical. I mean, what would a person actually do to a gas to get the second state at higher pressure, higher temperature and larger volume?
 
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  • #5
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Yes. Could that be used? More specifically, what would that look like? Polytropic processes are theoretical. I mean, what would a person actually do to a gas to get the second state at higher pressure, higher temperature and larger volume?
Well let’s model the problem and see. For the change you described , what would n have to be?
 
  • #6
davidwinth
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Well let’s model the problem and see. For the change you described , what would n have to be?
n would have to be negative, meaning work and heat transfer would have to take place. Such a process is not possible spontaneously. So something like a gas in a cylinder with the piston being pulled out while heat was added would work?
 
  • #7
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n would have to be negative, meaning work and heat transfer would have to take place. Such a process is not possible spontaneously. So something like a gas in a cylinder with the piston being pulled out while heat was added would work?
You would have to be adding enough heat for the piston to be moving outward on its own (since the pressure is increasing). I guess you don't want to solve this quantitativey?
 
  • #8
davidwinth
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Are you familiar with the term "polytropic process" in which ##PV^n=const##?

You would have to be adding enough heat for the piston to be moving outward on its own (since the pressure is increasing). I guess you don't want to solve this quantitativey?

If the piston were moving on its own, then the final pressure would not be greater than the initial pressure, correct? The goal is state 2 has all three IGL variables (p, V, T) greater than state 1. I guess we could add some stops to limit the travel before the heat addition stopped.
 
  • #9
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If the piston were moving on its own, then the final pressure would not be greater than the initial pressure, correct? The goal is state 2 has all three IGL variables (p, V, T) greater than state 1. I guess we could add some stops to limit the travel before the heat addition stopped.
All these speculations can be resolved if you just analyze the problem quantitatively. You’ve already spent more time speculating than it would have taken to solve the problem.
 
  • #10
davidwinth
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All these speculations can be resolved if you just analyze the problem quantitatively. You’ve already spent more time speculating than it would have taken to solve the problem.
But the problem just is speculative. I am wondering if it really is possible, assuming the IGL, to have such a state 2 because people say thing like, "By the IGL, if pressure goes up then temperature must go up or volume down or both." But my thinking is that a state equation doesn't impose such limits. So I ask about a counter example. This isn't homework!
 
  • #11
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But the problem just is speculative. I am wondering if it really is possible, assuming the IGL, to have such a state 2 because people say thing like, "By the IGL, if pressure goes up then temperature must go up or volume down or both." But my thinking is that a state equation doesn't impose such limits. So I ask about a counter example. This isn't homework!
Your original example already addresses these IGL issues.
 
  • #12
davidwinth
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Your original example already addresses these IGL issues.

But my original example was a two-step process. I wanted to see if there was a concrete practical example of a 1-step process that could achieve the same end. Perhaps one could always break any purported one-step process that would achieve the desired final state into two single step processes? For example, take the heat addition to a gas in a cylinder while it freely expands until it hits the stops before heat addition ceases. One could say this was really two steps: 1) add heat under free expansion, 2) add heat under constant volume. If so, then I still do not have a good example of one single process that goes from state 1 to state 2 where state 2 has higher P, V and T.
 
  • #13
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But my original example was a two-step process. I wanted to see if there was a concrete practical example of a 1-step process that could achieve the same end. Perhaps one could always break any purported one-step process that would achieve the desired final state into two single step processes? For example, take the heat addition to a gas in a cylinder while it freely expands until it hits the stops before heat addition ceases. One could say this was really two steps: 1) add heat under free expansion, 2) add heat under constant volume. If so, then I still do not have a good example of one single process that goes from state 1 to state 2 where state 2 has higher P, V and T.
What's wrong with my example of a polytropic process?
 
  • #14
davidwinth
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What's wrong with my example of a polytropic process?



Nothing, except I cannot say, "Do this to a gas." Do what, exactly? I am the kind who likes to visualize a thing happening. A mathematical relationship doesn't help me.
 
  • #15
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Nothing, except I cannot say, "Do this to a gas." Do what, exactly? I am the kind who likes to visualize a thing happening. A mathematical relationship doesn't help me.
Who said anything about stopping with mathematical relationships. Would graphs of P vs V, T vs V, and Q vs V at a selection of values for the polytropic index n (<0) work for you?
 
  • #16
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I think I might have a physical scenario that would give these results. ChesterMiller will correct me if I am wrong, but this is what I came up with for a single process.

The Setup: Consider a well-insulated, vertical cylinder with tight-fitting piston made from very high conductivity material. The piston is situated some distance from the "bottom" of the cylinder, and held up by the pressure in the ideal gas below. Above the piston is vacuum. Call this state 1, and here the gas has ##P_{1g}, \; V_{1g},## and ## T_{1g}##.

The Process: A very hot liquid is injected into the vacuum above the piston, partially filling the void (leaving plenty of room for expansion). This results in extra mass on the piston, so we know the pressure in the gas rises and therefore ##P_{2g}>P_{1g}##. As the gas and the liquid come to thermal equilibrium, the temperature of the gas will rise while that of the liquid falls, resulting in ##T_{2g}>T_{1g}##. The question is: under what conditions will the final volume of the gas be larger than the initial volume? Let's further say that the mass of the gas and liquid are both 1 unit.

The Math: Here is where I may make a mistake, as I don't have any of my thermo books around and my algebra skills are rusty. ChesterMiller will, hopefully, point out any mistakes and correct as needed.

##P_{1g} = \frac{m_p g}{A_p}##

##P_{2g} = \frac{( m_p+1) g}{A_p}##

Clearly ##P_{2g}>P_{1g} ##.

For the temperature changes,

##\Delta u_g = C_{vg} \Delta T_g##
##\Delta u_L = C_{vL} \Delta T_L##

Also, ##\Delta u_g = -\Delta u_L## and ##T_{2g}=T_{2L}##. If we define the ratio ##\frac{C_{vg}}{C_{vL}} = \gamma##, then we can rearrange the above relations into the following:

##\frac{\gamma T_{1g} + T_{1L}}{\gamma+1}=T_{2g}##

Clearly, ##T_{2g}>T_{1g}## when ##T_{1L}>T_{1g}##. So we only need to make the liquid hotter than the gas at the start, as expected.

Finally, we want to know what conditions will give ##V_{2g}>V_{1g}##, and whether they are possible to achieve.

So, when is ##V_{2g}>V_{1g}##? When ##\frac{V_{2g}}{V_{1g}}>1##, of course.

##\frac{V_{2g}}{V_{1g}} = \frac{T_{2g}P_{1g}}{T_{1g}P_{2g}}##

If we plug the previous results for pressure and temperature into the volume ratio requirement, we get (barring mistakes!) the following:

##\frac{\frac{T_{1L}}{T_{1g}}-1}{\gamma+1}>\frac{1}{m_p}##

This seems entirely possible. We only have to choose the temperature appropriately given the initial temperature of the gas and the mass of the piston. It seems this is a one step process that meets the requirement on the final state of the gas.

HTH
I'm unable to visualize the setup and process, so I'm stumped there. Can you please provide a diagram? Thanks.
 
  • #17
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For a polytropic expansion or compression of an ideal gas, $$n=-\frac{\ln{(P_2/P_1)}}{\ln{(V_2/V_1)}}$$$$\frac{P}{P_1}=\left(\frac{V}{V_1}\right)^{-n}$$$$\frac{T}{T_1}=\left(\frac{V}{V_1}\right)^{1-n}$$$$Q=mC_v\left[\frac{\gamma-n}{1-n}\right]T_1\left[\left(\frac{V}{V_1}\right)^{1-n}-1\right]$$$$=mC_v\left[\frac{\gamma-n}{1-n}\right](T-T_1)$$where m is the number of moles.

Plot these up in dimensionless form, with n as a parameter, and see what you get.
 
  • #18
davidwinth
99
8
I double checked the piston example and got the same results. My only concern is how to apply the second Law. It will change things so the end won't be the same.
 
  • #19
anuttarasammyak
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My only concern is how to apply the second Law. It will change things so the end won't be the same.
Entropy is quantity state so it does not depend on the path. Entropy of environment depends on the path e.g. reversible or irreversible to what extent.
 
  • #20
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I double checked the piston example and got the same results. My only concern is how to apply the second Law. It will change things so the end won't be the same.
Of course it will be the same. The entropy change depends only on the two end points.
 
  • #21
davidwinth
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But we have heat being converted to work. We can't have 100% conversion.
 
  • #22
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But we have heat being converted to work. We can't have 100% conversion.
What the hell are you talking about? Of course we have have 100% conversion of heat to work. Just not in a cyclic process.

True or false: Entropy is a function of state (meaning it is a physical property of the gas, and not the process).
 
  • #23
davidwinth
99
8
What the hell are you talking about? Of course we have have 100% conversion of heat to work. Just not in a cyclic process.

True or false: Entropy is a function of state (meaning it is a physical property of the gas, and not the process).
There's no need to curse at me.

Of course the change in entropy depends on the states. My question is: do we arrive at the correct end state if we ignore entropy change in the calculations? Don't we need to apply the second law to figure out the actual end state? I.e., won't the second law reduce the efficiency of the heat transfer such that the final state of the gas is not what we would expect from only applying the first Law and equation of state?
 
  • #24
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There's no need to curse at me.

Of course the change in entropy depends on the states. My question is: do we arrive at the correct end state if we ignore entropy change in the calculations? Don't we need to apply the second law to figure out the actual end state? I.e., won't the second law reduce the efficiency of the heat transfer such that the final state of the gas is not what we would expect from only applying the first Law and equation of state?
The 2nd law does not have to be applied at all to determine the final state. All that is needed is the 1st law (and the eos) even if the process is irreversible. Anyway, just like your original path, the polytropic path is also reversible. So the integral of dQ/T should be exactly the same.
 
  • #25
davidwinth
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  • #27
mfig
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I am not sure why my previous attempt was unclear to some. I may have made a mistake (or many) somewhere, and that could be the reason. Please let me know where the error is if that is the case.

This is meant to help with a "concrete" example. I believe all three conditions are possible to meet, thus the second state is reachable from the first state.


Process.jpg
 
  • #28
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I am not sure why my previous attempt was unclear to some. I may have made a mistake (or many) somewhere, and that could be the reason. Please let me know where the error is if that is the case.

This is meant to help with a "concrete" example. I believe all three conditions are possible to meet, thus the second state is reachable from the first state.


View attachment 289727
So you are neglecting any vapor from the liquid that is formed in the upper chamber? isn't this the same as putting a weight on top of the piston and placing the cylinder in contact with a constant temperature reservoir at the same temperature that the liquid would have?
 
  • #29
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I'm going to evaluate the entropy change for the 2-step process in the original post and then show that the integral of dQ/T for the polytropic process with negative n between the same two end states is exactly equal to the same entropy change. This will allow us to reach a definite conclusion about the reversibility of the polytropic process, and whether the polytonic process with negative n is consistent with the 2nd law of thermodynamics.

The 2-step process in the original post consists of an adiabatic reversible compression, followed by an isothermal reversible expansion, in which the final pressure, the final temperature, and the final volume are all greater then in the initial state before the compression. Let the pressure, volume, and temperature in the initial state be ##(P_1,V_1,T_1)##, let the pressure, volume, and temperature in the intermediate state after the compression be ##(P_C,V_C,T_C)##, and let the pressure, volume, and temperature in the final state after the isothermal expansion be ##(P_2,V_2,T_2)##. Since the expansion step is isothermal, we must have that ##T_C=T_2##.

For the initial adiabatic reversible compression step, we have that $$\frac{P_C}{P_1}=\left(\frac{V_C}{V_1}\right)^{-\gamma}$$
$$\frac{T_C}{T_1}=\frac{T_2}{T_1}=\left(\frac{V_C}{V_1}\right)^{1-\gamma}$$ From this, if follows that the volume in the intermediate compressed state can be expressed in terms of the temperature ratio for the overall two-step process by $$V_C=V_1\left(\frac{T_2}{T_1}\right)^{-\frac{1}{\gamma-1}}\tag{1}$$We will be using the value of this intermediate state compressed volume to determine the entropy change for the combined 2-step process.

In the present development, we are not only going to use the polytropic process relationships to represent a path from the initial state to the final state. We are also going to use it to establish the final state using the single polytropic process parameter n to guarantee that the final pressure and volume exceed the initial pressure and volume. For the direct polytropic path between the initial and final states, I indicated in post # 17 that the polytropic process path intersects the 2-step reversible path at $$\frac{P_2}{P_1}=\left(\frac{V_2}{V_1}\right)^{-n}$$This equation indicates that at n=0, the final pressure is equal to the initial pressure before compression while for ##n\rightarrow -\infty## the final volume approaches the initial volume before compression. Therefore, any negative value selected for n will guarantee that both the final pressure and final volume after the 2-step process are greater than their initial values before compression. In other words, the intersection of the P-V curve for a polytropic process with negative n will always intersect the P-V curve for the isothermal reversible expansion of the 2-step process at a point where the final pressure is greater than the initial pressure before compression and the final volume is greater than the initial volume before compression; and the full range of negative values for n from zero to ##-\infty## will span all the possible final states satisfying these requirements.

In addition to the previous relationship, we have for the polytropic process path $$\frac{V_2}{V_1}=\left(\frac{T_2}{T_1}\right)^{\frac{1}{1-n}}\tag{2}$$
Let's next evaluate the entropy change for the 2-step process in terms of the two parameters n and ##\gamma##, and the overall process temperature ratio ##T_2/T_1##. The entropy change for the adiabatic reversible compression step is zero, so the entropy change for the overall 2-step process is equal to that for the isothermal reversible expansion step: $$\Delta S=mR\ln{(V_2/V_C)}=mR\ln{\left(\frac{V_1}{V_C}\frac{V_2}{V_1}\right)}$$where m is the number of moles.
If we substitute Eqns. 1 and 2 into this relationship, we obtain:$$\Delta S=mR\ln{(T_2/T_1)^{\left[\frac{1}{\gamma-1}+\frac{1}{1-n}\right]}}$$$$=mR\ln{(T_2/T_1)^{\frac{\gamma-n}{(\gamma-1)(1-n)}}}$$$$=mR\frac{(\gamma-n)}{(\gamma-1)(1-n)}\ln{(T_2/T_1)}$$$$=mC_v\frac{(\gamma-n)}{(1-n)}\ln{(T_2/T_1)}\tag{3}$$

Next, let's consider the integral of dQ/T for the polytropic path, and see how it compares with the known entropy change between the initial and final states, as determined from the 2-step process. Unlike the 2-step process where all the heat transfer takes place from a single constant temperature reservoir at temperature ##T_2##, in the polytropic path, the heat transfer occurs from a continuous sequence of reservoirs (in differential increments) at temperatures running all the way from the initial temperature ##T_1## to the final temperature ##T_2##.

In post #17, I indicated that, for a polytropic process, the 1st law of thermodynamics tells us that the cumulative amount of heat received by the gas in going from the initial temperature ##T_1## to temperature T during the process is given by the equation $$Q=mC_v\frac{(\gamma-n)}{(1-n)}(T-T_1)$$Therefore, the differential amount of heat received by the gas in going from temperature T to temperature ##T+dT## is given by $$dQ=mC_v\frac{(\gamma-n)}{(1-n)}dT$$If we divide this by the temperature T at the interface with the reservoir from which the heat is received, we obtain:$$\frac{dQ}{T}=mC_v\frac{(\gamma-n)}{(1-n)}\frac{dT}{T}$$And integrating this over the entire polytropic path between temperatures ##T_1## and ##T_2## gives: $$\int_{T_1}^{T_2}{\frac{dQ}{T}}=mC_v\frac{(\gamma-n)}{(1-n)}\ln{(T_2/T_1)}$$
But this is exactly equal to the entropy change between the initial and final states as determined from the 2-step process. So we have that: $$\left[\int{\frac{dQ}{T}}\right]_{polytropic\ path}=\Delta S$$The only way this can happen is if the polytropic path is reversible. According to the Clausius inequality, to be consistent with the 2nd law of thermodynamics, the process path must be such that the integral of dQ divided by the boundary temperature at which the heat transfer occurs must be less than or equal to change in entropy between the initial and final states. In the case of a polytropic process, the equality condition is satisfied, even for negative values of the polytropic index n. Thus, the polytropic process (even with negative n) is consistent with the 2nd law of thermodynamics.
 
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  • #30
mfig
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So you are neglecting any vapor from the liquid that is formed in the upper chamber? isn't this the same as putting a weight on top of the piston and placing the cylinder in contact with a constant temperature reservoir at the same temperature that the liquid would have?

Yes. If you think neglecting the vapor is untenable, we could replace the liquid with a hot solid.

It is and it isn't the same. In the process you suggest, one would perform two actions: place a weight and add heat. In the process I suggest, these are accomplished in one action. I believe that was the requirement.
 
  • #31
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Here is my entropy change analysis for the irreversible path suggested by @mfig. If we apply the first law of thermodynamics to his path, we obtain $$mC_v(T_2-T_1)=Q-P_2(V_2-V_1)$$where ##P_2=(M_P+M_L)g/A## is the final pressure. Applying the ideal gas law to the work term in the above equation then gives: $$mC_v(T_2-T_1)=Q-mRT_2(1-\frac{V_1}{V_2})\tag{1}$$
Even though the present irreversible path bears little resemblance to the reversible polytropic path analyzed previously, as pointed out in post #29, we can still use the polytropic parameterization in terms of the polytropic parameter n to establish the final state of the gas, such that the range of values for the single parameter n from 0 to ##-\infty## span all possible final states of the gas for which the final pressure and final volume are greater than their values in the initial state. With this in mind, combining Eqn. 2 of post #29 with Eqn. 1 of the present develop gives:$$mC_v(T_2-T_1)=Q-mRT_2\left[1-\left(\frac{T_1}{T_2}\right)^{\frac{1}{1-n}}\right]\tag{2}$$From this, it follows that the total heat flow over the irreversible path is given by:$$Q=mC_v(T_2-T_1)+mRT_2\left[1-\left(\frac{T_1}{T_2}\right)^{\frac{1}{1-n}}\right]\tag{3}$$Since, for this irreversible path, all the heat transfer at the boundary takes place at the final temperature ##T_2##, the integral of dQ divided by the boundary temperature (i.e., the boundary where the heat flow occurs) is just equal to ##Q/T_2##: $$\int{\frac{dQ}{T_{boundary}}}=\frac{Q}{T_2}$$$$=mC_v\left(1-\frac{T_1}{T_2}\right)+mR\left[1-\left(\frac{T_1}{T_2}\right)^{\frac{1}{1-n}}\right]\tag{4}$$Calculations show that, for all values of (1) the "final state parameter" n in the range 0 to ##-\infty##, (2) the heat capacity ratio parameter ##1<\gamma<1.67##, and (3) the overall temperature ratio ##\frac{T_2}{T_1}>1## the integral of ##dQ/T_{boundary}## calculated from Eqn. 4 for this irreversible process path is less than the entropy change between the initial and final states calculated from Eqn. 3 of post #29 for the 2-step reversible path (or the reversible polytropic path). For example, for n = -1, ##\gamma=1.4##, and ##\frac{T_2}{T_1}=2##, we find that ##dQ/T_{boundary}=0.617mC_v## while ##\Delta S =0.832mC_v##. This is consistent with the Clausius inequality, which represents a mathematical statement of the 2nd law of thermodynamics.
 

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