Thermal Physics: Thermodynamic Identity

In summary: So the final equilibrium state of the system is: $$nC_v=0$$Now, we can use the first law of thermodynamics to calculate the change in entropy. The entropy change is:$$S=nC_v\Delta S$$So, in summary, the process is reversible, the entropy change is quasistatic, and the final equilibrium state is equilibrium.
  • #1
WWCY
479
12

Homework Statement


Screen Shot 2017-10-23 at 7.09.55 PM.png


Homework Equations



##dS = \frac{1}{T} (dU - PdV)## assuming dN = 0

The Attempt at a Solution


I have actually managed to solve all 4 parts correctly, except for the fact that I solved Part d) with the Sackur-Tetrode equation rather than the thermodynamic identity.

I don't understand 1) why and 2) how I can use the identity in this scenario. Was the change in the system not quasi-static? My current understanding of the thermodynamic identity is that it can only be used during quasi-static changes where variables like T and P are well-defined.

Help is greatly appreciated!
 

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  • #2
WWCY said:

Homework Statement


View attachment 213601

Homework Equations



##dS = \frac{1}{T} (dU - PdV)## assuming dN = 0

The Attempt at a Solution


I have actually managed to solve all 4 parts correctly, except for the fact that I solved Part d) with the Sackur-Tetrode equation rather than the thermodynamic identity.

I don't understand 1) why and 2) how I can use the identity in this scenario. Was the change in the system not quasi-static? My current understanding of the thermodynamic identity is that it can only be used during quasi-static changes where variables like T and P are well-defined.

Help is greatly appreciated!
The actual process is irreversible (non-quasistatic). So to determine the change in entropy of the system, since entropy is a thermodynamic function of state, one must, first of all, immediately forget about the irreversible path for the actual process, and, instead, focus on an alternate reversible process path between the same two end states for which dS=dQrev/T. The Sackur-Tetrode equation has done the integration for you for a reversible path, and integration of the identity will automatically do the same.
 
  • #3
Chestermiller said:
The actual process is irreversible (non-quasistatic). So to determine the change in entropy of the system, since entropy is a thermodynamic function of state, one must, first of all, immediately forget about the irreversible path for the actual process, and, instead, focus on an alternate reversible process path between the same two end states for which dS=dQrev/T. The Sackur-Tetrode equation has done the integration for you for a reversible path, and integration of the identity will automatically do the same.

Let me try to rephrase what your explanation to see if I understood it. The process that was described here was irreversible, and consisted of changes ##\Delta U, \Delta V##. Because S is a state variable, it doesn't matter how the system transforms as long as it reaches the same end state. Therefore the change in entropy can also be described a quasistatic process that consists of the same changes.

If that is the case, ##\Delta U## suggests a change in T, and ##\Delta V## suggests a change in P. Yet, the final answer shows that ##\Delta S = \frac{1}{T_0} (\Delta U + P_0\Delta V)## where ##T_0## and ## P_0## are the temperature and pressure of the gas before the compression. Why is that?

Also, if I took a reversible path to the end-state of the system, shouldn't there be no new entropy created, since quasistatic, reversible processes don't create entropy?

Thank you for your patience and assistance.
 
  • #4
WWCY said:
Let me try to rephrase what your explanation to see if I understood it. The process that was described here was irreversible, and consisted of changes ##\Delta U, \Delta V##. Because S is a state variable, it doesn't matter how the system transforms as long as it reaches the same end state. Therefore the change in entropy can also be described a quasistatic process that consists of the same changes.
Correct
If that is the case, ##\Delta U## suggests a change in T, and ##\Delta V## suggests a change in P. Yet, the final answer shows that ##\Delta S = \frac{1}{T_0} (\Delta U + P_0\Delta V)## where ##T_0## and ## P_0## are the temperature and pressure of the gas before the compression. Why is that?
The answer is not quite correct, but it is a good approximation for this specific problem (since the volume change is very small). So, I am going to do the problem over for you so that we can compare "the final answer" with the exact answer.
Also, if I took a reversible path to the end-state of the system, shouldn't there be no new entropy created, since quasistatic, reversible processes don't create entropy?
Quasistatic reversible processes don't create entropy. But, entropy can be transferred between the system and surroundings if the reversible path is not adiabatic.

Now, let's solve the problem the right way. We are going to assume that the irreversible path is adiabatic. The first step is to determine the final thermodynamic equilibrium state of the system using the first law of thermodynamics. For our irreversible process, we have:
$$nC_v\Delta T=-P_{ext}\Delta V$$where, for this constant-external-force problem (with ##P_{ext}## constant) with $$P_{ext}=2000N/0.01m^2=200000Pa$$ The number of moles of air is given by the ideal gas law: $$n=\frac{P_0V_0}{RT_0}=\frac{10^5(1)}{(8314.)(300)}=0.0401\ moles$$ The molar heat capacity of the air is: $$C_v=2.5R=(2.5)(8.314)=20.785\ \frac{J}{mole.K}$$The work done by the gas on the surroundings is:$$W=P_{ext}\Delta V=-(200000)(0.001)(0.01)=-2 J$$So the temperature rise is:
$$\Delta T=\frac{2}{nC_v}=\frac{2}{(0.0401)(20.785)}=2.40 C$$
The final temperature is then $$T_f=300+2.4=302.4\ K$$and the final volume is $$V_f=1 - 0.01 = 0.99\ liters$$After the gas has equilibrated in the cylinder, its final pressure can be determined from the ideal gas law: $$P_f=\frac{nRT_f}{V_f}=\frac{(0.0401)(0.08314)(302.4)}{0.99}=1.0184\ bars=1.0184\times 10^5 Pa$$

Now let's get the exact answer for the entropy change. If we evaluate the entropy change either by following a reversible path or, equivalently, by integrating the thermodynamic identity between the initial and final states, we obtain:
$$\Delta S=nC_v\ln{\frac{T_f}{T_0}}+nR\ln{\frac{V_f}{V_0}}=nC_v\ln{\left(1+\frac{\Delta T}{T_0}\right)}+nR\ln{\left(1+\frac{\Delta V}{V_0}\right)}$$But we know from what we did above that:
$$nC_v=\frac{\Delta U}{\Delta T}$$ and $$nR=\frac{P_0V_0}{T_0}$$
If we substitute these relationships into our equation for the exact entropy change, we obtain:
$$\Delta S=\frac{\Delta U}{\Delta T}\ln{\left(1+\frac{\Delta T}{T_0}\right)}+\frac{P_0V_0}{T_0}\ln{\left(1+\frac{\Delta V}{V_0}\right)}$$
Note the similarity now to the approximate answer which was portrayed in your course as the exact answer. We know that in this problem, ##\frac{\Delta T}{T_0}<<1## and ##\frac{\Delta V}{V_0}<<1##. If we make use of these conditions to expand the logarithmic terms in our exact answer in a Taylor series, and retain only the first terms in the expansions, we obtain: $$\ln{\left(1+\frac{\Delta T}{T_0}\right)}\approx \frac{\Delta T}{T_0}$$and $$\ln{\left(1+\frac{\Delta V}{V_0}\right)}\approx \frac{\Delta V}{V_0}$$If we substitute these relationships into our exact equation for ##\Delta S##, we obtain the following approximate equation for ##\Delta S##:

$$\Delta S\approx \frac{\Delta U+P_0\Delta V}{T_0}$$

So the equation they gave in your course is only an approximation to the exact entropy change.
 
  • #5
Thank you @Chestermiller for your effort and patience in laying the working out, it's helped clear things up a lot!

Chestermiller said:
Quasistatic reversible processes don't create entropy. But, entropy can be transferred between the system and surroundings if the reversible path is not adiabatic.

Is it right to say that this transfer of entropy doesn't make the path irreversible, and its only when entropy is created that a path is irreversible?

Chestermiller said:
If we evaluate the entropy change either by following a reversible path or, equivalently, by integrating the thermodynamic identity between the initial and final states

Since we can take a quasistatic path using the Sakur-Tetrode equation, why will we still end up creating entropy? Is this something to do with the lack of a ##VT^{\frac{1}{\gamma - 1}} = C## relation?

Also, is the integration of the Thermodynamic Identity representative of the non-quasistatic path, and if so/not why?

Chestermiller said:
$$\Delta S=nC_v\ln{\frac{T_f}{T_0}}+nR\ln{\frac{V_f}{V_0}}=nC_v\ln{\left(1+\frac{\Delta T}{T_0}\right)}+nR\ln{\left(1+\frac{\Delta V}{V_0}\right)}$$

Does this come from doing this?

##\int_{S_i}^{S_f} dS = \int_{T_i}^{T_f} \frac{nC_v}{T}dT + \int_{V_i}^{V_f} \frac{nR}{V}dV##

Thanks very much again for your guidance.
 
  • #6
WWCY said:
Thank you @Chestermiller for your effort and patience in laying the working out, it's helped clear things up a lot!
Is it right to say that this transfer of entropy doesn't make the path irreversible, and its only when entropy is created that a path is irreversible?
Yes.
Since we can take a quasistatic path using the Sakur-Tetrode equation, why will we still end up creating entropy? Is this something to do with the lack of a ##VT^{\frac{1}{\gamma - 1}} = C## relation?
I don't quite understand this question.
Also, is the integration of the Thermodynamic Identity representative of the non-quasistatic path, and if so/not why?
The change in entropy for a system can only be obtained by integrating dq/T along a reversible path. If the initial and final states of the system are the same for a non-quasistatic path as for a reversible path between the same two states, then the change in entropy for the reversible path gives the change in entropy for the non-quasistatic path. The Thermodynamic identity describes the relationship between changes in dU, dS, and dV for closely neighboring thermodynamic equilibrium states of a system. Since a reversible path consists of a continuous sequence of thermodynamic equilibrium states, integration of the identity between two end states gives the entropy change for both a reversible path and an irreversible path between the same two states.
Does this come from doing this?
##\int_{S_i}^{S_f} dS = \int_{T_i}^{T_f} \frac{nC_v}{T}dT + \int_{V_i}^{V_f} \frac{nR}{V}dV##
Yes.

I wrote a cookbook primer a while ago on how to determine the change in entropy for an irreversible process. This Physics Forums Insights article gives the step-by-step on how to proceed: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
 
  • #7
Chestermiller said:
WWCY said:
Since we can take a quasistatic path using the Sakur-Tetrode equation, why will we still end up creating entropy? Is this something to do with the lack of a VT1γ−1=CVT1γ−1=CVT^{\frac{1}{\gamma - 1}} = C relation?

I don't quite understand this question.

Apologies: What I meant to ask was this,

1. Am I right to say that quasistatic processes do not create new entropy?
2. If so, why, when we follow a quasistatic path from state 1 to state 2 via either Sakur-Tetrode or integrating the Thermodynamic Identity, do we still see an increase in entropy (in this question)?

Chestermiller said:
I wrote a cookbook primer a while ago on how to determine the change in entropy for an irreversible process. This Physics Forums Insights article gives the step-by-step on how to proceed: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Thank you! I'll give it a read.

Also, apologies if my questions aren't clear. These concepts really don't come naturally to me.

Thank you for your patience.
 
  • #8
WWCY said:
1. Am I right to say that quasistatic processes do not create new entropy?
Yes.
2. If so, why, when we follow a quasistatic path from state 1 to state 2 via either Sakur-Tetrode or integrating the Thermodynamic Identity, do we still see an increase in entropy (in this question)?
There are two ways in which the entropy of a system can change: by entropy generation within the system and by entropy transfer across its boundary with another system. In a reversible process, there is no entropy generation within the system; but entropy can be transferred across its boundary with its surroundings (via heat flow).

If we are dealing with an adiabatic irreversible process like the present question, there is entropy generated within the system, but no entropy exchange with its surroundings. However, all alternate reversible paths for the system between the same two end states are not adiabatic, involve heat exchange with its surroundings, and involve entropy transfer from the surroundings (but no entropy generation within the system). It is impossible to devise an alternate adiabatic reversible path for a system that has experienced an adiabatic irreversible process path.

In the present problem, all adiabatic reversible paths between the two end states involve heat and entropy exchange with the surroundings. Can you conceive of a reversible process path that can bring about the desired change?
 
  • #9
Chestermiller said:
If we are dealing with an adiabatic irreversible process like the present question, there is entropy generated within the system, but no entropy exchange with its surroundings. However, all alternate reversible paths for the system between the same two end states are not adiabatic, involve heat exchange with its surroundings, and involve entropy transfer from the surroundings (but no entropy generation within the system). It is impossible to devise an alternate adiabatic reversible path for a system that has experienced an adiabatic irreversible process path.

Let me attempt to explain this to test my understanding:

An adiabatic reversible path has the following relation between temperature and volume: ##VT^{\frac{1}{\gamma - 1}} = constant##. When we change variables ##V## and ##T## non-quasistatically as in this problem, the ##V-T## relation ceases to hold even though it's adiabatic. When we input these non-quasistatic changes into say, the Sakur-Tetrode equation, we are actually formulating a reversible path that isn't adiabatic (since ##VT^{\frac{1}{\gamma - 1}} = constant## doesn't hold). This means that while this path doesn't create entropy, it "allows" heat to flow into increase the system's entropy to reach our desired final state.

I hope this makes at least some sense!

Thank you again for your help.
 
  • #10
WWCY said:
Let me attempt to explain this to test my understanding:

An adiabatic reversible path has the following relation between temperature and volume: ##VT^{\frac{1}{\gamma - 1}} = constant##. When we change variables ##V## and ##T## non-quasistatically as in this problem, the ##V-T## relation ceases to hold even though it's adiabatic. When we input these non-quasistatic changes into say, the Sakur-Tetrode equation, we are actually formulating a reversible path that isn't adiabatic (since ##VT^{\frac{1}{\gamma - 1}} = constant## doesn't hold). This means that while this path doesn't create entropy, it "allows" heat to flow into increase the system's entropy to reach our desired final state.

I hope this makes at least some sense!

Thank you again for your help.
Yes. All correct.
 
  • #11
Thanks for all your guidance. It's helped a great, great deal.
 

1. What is the thermodynamic identity?

The thermodynamic identity is a mathematical expression that relates the changes in internal energy, temperature, and entropy of a system. It is derived from the first and second laws of thermodynamics and is often used to analyze and understand the thermodynamic properties of a system.

2. How is the thermodynamic identity derived?

The thermodynamic identity is derived by applying the first and second laws of thermodynamics to a closed system. It involves manipulating the equations for internal energy, heat, and work to arrive at the final form of the identity, which is dU = TdS - PdV.

3. What does each term in the thermodynamic identity represent?

The term dU represents the change in internal energy of the system, TdS represents the heat added or removed from the system, and PdV represents the work done by or on the system. These terms can also be interpreted as the energies associated with the system's temperature, entropy, and volume, respectively.

4. How is the thermodynamic identity used in practical applications?

The thermodynamic identity is used in a variety of practical applications, including analyzing and predicting the behavior of gases, understanding the efficiency of heat engines, and determining the conditions for phase transitions. It is also used in thermodynamic calculations and experiments to accurately measure and predict the properties of a system.

5. What are some limitations of the thermodynamic identity?

While the thermodynamic identity is a useful tool in thermodynamics, it does have some limitations. It assumes that the system is in a state of equilibrium and that all processes are reversible. In real-world situations, these assumptions may not always hold, and the identity may not accurately describe the behavior of the system.

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