# Homework Help: Need clarification on adiabatic process and heat capacity

1. Feb 6, 2015

### worryingchem

For reversible process, we got dq=0, so dU = dw = -P*dV = Cv*dT.
What I don't get is where did the Cv*dT come from?
I remember q=C*dT, but dq = 0 so I'm not sure what that means?

Also, where does Cv*dT apply? In irreversible process, is dU also equal to Cv*dT? And when would we ever use Cp?

Any help clarifying this is appreciable. Thank you.

2. Feb 6, 2015

### Quantum Defect

For an ideal gas, Delta U = Cv Delta_T [i.e. the energy is a function of temperature only (no intermolecular forces)]

You use Cp when you want to find the heat at constant pressure. Also, for an ideal gas, Delta_H = Cp Delta_T.

3. Feb 6, 2015

### rude man

dQ = dU + pdV ... quasi-static process
CV ={∂Q/∂T}V = {∂U/∂T}V since dV = 0,
but for an ideal gas, U = U(T) only, so ∂ → d and
CV = dU/dT.
"Quasi-static" means process is done very slowly.
"Reversible" means a system, going from thermodynamic coordinates A to B, can be reversed without any changes to the external environment (external to the system). A reversible process is always quasi-static, plus there must be no dissipative forces like friction involved.
So therefore, dQ = dU + pdV can apply to an irreversible process if irreversibility is due to dissipative forces like friction. But, it must be quasi-static.

You can derive dQ = CpdT - Vdp for an ideal gas undergoing a quasi-static process.

4. Feb 6, 2015

### Staff: Mentor

See the definitions of Cv and Cp in the officially recommended Physics Formulary (officially recommended by Physics Forums) at this link:

The definitions you are looking for are in Section 8.3. Note that these definitions are functions of the material properties U (internal energy) and H (enthalpy), and not in terms of the process-path-dependent quantity Q (which is not a material property). So the heat capacities are thermodynamic properties of the material, and not features of any specific process.

Chet

5. Feb 7, 2015

### rude man

My good friend Chet, who knows lots more about thermodynamics than I do, and I nevertheless disagree on this punctilio. Not that what he says is wrong of course, it's just that very few introductory physics students have even encountered enthalpy. I can cite sources like Mark Zemansky, an internationally recognized thermodynamics professor, an MIT source, my Resnick & Halliday text, many others that define CV and CP the way I have. And I'll bet your textbook does the same. Chet's formulas can easily be derived directly from my definitions.

6. Feb 7, 2015

### Staff: Mentor

Hi Rude man. Yes, we belong to a mutual admiration society and, yes, we have agreed to disagree on this point. With all due respect to these famous texts (and others), my gripe is the way they teach the concept of heat capacity to their fledgling students. If there were not problems with the way it is taught, then there would not be as much confusion on the part of the students, as evidenced by the huge number of past posts on this issue in Physics Forums. There has got to be a better way.

Chet

7. Feb 13, 2015

### worryingchem

Thank you for all your replies. I forgot about the properties of perfect gases.

So because it's an ideal gas, where they are point particles with no volumes, you can use the equation Cv =(dQ/dT)V = (dU/dT)V because dU= dq + PextdV. And got the result dU=CvdT, meaning that U is temperature dependent only for perfect gas. I don't understand the difference between ∂ and d that you pointed out. Can you explain what the differences between them are?

And thank you Chestermiller for the link, I got that dH=dq for constant pressure so dH = CpdT. Meaning H is temperature dependent only for perfect gas?

I also looked at the entropy part in 8.4 of the link you sent since we've started learning about entropy. I don't quite get what entropy is suppose to be and what kind of information does it give us so can someone clarify that?

In class, we got dS = dq/T = dU/T + dw/T.
So for perfect gases and reversible processes, dS = (Cv/T)dT + (nR/V)dV.
And ΔS=Cvln(Tf/Ti) + nR ln(Vf/Vi). We looked at the Carnot cycle and got that ΔS=0.

And for perfect gas and irreversible processes, dS = (Cv/T)dT + PexternaldV.
So ΔS = Cvln(Tf/Ti) + Pexternal(Vf - Vi)?
How do I show that ΔS is less than 0 for this?
I thought about the irreversible example where you quickly pushes a piston until it won't budge. So then PΔV is negative because of decrease in volume. But then how do I show that abs{PΔV} is greater than Cvln(Tf/Ti)?

Any help explaining these to me will be greatly appreciated. Thank you.

8. Feb 13, 2015

### Staff: Mentor

For real gases beyond the ideal gas region (i.e., low pressures), H also depends on pressure.
This contains the short set of notes I have written on entropy. I have been told that it is very understandable.

This relationship gives the change in entropy of an ideal gas irrespective of whether the process is reversible or irreversible.

This relationship is not correct for getting ΔS for an irreversible process. To get ΔS for an irreversible process, you need to dream up a reversible path between the same initial and final equilibrium states and calculate the integral of dQ/T for that reversible path. But, for an ideal gas, you don't even need to do this because the equation you presented in the previous paragraph applies to any process, either reversible or irreversible.

Chet

9. Feb 20, 2015

### worryingchem

I got that the ΔSreversible ≥ ΔSirreversible through Clausian inequality.

But I don't get why, in isothermal condition, is ΔStotal > 0 for irreversible process.
If ΔSsys is lowered for the irreversible process, then wouldn't that mean ΔSsurrounding will be of an even lesser magnitude, and I don't get how that happen for the surrounding.

Can you could explain how the ΔSsurrounding was even less than the ΔSsys for irreversible? Thank you.

10. Feb 20, 2015

### Staff: Mentor

I never said this, and it is not correct. For a change from one equilibrium thermodynamic state of a system to another, the change in entropy is exactly the same for all reversible and irreversible paths between the two states. The entropy change can be calculated by first identifying a reversible path between the initial and final equilibrium states, and then calculating the integral of dQ/T for that reversible path. This will also be the change in entropy for any of the irreversible paths. If one calculates the integral of dQ/TI (where TI is the temperature at the interface with the surroundings where the heat transfer dQ is occurring) for any of the irreversible paths, one obtains a value that is lower than the value for the reversible path.
I don't really follow any of this.

Chet

11. Mar 5, 2015

### worryingchem

In the Physics Formulary, it shows that dS ≥ dQ/T and that for an irreversible cycle, ∫ dqirrev/T < 0 while a reversible is 0.
I got to that point by using
dwirrev ≥ dwrev
dq ≤ dqrev
so dq/T ≤ dS and then can't I say ΔSirreversible ≤ ΔSrev?

For my second question about ΔStotal > 0 for irreversible process during isothermal condition.
Starting with
ΔStotal = ΔSsys + ΔSsurrounding.
In isothermal condition ΔSsys = nR*ln(Vf/Vi)
So in gas compression, ΔSsys would be negative. To make ΔStotal > 0 true, ΔSsurrounding would have to be greater in magnitude than ΔSsys.
This is where I don't understand how ΔSsurrounding will be greater than ΔSsys. Is there an example you can give for this?

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12. Mar 6, 2015

### Staff: Mentor

You're obviously very confused, but I am going to help you. Let me start out by saying that those equations in the Physics Formulary are correct, but there is a constraint that they don't mention (and is not mentioned in most thermodynamics books) that I'm going to make you aware of. But, before I try to address your questions, I want to take a step backwards (temporarily). I have prepared a short tutorial on the first and second laws of thermodynamics as a supplement for students who are struggling with the subject. It presents the material in a little different way than it has been taught to you so far, and is designed to address just the kind of questions you are asking. I'm going to append it to this response.

Now, I have a couple of questions for you to stimulate your thinking.

1. When you heat a turkey in the oven (an irreversible process), is the temperature the same at all locations throughout the turkey? If not, and you are trying to apply the Clausius inequality to the turkey for this irreversible process, what temperature T do you plug into the inequality?

2. If you expand an ideal gas in a cylinder irreversibly (say by suddenly removing a weight from the top of the piston) and you are trying to carry out the process isothermally (say by having the cylinder immersed in a constant temperature bath), the bath is going to have to supply heat to the gas to make up for the work that the gas does in expanding against the piston. Is the temperature of the gas the same at all locations throughout the gas during this irreversible expansion? If not, and you are trying to apply the Clausius inequality to the gas for this irreversible process, what temperature T do you plug into the inequality?

Now for the brief tutorial.

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 $\dot{q}(t)$ represent the rate of heat addition across the interface between the system and the surroundings at time t, and let $\dot{w}(t)$ 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),
$$\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W$$
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 $\dot{q}(t)$ and $\dot{w}(t)$ 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 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 $\dot{q}(t)$ and $\dot{w}(t)$).

Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:
$$\dot{w}(t)=P_I(t)\dot{V}(t)$$
where, again, PI is the pressure at the interface with the surroundings, and $\dot{V}(t)$ 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

$P_I(t)=P(t)$ (reversible process path)

Therefore, $\dot{w}(t)=P(t)\dot{V}(t)$ (reversible process path)

Another feature of reversible process paths is that they are carried out very slowly, so that $\dot{q}(t)$ and $\dot{w}(t)$ 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 $\dot{q}(t)$ and the rate of doing work $\dot{w}(t)$ 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:
$$Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}$$
$$W=\int_{t_i}^{t_f}{\dot{w}(t)dt}$$
In the present section, we will be introducing a third integral of this type (involving the heat transfer rate $\dot{q}(t)$) 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:
$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}$$
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 dream up 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:

$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq\Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}$$
where $\dot{q}_{rev}(t)$ 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.

Last edited: Mar 6, 2015
13. Mar 20, 2015

### worryingchem

Hi, for the turkey and perfect gas expansion example, since they're irreversible, the temperature is not homogenous and so the temperature of the interface have to be used.

dS = dqrev/T as the reversible process give the maximum value of this "I" variable that was then termed S.
So when calculating the irreversible process it gives a value that is less than the maximum value, dS.
So dS > dqirrev/TI

To hopefully clarify my question:
Using the example of a perfect gas isothermally expanding and trying to calculate the different parts of the equation ΔStotal = ΔS + ΔSsurrounding.

For reversible ΔStotal = 0
So ΔS = -ΔSsurrounding
ΔS = nR ln(Vf/Vi). So the amount of volume the gas expanded is the same amount of volume the surrounding lost.

For irreversible ΔStotal > 0
so ΔS > -ΔSsurrounding
ΔS would still be nR ln(Vf/Vi).
So what would ΔSsurrounding be? Is it dqirrev/TI? Or is the equation← ΔStotal = ΔS + ΔSsurrounding not clear?

Last edited: Mar 20, 2015
14. Mar 20, 2015

### Staff: Mentor

Yes.
A reversible change in the system can be brought about without specifically specifying what happens in the surroundings. All that is required to assure that the integral of dq/TI for the system is equal to the change in entropy of the system is for the system to be subjected to a path in which the system is only slightly removed from thermodynamic equilibrium over the entire path. This can be accomplished by many different methods with regard to the surroundings, such as, for example, moving the piston manually by hand. In some cases, the change in the surroundings does not even have to be reversible (as in the case of moving the piston by hand). So great flexibility is possible with regard to what is done in the surroundings.

However, if we wish to consider only situations in which it is possible to return both the system and the surroundings reversibly to their original states (without significantly affecting anything else), then I like to restrict the tools available to us with regard to the surroundings to just two kinds: 1. Very small weights that can be added or removed from the piston to do work (reversibly) and 2. An infinite array of constant temperature baths at different temperatures to exchange heat with the system (reversibly). Restricting ourselves to only these sets of tools allows us to constrain much more precisely what the surroundings are doing and guarantee that the total change in entropy for the combination of system and surroundings is equal to zero.

This is one of those situations in which we can't allow too much flexibility for what happens in the surroundings. If the volume of the gas expands at constant temperature, its pressure has to decrease. This requires the force per unit area imposed by the surroundings on the piston to also decrease. If there is gas imposing pressure by the surroundings, then its force per unit area must decrease, but it is not going to decrease on its own. So there is going to have be some additional unspecified mechanical equipment to do this, and this equipment is also going to have to be taken into account in determining the change in entropy of the surroundings. Things get pretty complicated pretty fast.

It is much easier to restrict ourselves to the two types of equipment I discussed above: weights and constant temperature baths.

In the isothermal reversible expansion case under consideration, we put the cylinder containing the gas into a constant temperature bath at the same temperature as the gas, and we gradually remove tiny weights from the top of the piston in order allow the gas to expand reversibly. The heat absorbed from the constant temperature bath is exactly equal to the work done in raising the weights (as weights are removed from the top of the piston). So the change in entropy of the system is Q/T, and the change in entropy of the surroundings (the reservoir) is -Q/T. So the total change in entropy of system plus surroundings is zero.
Well, if we work with our same tool kit, then, to bring about an irreversible expansion of the gas, rather than removing weights gradually from the piston, we will remove a substantial amount of weight suddenly, and then allow that gas to expand spontaneously until it re-equilibrates with the remaining weights on the piston. If the final gas volume is equal the same value as in the reversible case, so also will be the final temperature and pressure. (This can be guaranteed by removing the same amount of weight as in the reversible case, except suddenly). So the change in entropy of the system will be the same as in the reversible case. But, in the irreversible case, less heat will flow from the surroundings to the system, so the change in entropy for the combination of system and surroundings will be greater than zero.

If you would like to solve this specific problem, I can help you do it. In that way you can see exactly how the irreversible case plays out. Just say the word, and we can begin (if you are interested).

Chet

15. Mar 23, 2015

### worryingchem

Yes, please. Where would we start for the irreversible case?

16. Mar 23, 2015

### Staff: Mentor

Don't worry about that. I'll specify the problem, and I'll ask you leading questions to help us move along. We are going to consider two different cases: adiabatic irreversible expansion and isothermal irreversible expansion. Which would you like to work on first?

Chet

17. Mar 25, 2015

### worryingchem

18. Mar 25, 2015

### Staff: Mentor

We have n moles of an ideal gas contained within a vertical cylinder of cross sectional area A by a frictionless piston of mass M. On top of the piston sits a mass m. The cylinder sits within an evacuated chamber (vacuum), so that the absolute pressure outside the cylinder is zero.

The cylinder is partially immersed in a constant temperature bath which maintains the gas temperature at its interface with the cylinder wall equal to T for all time (at least in the region where the cylinder is immersed). In the region where the cylinder is not immersed, as well as at the piston, the cylinder and piston are perfectly insulated, so that no heat can enter or leave over that part of the boundary. (During the process we are going to describe, the gas temperature can vary with time and position throughout the gas, but not on the portion of the boundary that is maintained in the bath at constant temperature T).

At time t = 0, the mass m is suddenly removed from the top of the piston; this allows the gas to expand rapidly and spontaneously, eventually attaining a new thermodynamic equilibrium state. For this irreversible expansion process, we are to determine:

1. The physical conditions at the final thermodynamic equilibrium state
2. The work W done by the gas on the piston during the irreversible expansion process
3. The heat Q transferred from the constant temperature bath to the gas during the irreversible expansion process
4. The change in Entropy of the gas between the initial and final equilibrium states
5. The heat transferred Q divided by the temperature T at the boundary where the heat transfer takes place during the irreversible expansion process

What is the gas pressure in the initial equilibrium state before the mass m is removed?
What is the initial volume of the gas?
What is the gas pressure in the final equilibrium state?
What is the final volume of the gas?

That's it for now. I want to give you a chance to read the problem specification and to ask questions. After you have provided answers to the four questions I asked, we will continue further.

Chet

19. Mar 31, 2015

### worryingchem

So for the 4 questions:
A.) P = F/A. So $P_i = \frac{9.81(M + m)}{A}$
B.) Using PV=nRT. $V_i = nRT \frac{A}{9.81(M+m)}$
C.) $P_f = \frac{9.81M}{A}$
D.) $V_f = nRT \frac{A}{9.81(M)}$

I have a question about the temperature of the gas. For the temperature of the gas, while the temperature at the boundary is constant throughout the process, the temperature of the gas inside the cylinder is relative to the time and position. But at equilibrium the temperature of the gas is the same throughout (inside and at boundary)?

20. Mar 31, 2015

### Staff: Mentor

Excellent job. I think that, for conciseness, instead of the 9.81, we should simply put a g there.
Yes. This is correct. Very well said.

Let's now move on to item 2 on our to-do list: determining the work that the gas (our system) does on its surroundings. Let FI(t) represent the force that the gas exerts on the piston (i.e., at the interface between the gas and the piston) at any arbitrary time t during the irreversible expansion. Going back to introductory physics, please apply Newton's 2nd law to obtain a force balance equation on the piston. We will be using this force balance to determine the work that the gas does on its surroundings (i.e., the piston) during the irreversible expansion between the initial and final equilibrium states. Although this may sound kind of unorthodox to you, please bear with me.

Chet

21. Mar 31, 2015

### worryingchem

For this is it, FI(t) - Mg = M*a(t)

Last edited: Mar 31, 2015
22. Mar 31, 2015

### Staff: Mentor

Yes. This is correct. So,
$$F_I(t)= Mg+M\frac{dv}{dt}$$
where $v=\frac{dx}{dt}$ is the velocity of the piston and x(t) is the upward displacement of the piston from its initial equilibrium position. If we multiply this equation by the velocity v, we obtain:
$$F_I(t)\frac{dx}{dt}= Mg\frac{dx}{dt}+Mv\frac{dv}{dt}=Mg\frac{dx}{dt}+\frac{1}{2}M\frac{dv^2}{dt}$$
If we integrate this equation from t = 0 to t = arbitrary t, we obtain:
$$\int_0^{x(t)}{F_Idx}=Mgx(t)+\frac{1}{2}Mv^2(t)$$
The left hand side of this equation is the cumulative amount of work W(t) that the gas does on the piston (i.e., the surroundings) up to time t. So,
$$W(t)=Mgx(t)+\frac{1}{2}Mv^2(t)$$
The first term on the right hand side is the change in potential energy of the piston, and the second term is the change in kinetic energy of the piston.

I'm going to stop here to make sure you are comfortable with this so far. Next I want to talk about what happens with regard to the motion of the frictionless piston.

Chet

23. Apr 3, 2015

### worryingchem

24. Apr 3, 2015

### Staff: Mentor

OK. We know that the piston is frictionless. We also know that, when we remove the mass m, the piston will no longer be in equilibrium and it will accelerate upwards gaining kinetic energy (and potential energy). The question is, "what is the subsequent motion of the piston?" Here are some possibilities:

1. The piston will initially speed up, but then will slow down as it approaches the final equilibrium position, and will stop at the final equilibrium position.

2. The piston will overshoot the equilibrium position and will then exhibit undamped oscillation about the equilibrium position forever.

3. The piston will overshoot the equilibrium position, but will then exhibit damped oscillation about the equilibrium position (even though it is frictionless) until the oscillation dies out.

What do you think?

We need to be able to answer this question so that we can determine the cumulative amount of work that the gas does on the piston up to infinite time, W(t = ∞), when the final thermodynamic equilibrium state of the gas has been attained.

Chet

25. Apr 3, 2015

### worryingchem

The piston would initially speed up and then slow down and stop at the final equilibrium since the piston mass M will act as the opposing force to FI which decreases over time as the gas volume increases. I don't think there is any oscillation since the piston isn't hanged on a spring.