Heat capacity under constant pressure or volume question

[Joker93]

HOMEWORK POSTED IN WRONG FORUM, SO NO TEMPLATE

I have encountered a problem at the university in which there is a thermally isolated container of constant volume in which the number of particles and temperature change with time(the temperature increases). The change in particle number ensures constant pressure. The question was to find the heat that is needed to change that temperature.
So, naturally I would conclude from the beginning that δQ=0 since it's thermally isolated but then again, particles flow out of the system. So, I used δQ=C*dT.

Our professor said that C(the heat capacity) should be that of a process of constant pressure. But, I don't quite understand why. His reasoning was that the system in reality is not in constant volume since particles flow out of it. But, my reasoning is that while particles do flow out, the remaining particles spread out in the original volume of the container and it's just the density of the gas that changes.

Could someone explain to me what is the correct reasoning here?
Thanks in advance!.

 

[Chestermiller]

Please provide the exact wording of the problem.

 

[hmmm27]

.

 

[Joker93]

Please provide the exact wording of the problem.

A thermally isolated container contains ideal has of increasing temperature. Its volume is constant and the process is adiabatic. Also, there is a tiny hole so that the inside of the container is in contact with the outside of it which has pressure of 1 atm. We need to find the heat that causes that temperature change.
So, to solve the problem I essentially used that the pressure inside the container is always 1atm and that particles are exchanged in order for the temperature to change. I think there is no other way since the process is adiabatic and isochoric.
But, as my professor pointed out, by solving it through the relation δQ=C*dT with C the heat capacity of the gas. I have solved the problem but my professor pointed out that C should be of constant pressure rather than constant volume. I don't understand that. Since both pressure and volume are constant, why should I use the heat capacity under constant pressure? His argument was that since particles are flowing out of the system, its volume changes. But, the volume is constant as stated in the problem.

 

[Joker93]

Assumedly it's a tank with a baggie on the end for overflow gas.

In which case you'll probably be heating all the gas, unless there's a one-way valve. So - constant pressure: yes; constant volume, no

Constant volume is stated from the beginning. It's a given of the problem. Also given is that it's adiabatic and the container has a hole that allows contact with air of P=1atm.

 

[Chestermiller]

A thermally isolated container contains ideal has of increasing temperature. Its volume is constant and the process is adiabatic. Also, there is a tiny hole so that the inside of the container is in contact with the outside of it which has pressure of 1 atm. We need to find the heat that causes that temperature change.

If the process is adiabatic, how can the temperature be increasing? Also, if gas is escaping through a hole, the number of moles of gas in the container is changing.

 

[Joker93]

If the process is adiabatic, how can the temperature be increasing? Also, if gas is escaping through a hole, the number of moles of gas in the container is changing.

Well, that's the whole point of the problem. The number of particles is changing and so the temperature changes.

 

[hmmm27]

It makes at least a bit of sense if it isn't described as adiabatic. Heat the tank, the air warms, some warm air escapes, apply calculus.

 

[Chestermiller]

It makes at least a bit of sense if it isn't described as adiabatic. Heat the tank, the air warms, some warm air escapes, apply calculus.

Adiabatic means that there is no heat transfer. So, it makes no sense.

 

[Joker93]

Adiabatic means that there is no heat transfer. So, it makes no sense.

Well, that's the problem I was given..

 

[Joker93]

Adiabatic means that there is no heat transfer. So, it makes no sense.

It makes at least a bit of sense if it isn't described as adiabatic. Heat the tank, the air warms, some warm air escapes, apply calculus.

PV=k(NT). Because PV is constant, while T is changing, so does N. So, also (NT) is constant and so is the energy of the container because its described as E=3/2k(NT).

 

[hmmm27]

Oh, I see... start off with a cold chamber... still not adiabatic unless there's an outer chamber.

 

[Chestermiller]

PV=k(NT). Because PV is constant, while T is changing, so does N. So, also (NT) is constant and so is the energy of the container because its described as E=3/2k(NT).

I agree with you and not your professor. You should be using Cv and not Cp, but that is because the focus should be on the internal energy of the remaining gas in the container, and not the enthalpy. Also, the gas which leaves the tank takes energy with it, and the gas that has left was at a lower temperature than the gas that has remained. I will be back later to provide a more complete analysis of this of this problem.

Chet

 

[Chestermiller]

The open system (control volume) version of the first law of thermodynamics applied to this particular problem gives:
$$\frac{dU}{dt}=\frac{dQ}{dt}-\dot{m}h\tag{1}$$where U is the internal energy of the gas remaining in the container, Q is the cumulative amount of heat added, $\dot{m}$ is the molar flow rate out of the container, and h is the enthalpy per mole of the stream exiting the container. The internal energy U is given by:$$U=nC_v(T-T_0)\tag{2}$$ where n(t) is the number of moles of gas remaining in the container at time t, T(t) is the gas temperature in the tank at time t, and $T_0$ is the arbitrary datum temperature for zero internal energy. The number of moles in the container at time t is given by:$$n=\frac{PV}{RT}\tag{3}$$, where P is the the pressure (1 atm.) and V is the volume of the container. The enthalpy per mole of the stream leaving the container at time t is given by: $$h=C_v(T-T_0)+Pv=C_v(T-T_0)+RT\tag{4}$$ where v is the molar volume of the exit stream, which is equal to RT/P. The molar flow rate $\dot{m}$ out of the container is equal to minus the rate of change of the number of moles of gas inside the container:$$\dot{m}=-\frac{dn}{dt}=\frac{PV}{RT^2}\frac{dT}{dt}\tag{5}$$If we substitute Eqns. 2-5 into Eqn. 1, we obtain:
$$\frac{dQ}{dt}=C_p\frac{PV}{RT}\frac{dT}{dt}\tag{6}$$where $C_p$ is the molar heat capacity at constant pressure $(=C_v+R)$. We can immediately integrate Eqn. 6 subject to the initial condition $T=T_i$ at t = 0 to obtain:
$$Q=C_p\frac{PV}{R}\ln{(T/T_i)}\tag{7}$$

 

[Joker93]

The open system (control volume) version of the first law of thermodynamics applied to this particular problem gives:
$$\frac{dU}{dt}=\frac{dQ}{dt}-\dot{m}h\tag{1}$$where U is the internal energy of the gas remaining in the container, Q is the cumulative amount of heat added, $\dot{m}$ is the molar flow rate out of the container, and h is the enthalpy per mole of the stream exiting the container. The internal energy U is given by:$$U=nC_v(T-T_0)\tag{2}$$ where n(t) is the number of moles of gas remaining in the container at time t, T(t) is the gas temperature in the tank at time t, and $T_0$ is the arbitrary datum temperature for zero internal energy. The number of moles in the container at time t is given by:$$n=\frac{PV}{RT}\tag{3}$$, where P is the the pressure (1 atm.) and V is the volume of the container. The enthalpy per mole of the stream leaving the container at time t is given by: $$h=C_v(T-T_0)+Pv=C_v(T-T_0)+RT\tag{4}$$ where v is the molar volume of the exit stream, which is equal to RT/P. The molar flow rate $\dot{m}$ out of the container is equal to minus the rate of change of the number of moles of gas inside the container:$$\dot{m}=-\frac{dn}{dt}=\frac{PV}{RT^2}\frac{dT}{dt}\tag{5}$$If we substitute Eqns. 2-5 into Eqn. 1, we obtain:
$$\frac{dQ}{dt}=C_p\frac{PV}{RT}\frac{dT}{dt}\tag{6}$$where $C_p$ is the molar heat capacity at constant pressure $(=C_v+R)$. We can immediately integrate Eqn. 6 subject to the initial condition $T=T_i$ at t = 0 to obtain:
$$Q=C_p\frac{PV}{R}\ln{(T/T_i)}\tag{7}$$

Yes, his was also the solution. From your previous comment, the temperature is rising so we are supposed to conclude that particles are added in the system.
Also, as I do not know about enthalpy yet, I did not understand your argument about using Cv over Cp. I don't suggest using Cv instead of Cp, I am just saying that I do not understand why distinguish between the two in a problem where both pressure and volume remain constant.
Also, thanks for the detailed analysis.

 

[Chestermiller]

Actually, as the contents of the container is heated, molecules are leaving the container, not being added.

Eqn. 6 of my previous post really tells an important story. Written is a slightly different way, it reads:
$$dQ=\frac{PV}{RT}C_pdT=nC_pdT$$But this is the equation that you would get if the gas was being heated at constant pressure (as your professor contended), not at constant volume. But, how can this be if the container is constant volume? There is a simple physically intuitive way of resolving this issue. Imagine that, at time t, the temperature of the gas in the container is T and the number of moles is n. Picture an imaginary membrane surrounding the gas within the container at this time. Now imagine that, during the time interval between t and t + dt, you add a small amount of heat dQ to the gas. The temperature of the gas in the container will rise by dT, and, because the pressure is constant (i.e., the gas is in contact with the atmosphere through the hole in the container), the membrane that contains the n moles of gas will have to bulge out through the hole in the container to allow the volume to increase slightly. So the net effect is that, over the time interval between t and t + dt, the gas that was in the container at time t expands at constant pressure as a result of adding dQ. Once the time interval dt is over, we put a new membrane around the gas in the container at time t + dt, and start over.

Hope this helps.

 

[Joker93]

Actually, as the contents of the container is heated, molecules are leaving the container, not being added.

Eqn. 6 of my previous post really tells an important story. Written is a slightly different way, it reads:
$$dQ=\frac{PV}{RT}C_pdT=nC_pdT$$But this is the equation that you would get if the gas was being heated at constant pressure (as your professor contended), not at constant volume. But, how can this be if the container is constant volume? There is a simple physically intuitive way of resolving this issue. Imagine that, at time t, the temperature of the gas in the container is T and the number of moles is n. Picture an imaginary membrane surrounding the gas within the container at this time. Now imagine that, during the time interval between t and t + dt, you add a small amount of heat dQ to the gas. The temperature of the gas in the container will rise by dT, and, because the pressure is constant (i.e., the gas is in contact with the atmosphere through the hole in the container), the membrane that contains the n moles of gas will have to bulge out through the hole in the container to allow the volume to increase slightly. So the net effect is that, over the time interval between t and t + dt, the gas that was in the container at time t expands at constant pressure as a result of adding dQ. Once the time interval dt is over, we put a new membrane around the gas in the container at time t + dt, and start over.

Hope this helps.

But,shouldn't we analyze what is happening in the system using something like a control volume that is inside the container since we are concerned only for what is inside of it? Also, from PV=NkT, because PV is constant, isn't also NT? So, as the temperature rises, the number of molecules decreases.

 

[Chestermiller]

But,shouldn't we analyze what is happening in the system using something like a control volume that is inside the container since we are concerned only for what is inside of it?

That's what my previous analysis did. It treated the container as a control volume, and allowed for gas escaping. But, it the end, it gave exactly the same answer.

Also, from PV=NkT, because PV is constant, isn't also NT? So, as the temperature rises, the number of molecules decreases. w

The number of molecules inside the container decreases, but the number of molecules that exited the container is contained within the bulge in the imaginary membrane outside the container. The imaginary membrane allows us to focus on a closed system in which the total number of molecules is constant during the time interval dt.

 

[Chestermiller]

In my judgment, this problem involves way too much of a conceptual leap for a novice student such as yourself to be subjected to. You should not feel bad that you have been struggling with it. I think your professor should have held off until you had learned about the control volume version of the first law.

Chet

 

[Joker93]

That's what my previous analysis did. It treated the container as a control volume, and allowed for gas escaping. But, it the end, it gave exactly the same answer.

The number of molecules inside the container decreases, but the number of molecules that exited the container is contained within the bulge in the imaginary membrane outside the container. The imaginary membrane allows us to focus on a closed system in which the total number of molecules is constant during the time interval dt.

So, we can solve the problem with a moving control volume where the particle number does not change and we can also solve it by considering a static control volume(that of the inside of the container) with changing particle number, right? But, because in reality the volume does change(like in the moving control volume case), we must use Cp rather than Cv. Did I get it right?

In my judgment, this problem involves way too much of a conceptual leap for a novice student such as yourself to be subjected to. You should not feel bad that you have been struggling with it. I think your professor should have held off until you had learned about the control volume version of the first law.

Chet

Well, the course is considered as being an advanced undergraduate thermodynamics and statistical physics course, but the lectures give too simple examples for us to be able to solve problems like this. In any way, in the end I learned a lot..

 

[Chestermiller]

So, we can solve the problem with a moving control volume where the particle number does not change and we can also solve it by considering a static control volume(that of the inside of the container) with changing particle number, right? But, because in reality the volume does change(like in the moving control volume case), we must use Cp rather than Cv. Did I get it right?

Yes. Very nice.

 

[Joker93]

Yes. Very nice.

Thanks a lot for the help!

 

[Joker93]

Yes. Very nice.

Can I also ask something else? When particles are extracted from the system, why is work equal to zero? Or, put differently, when the control volume deforms, isn't the work being done on the system not equal to zero?
In your solution, I think that you implied that the work being done on the system is zero. Also, my professor's solution also uses that because the full solution uses the fact that δQ=Cp*dT.

 

[Joker93]

@Chestermiller I think I got it: Because we are talking about an ideal gas where its particles are not interacting(the forces between them are nonexistent), the work that the particles do to each other as they get out of the container is zero. Thus, the change in temperature is due to the flow of particles that flow outside the container which causes a heat flow(or just heat) outside the container.
Did I get it right?
If I did, then if the gas was not ideal, then we should have also considered changes in work of the system. And because (I think) the process is irreversible, we couldn't find exactly how much heat was exchanged, right?

 

[Chestermiller]

Can I also ask something else? When particles are extracted from the system, why is work equal to zero? Or, put differently, when the control volume deforms, isn't the work being done on the system not equal to zero?
In your solution, I think that you implied that the work being done on the system is zero. Also, my professor's solution also uses that because the full solution uses the fact that δQ=Cp*dT.

This equation does not imply that the work being done is zero. For that to be the case, there would have to be a Cv in the equation, not a Cp. The gas in the container is doing work to force the gas ahead of it out of the container. So most of the gas in the container is doing work at constant pressure. This is all easier to visualize if you imagine that the gas is a continuum rather then being comprised of individual molecules (particles). Do you think you can do that?

 

[Chestermiller]

@Chestermiller I think I got it: Because we are talking about an ideal gas where its particles are not interacting(the forces between them are nonexistent), the work that the particles do to each other as they get out of the container is zero. Thus, the change in temperature is due to the flow of particles that flow outside the container which causes a heat flow(or just heat) outside the container.
Did I get it right?
If I did, then if the gas was not ideal, then we should have also considered changes in work of the system. And because (I think) the process is irreversible, we couldn't find exactly how much heat was exchanged, right?

In my judgment, none of this is right. As I said, it is easier to visualize if you imagine the gas to be a continuum.

 

[Joker93]

This equation does not imply that the work being done is zero. For that to be the case, there would have to be a Cv in the equation, not a Cp. The gas in the container is doing work to force the gas ahead of it out of the container. So most of the gas in the container is doing work at constant pressure. This is all easier to visualize if you imagine that the gas is a continuum rather then being comprised of individual molecules (particles). Do you think you can do that?

So, if it is work that "pushes" the particles out of the container, then why do we calculate heat instead of work?

 

[Chestermiller]

So, if it is work that "pushes" the particles out of the container, then why do we calculate heat instead of work?

The work is included in the enthalpy change at constant pressure.

 

[Joker93]

The work is included in the enthalpy change at constant pressure.

To be honest with you, we have not covered the concept of enthalpy yet. I just saw the answer that you got from your way and compared it to mine. We agree on the answer. I calculated Q directly from δQ=Cp*dT and PV=NkT. So, I have not used the concept of enthalpy. Maybe I am getting something wrong here but isn't δQ=Cp*dT used when δW=0(so dE=δQ)? If so, then using δQ=Cp*dT I implied that δW=0. Μaybe I am misinterpreting something here?

 

[Chestermiller]

To be honest with you, we have not covered the concept of enthalpy yet. I just saw the answer that you got from your way and compared it to mine. We agree on the answer. I calculated Q directly from δQ=Cp*dT and PV=NkT. So, I have not used the concept of enthalpy. Maybe I am getting something wrong here but isn't δQ=Cp*dT used when δW=0(so dE=δQ)? If so, then using δQ=Cp*dT I implied that δW=0. Μaybe I am misinterpreting something here?

W is equal to zero when V is constant, not P.

 

[Joker93]

W is equal to zero when V is constant, not P.

Sorry for being such a pain, but I don't understand what you are implying.

 

[Chestermiller]

Sorry for being such a pain, but I don't understand what you are implying.

No problem. If the volume of a parcel of gas containing a constant number of molecules is constant, then it is not doing any work on the surrounding gas. If the volume of a parcel of gas containing a constant number of molecules is increasing, the parcel is doing work on the surrounding gas. So, at constant volume, work is being done, and, at constant pressure, work is being done.

 

[Joker93]

No problem. If the volume of a parcel of gas containing a constant number of molecules is constant, then it is not doing any work on the surrounding gas. If the volume of a parcel of gas containing a constant number of molecules is increasing, the parcel is doing work on the surrounding gas. So, at constant volume, work is being done, and, at constant pressure, work is being done.

In your last sentence, you meant "at constant volume, work is not being done", right?

 

[Chestermiller]

In your last sentence, you meant "at constant volume, work is not being done", right?

Yes. Thanks.

 

[Joker93]

@Chestermiller I tried to solve it in 2 ways: first considering a control volume inside the container so that ΔV=0 and ΔΝ not equal to zero. Second, I considered the whole control volume(also outside the container) so that ΔV not equal to zero and ΔΝ=0. Aren't both ways of doing it valid(I do not know the concept of enthalpy so I can not do it your way)?
Doing these two ways, I do not obtain the same answer. In the second case, for N=constant, I got that ΔQ and T are related linearly and not logarithmically.

 

[Chestermiller]

@Chestermiller I tried to solve it in 2 ways: first considering a control volume inside the container so that ΔV=0 and ΔΝ not equal to zero. Second, I considered the whole control volume(also outside the container) so that ΔV not equal to zero and ΔΝ=0. Aren't both ways of doing it valid(I do not know the concept of enthalpy so I can not do it your way)?
Doing these two ways, I do not obtain the same answer. In the second case, for N=constant, I got that ΔQ and T are related linearly and not logarithmically.

Both ways of doing it are valid. However, you have to take into account that, during the process, once a parcel of mass has left the container, it no longer receives any of the heat the is still being added to the gas within the container.

 

[Joker93]

Both ways of doing it are valid. However, you have to take into account that, during the process, once a parcel of mass has left the container, it no longer receives any of the heat the is still being added to the gas within the container.

But, the container is adiabatically isolated. So, the only heat flow in and out of the container(fixed control volume case) is due to mass flowing in or out of it.

 

[Joker93]

@Chestermiller Can I write my two solutions so you can see what's wrong with them?

 

[Chestermiller]

But, the container is adiabatically isolated. So, the only heat flow in and out of the container(fixed control volume case) is due to mass flowing in or out of it.

What do the words "The question was to find the heat that is needed to change that temperature." mean to you?

 

[Chestermiller]

@Chestermiller Can I write my two solutions so you can see what's wrong with them?

Sure. No problem.

 

[Joker93]

What do the words "The question was to find the heat that is needed to change that temperature." mean to you?

That I have to find the heat that is being transferred out of the system by the particles leaving it.

 

[Chestermiller]

That I have to find the heat that is being transferred out of the system by the particles leaving it.

That doesn't count as heat. The term heat is reserved for energy transferred across the boundary enclosing the system (excluding mass flow, which, of course, carries internal energy).

 

[Joker93]

Sure. No problem.

Thanks.

1st approach-The whole control volume(also outside the container):
ΔΝ=0 and ΔV not equal to zero.
dE=f/2*P*dV (f are the degrees of freedom of the molecules)
also dE=δQ-P*dV
So we get δQ=(1+f/2)*P*dV

Also, using PV=kNT--> P*dV=k*d(N*T)=k*(NdT+TdN)
we get δQ=k(1+f/2)*(NdT+TdN) and because dN=0
we get δQ=N*k*(1+f/2)*dT with N being constant across the whole control volume.
By integrating:
ΔQ=N*k*(1+f/2)*ΔΤ which is a linear relationship between ΔQ and T.
Here, I think that ΔQ is the heat that is flowing in or out of the whole control volume, so I think that this might not be the heat that the exercise needs us to find("the heat that is needed to change that temperature"), but I am not so sure.

2nd approach-Control volume inside the container:
ΔV=0 and ΔΝ not equal to zero.
dE=f/2*P*dV=0 because dV=0.
So, by the first thermodynamic law: δQ+δW=0
Now, dE=f/2*k*d(NT)=f/2*k*(NdT+TdN)
and thus:
δQ+δW=f/2*k*NdT+f/2*k*TdN
Now, I have a suspicion that f/2*k*T of the second term on the right hand side of the above equation must be related to chemical potential. So, if the process is reversible(and it is-I think- because the exercise says that the temperature is changing very slowly implying a pseudostatic, reversible process)
then we can say that δW=f/2*k*TdN
and δQ=f/2*k*NdT
and thus we finally get
ΔQ=fk/2*(PV)*ln[T/To]
which is similar to your result with the difference that mine has a factor of f/2 in front while yours has a factor of (f/2+1) in front (since Cp=(f/2+1)*k*N).
I also think that this is the heat that the exercise wants us to compute.

So, both methods give me an answer, but both answers are different and also are different from yours(and the one proposed by my professor). So, there must be a logical mistake in both of these "derivations".

Thanks.

 

[Joker93]

That doesn't count as heat. The term heat is reserved for energy transferred across the boundary enclosing the system (excluding mass flow, which, of course, carries internal energy).

So, heat here is the heat that flows through the small hole through which the particles can pass?

 

[Chestermiller]

So, heat here is the heat that flows through the small hole through which the particles can pass?

No. The whole tank is being heated.

 

[Joker93]

No. The whole tank is being heated.

But it says that the walls are adiabatically isolated.

 

[Chestermiller]

But it says that the walls are adiabatically isolated.

Can you please provide the exact wording of the problem statement?

 

[Joker93]

Can you please provide the exact wording of the problem statement?

Of course. Here it goes:
A quantity of air in equilibrium has pressure of 1atm in temperature of 0 Celcius and volume 27m^3. Calculate the heat needed to heat up the same quantity of air in the following conditions:
The air is contained in a thermally insulated container of constant volume. A wall of the container has a very small hole on it that allows the "communication" of the air inside it with the outside air. The outside air has a pressure of 1atm. The air changes from 0C to 20C very slowly.
(I think that the last sentence might imply that the air is heated inside the container).

 

[Chestermiller]

Of course. Here it goes:
A quantity of air in equilibrium has pressure of 1atm in temperature of 0 Celcius and volume 27m^3. Calculate the heat needed to heat up the same quantity of air in the following conditions:
The air is contained in a thermally insulated container of constant volume. A wall of the container has a very small hole on it that allows the "communication" of the air inside it with the outside air. The outside air has a pressure of 1atm. The air changes from 0C to 20C very slowly.
(I think that the last sentence might imply that the air is heated inside the container).

This is a very ambiguous problem statement. Maybe they are thinking that there is an electric heater inside the container to heat the air and that no heat can be transferred to or through the walls of the container. There is certainly no way that the air in the container can have its temperature raised from 0 to 20 (at constant container volume) without adding any heat to the air. The given solution also certainly implies that the heat is added to the gas in the container.

 

[Joker93]

This is a very ambiguous problem statement. Maybe they are thinking that there is an electric heater inside the container to heat the air and that no heat can be transferred to or through the walls of the container. There is certainly no way that the air in the container can have its temperature raised from 0 to 20 (at constant container volume) without adding any heat to the air. The given solution also certainly implies that the heat is added to the gas in the container.

Okay, so assuming that this is what happens then, what is wrong with the 2 solutions I gave above?