Heat capacity under constant pressure or volume question

[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?

 

[Chestermiller]

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

In method 1, you neglected the fact that, once a parcel of mass has left the container, it no longer receives heat. Only the gas that, at any instant of time still remains inside the container, receives heat.

I don't understand what you did in method 2. Certainly, if the temperature of the gas in the container is increasing, its internal energy E is increasing. After that step, I'm lost.

 

[Joker93]

In method 1, you neglected the fact that, once a parcel of mass has left the container, it no longer receives heat. Only the gas that, at any instant of time still remains inside the container, receives heat.

I don't understand what you did in method 2. Certainly, if the temperature of the gas in the container is increasing, its internal energy E is increasing. After that step, I'm lost.

So, how do I include that effect?

In the second method, dV=0(the change in the volume of the control volume), so dE=0. And for this to happen: E=f/2*k*(NT)=const. so (NT) must be constant. So, while the temperature is rising, the number of particles is decreasing.

 

[Chestermiller]

So, how do I include that effect?

You are only heating the gas that is remaining within the tank. If n is the number of moles of gas remaining in the container at any time t, then $$dQ=nC_pdT$$This is combined with $$n=\frac{PV}{RT}$$ to give:$$dQ=\frac{PV}{RT}C_pdT$$

In the second method, dV=0(the change in the volume of the control volume), so dE=0.

dV equal to zero does not mean that dE = 0. There is heat being added.

 

[Joker93]

You are only heating the gas that is remaining within the tank. If n is the number of moles of gas remaining in the container at any time t, then $$dQ=nC_pdT$$This is combined with $$n=\frac{PV}{RT}$$ to give:$$dQ=\frac{PV}{RT}C_pdT$$

dV equal to zero does not mean that dE = 0. There is heat being added.

for the second method, i only used that PV=NkT and E=f/2*NkT. So, we get E=f/2*PV. Taking differentials: dE=f/2*PdV. I did not use the concept of work here.

Also, before proceeding, I want to ask a conceptual question first. Is the temperature in the whole control volume the same as the temperature in the control volume that only covers the inside of the container?

 

[Chestermiller]

for the second method, i only used that PV=NkT and E=f/2*NkT. So, we get E=f/2*PV. Taking differentials: dE=f/2*PdV. I did not use the concept of work here.

Also, before proceeding, I want to ask a conceptual question first. Is the temperature in the whole control volume the same as the temperature in the control volume that only covers the inside of the container?

Ah. I see what you did in Method 2 now. It is very similar to what I did in post #14 (believe it or not). Your only mistake was leaving out the work taking place at the exit hole. Even though the rest of the tank is rigid, the gas in the tank is doing work to push gas out the exit hole (which is not a rigid boundary). If dn is the number of moles going out the exit hole between time t and time t + dt, the amount of work required to push this gas out the exit hole is Pvdn, where v is the molar volume of the gas at time t: $v=RT/P$. So, the amount of work is RTdn.

 

[Joker93]

for the second method, i only used that PV=NkT and E=f/2*NkT. So, we get E=f/2*PV. Taking differentials: dE=f/2*PdV. I did not use the concept of work here.

Ah. I see what you did in Method 2 now. It is very similar to what I did in post #14 (believe it or not). Your only mistake was leaving out the work taking place at the exit hole. Even though the rest of the tank is rigid, the gas in the tank is doing work to push gas out the exit hole (which is not a rigid boundary). If dn is the number of moles going out the exit hole between time t and time t + dt, the amount of work required to push this gas out the exit hole is Pvdn, where v is the molar volume of the gas at time t: $v=RT/P$. So, the amount of work is RTdn.

But, shouldn't this extra term come out naturally? In the solution I provided, I ended up with δQ+δW=f/2*k*NdT+f/2*k*TdN by only using that δQ+δW=0
and dE=f/2*k*d(NT)=f/2*k*(NdT+TdN). So, where would his extra term come up?

Also, before proceeding, I want to ask a conceptual question first. Is the temperature in the whole control volume the same as the temperature in the control volume that only covers the inside of the container?

 

[Chestermiller]

But, shouldn't this extra term come out naturally? In the solution I provided, I ended up with δQ+δW=f/2*k*NdT+f/2*k*TdN by only using that δQ+δW=0
and dE=f/2*k*d(NT)=f/2*k*(NdT+TdN). So, where would his extra term come up?

The equation should read: $dQ=C_VTdn+Pvdn$. The first term represents the internal energy leaving the tank in the parcel dn, and the second term represents the work done to push the parcel dn out of the tank.

Also, before proceeding, I want to ask a conceptual question first. Is the temperature in the whole control volume the same as the temperature in the control volume that only covers the inside of the container?

No. In the latter case, once a parcel leaves the tank, its temperature stops changing (assuming that it doesn't exchange heat with the outside air). So, for the whole control volume, the temperature in the tank is uniform (and equal to the temperature that we are solving for), but, for the parcels that have already exited the tank, their temperatures vary over the range from the temperature of the first parcel to leave the tank to the temperature of the parcel that just most recently left the tank.

 

[Joker93]

The equation should read: $dQ=C_VTdn+Pvdn$. The first term represents the internal energy leaving the tank in the parcel dn, and the second term represents the work done to push the parcel dn out of the tank.

I've understood that, but my problem is with how does it arise from the method of solution I am using. I mean, following the steps I am taking in method 2, where does the extra term show up that I neglected? I only used dE=δQ+δW and E=f/2*k*NT to arrive at the seemingly wrong δQ+δW=f/2*k*NdT+f/2*k*TdN

 

[Chestermiller]

I've understood that, but my problem is with how does it arise from the method of solution I am using. I mean, following the steps I am taking in method 2, where does the extra term show up that I neglected? I only used dE=δQ+δW and E=f/2*k*NT to arrive at the seemingly wrong δQ+δW=f/2*k*NdT+f/2*k*TdN

I can't go through the entire fixed control volume version of the first law, which is covered in detail in every thermodynamics book. But, for this problem, if we do an energy balance on the contents of the tank between time t and t + dt, we obtain:$$dE=dQ-dW-edn$$ where $E=(nC_vT)$, dW is the work done to push the gas of mass dn out of the tank, and edn is the internal energy exiting the tank in the exit stream (e is the internal energy per mole of the exit stream). You have already shown that dE = 0. So, $$dQ=dW+edn$$The work to push the parcel of gas dn out of the tank is given by:$$dW=Pvdn=RTdn$$ So,$$dQ=(e+RT)dn=(C_vT+RT)dn=(C_v+R)Tdn$$

I'm running out of ideas on how to show this better. How does your analysis compare with this analysis?

 

[Joker93]

I can't go through the entire fixed control volume version of the first law, which is covered in detail in every thermodynamics book. But, for this problem, if we do an energy balance on the contents of the tank between time t and t + dt, we obtain:$$dE=dQ-dW-edn$$ where $E=(nC_vT)$, dW is the work done to push the gas of mass dn out of the tank, and edn is the internal energy exiting the tank in the exit stream (e is the internal energy per mole of the exit stream). You have already shown that dE = 0. So, $$dQ=dW+edn$$The work to push the parcel of gas dn out of the tank is given by:$$dW=Pvdn=RTdn$$ So,$$dQ=(e+RT)dn=(C_vT+RT)dn=(C_v+R)Tdn$$
=
I'm running out of ideas on how to show this better. How does your analysis compare with this analysis?

From your last help, I tried something more familiar to me. I again wrote dE=dQ+dW but this time I wrote dW=μ*dN - W where W is the work done to push the gas out but in a form that implies a change in volume. To clarify, I wrote W=Pv*dN(as you did) which is like pushing small volumes v out of the fixed control volume. Also, μ is the chemical potential(like you said, the energy that gets transported outside the control volume) and it is equal to μ=dE/dN which gives μ=f/2*kT. So, dW=f/2*kT*dN + Pv*dN with v=V/N (note that I wrote +PvdN rather than -PvdN because when dN>0 then work is being done ON the system so dW>0 as opposed to the typical dW=-PdV).
These, along with dE=dQ+dW and dE=0(due to dV=0) and thus dE=f/2*k*(NdT+TdN) gives:
dQ+f/2*kT*dN +Pv*dN=f/2*k*(NdT+TdN) and so
dQ=-PvdN+f/2*k*NdT and because TdN=-NdT and PV=kNT
dQ=(f/2+1)*k*PV*(dT/T)
which is the correct answer.

I think this is correct. What really annoys me is that our professor thought that it was an easy task of writing the extra term of PvdN. This is conceptually very tricky and I think I would have never think of this if it wasn't for you..

 

[Chestermiller]

From your last help, I tried something more familiar to me. I again wrote dE=dQ+dW but this time I wrote dW=μ*dN - W where W is the work done to push the gas out but in a form that implies a change in volume. To clarify, I wrote W=Pv*dN(as you did) which is like pushing small volumes v out of the fixed control volume. Also, μ is the chemical potential(like you said, the energy that gets transported outside the control volume) and it is equal to μ=dE/dN which gives μ=f/2*kT. So, dW=f/2*kT*dN + Pv*dN with v=V/N (note that I wrote +PvdN rather than -PvdN because when dN>0 then work is being done ON the system so dW>0 as opposed to the typical dW=-PdV).
These, along with dE=dQ+dW and dE=0(due to dV=0) and thus dE=f/2*k*(NdT+TdN) gives:
dQ+f/2*kT*dN +Pv*dN=f/2*k*(NdT+TdN) and so
dQ=-PvdN+f/2*k*NdT and because TdN=-NdT and PV=kNT
dQ=(f/2+1)*k*PV*(dT/T)
which is the correct answer.

I think this is correct. What really annoys me is that our professor thought that it was an easy task of writing the extra term of PvdN. This is conceptually very tricky and I think I would have never think of this if it wasn't for you..

Nice job.

As I said in a previous post, I think that your professor did you a disservice by giving you this problem, which is waaayyyy too advanced for you right now. The open system (fixed control volume) version of the first law is a little difficult to relate to at first, because of the very term you identified. Of course, once you have learned about it and got some experience applying it, it seems much easier.

I was very impressed with how you attacked this problem and with the questions you asked. I think you have a bright future in science.

Chet

 

[Joker93]

Nice job.

As I said in a previous post, I think that your professor did you a disservice by giving you this problem, which is waaayyyy too advanced for you right now. The open system (fixed control volume) version of the first law is a little difficult to relate to at first, because of the very term you identified. Of course, once you have learned about it and got some experience applying it, it seems much easier.

I was very impressed with how you attacked this problem and with the questions you asked. I think you have a bright future in science.

Chet

Wow, thanks! And I thought that you would think "ah, this kid did not understand the basics that I was saying to him"!
Thanks for your patience(I know I have taken a lot of your time; it's just that if I don't understand something exactly, then I can't connect the dots and find a solution that is also conceptually valid in my head).
Thanks again!