Heat capacity under constant pressure or volume question

[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!