How are the adiabatic and isobaric expansions of an ideal gas related?

[orthovector]

I'm deriving this formula for the adiabatic expansion of an ideal gas.

$$PV^{\gamma} = Constant_2$$

there are 3 ways to expand to end up at same internal energy dU.

1. this is the direct adiabatic expansion from T1 to T3
$$dU_{systA} = -dW_{systA} = -P_{systA} dV$$ where $$P_{systA}$$ is a function of Volume and Temperature of the gas

2. you can decrease the internal energy of the gas at constant Volume with the same change in temperature as #1. $$dU_{syst V} = C_{V}ndT$$

so, $$dU_{systA} = dU_{systV} = C_VndT$$
$$C_VndT = -P_{systA} dV$$
so
$$C_VndT + P_{systA} dV = 0$$

3. you can also take the isochoric then isobaric pathway to end up from T1 to T2 to T3. The total energy for this 2 step pathway is
$$PdV + VdP = nRdT$$ where dT is the same dT as #1 and #2

if you isolate dT from #3 and plug it into #2, you obtain this expression.
$$C_V VdP + C_V PdV + R P_{systA} dV = 0$$

THIS IS WHERE I AM STUCK!

I DO NOT THINK THAT
$$PdV = P_{systA} dV$$ where $$P_{systA} dV =$$ work involved to expand the gas ADIABATICALLY and $$PdV =$$ isobaric expansion of the gas to get to T3 from #3 above!

why are these two integrals the same?? they are not the same!

HELP HELP HELP!

 

[orthovector]

does anybody have any clues please?

 

[Andrew Mason]

The work done by the system and the work done on the surroundings are always the same magnitude (opposite sign) in a reversible process. It is only if the process is non-reversible (where there is more than an infinitessimal difference in internal and external pressure) that the two are different. $PV^\gamma$ is constant for a reversible adiabatic process only. For example, it does not hold if there is a free expansion

AM

 

[orthovector]

The work done by the system and the work done on the surroundings are always the same magnitude (opposite sign) in a reversible process. It is only if the process is non-reversible (where there is more than an infinitessimal difference in internal and external pressure) that the two are different. $PV^\gamma$ is constant for a reversible adiabatic process only. For example, it does not hold if there is a free expansion

AM

thank you , but you haven't answered my question. why is the work done by an isobaric pathway equal to the work done by the adiabatic pathway for the expansion of an ideal gas?

 

[Redbelly98]

Those shouldn't be equal, should they?

In general, the work done depends on the path taken between two points in a PV diagram. Since work is the integral of P·dV, and the area under the PV curve will be different for the different paths.

 

[orthovector]

Those shouldn't be equal, should they?

In general, the work done depends on the path taken between two points in a PV diagram. Since work is the integral of P·dV, and the area under the PV curve will be different for the different paths.

yes, that's my point! but, according to the derivation of $$PV^{\gamma}$$ the work done isobarically and the work done adiabatically to expand the gas to T3 and V3 are the same!

please look at the 3 ways you can expand the gas adiabatically above (#1,2,3)

we are missing a CRUCIAL POINT here...!

 

[Redbelly98]

... you obtain this expression.

$$C_V VdP + C_V PdV + R P_{systA} dV = 0$$

I agree with that, except that I am going to write what happens when you actually do the integrals (because it's easier for me to think about it that way):

C_v V₁ ΔP + C_v P₃ ΔV + R ∫ P_systA dV = 0 ​

where V₁ is initial volume, P₃ is final pressure

THIS IS WHERE I AM STUCK!

I DO NOT THINK THAT
$$PdV = P_{systA} dV$$ where $$P_{systA} dV =$$ work involved to expand the gas ADIABATICALLY and $$PdV =$$ isobaric expansion of the gas to get to T3 from #3 above!

This is where I lose you. How does the previous equation imply what you are saying here? I don't see it.

 

[orthovector]

I agree with that, except that I am going to write what happens when you actually do the integrals (because it's easier for me to think about it that way):

C_v V₁ ΔP + C_v P₃ ΔV + R ∫ P_systA dV = 0 ​

where V₁ is initial volume, P₃ is final pressureThis is where I lose you. How does the previous equation imply what you are saying here? I don't see it.

please note that the pressure in $$P_{systA} dV$$ is varying as the gas is being expanded adiabatically. However, there is a derivation in my textbook where the $$P_3 dV = P_{systA} dV$$

and when you take it out, you are left with $$C_v + R = C_p$$ that is substituted into the equation to get $$PV^{\gamma} = Constant_2$$

if you differentiate the ideal gas law, you get
$$PdV + VdP = nRdT$$ this is the total energy involved in the isochoric and then the isobaric pathway. Am I wrong? While my textbook used this differential equation, I followed the pressures, volumes, and temperature as you go to the isochoric and then the isobaric pathway to finally expand the gas adiabatically. If I blindly accept the differential equation of the ideal gas, then there is no problem in getting to the final derivation. HOWEVER, if you take the isochoric and then the isobaric pathway, I get stuck at the above problem because the work involved in the Isobaric pathway is NOT EQUAL to the work involved in the adiabatic expansion.

I hope I cleared it up some more.

please try to derive the equation yourself...tell me if you come across the same problem.

 

[Mapes]

The total energy for this 2 step pathway is
$$PdV + VdP = nRdT$$ where dT is the same dT as #1 and #2

$$PdV + VdP = nRdT$$ this is the total energy involved in the isochoric and then the isobaric pathway.

This is not correct. Does this resolve the problem?

 

[Dadface]

The work done in the isochroic pathway is zero since V remains constant.

 

[Redbelly98]

This is not correct. Does this resolve the problem?

Which isn't correct, the equation itself or the statement that it gives the change in energy for the path?

The work done in the isochroic pathway is zero since V remains constant.

That pathway consists of an isochoric and isobaric section; work is done in the isobaric part.

 

[Dadface]

That pathway consists of an isochoric and isobaric section; work is done in the isobaric part.[/QUOTE](from Redbelly)

And to get to the same final condition heat must enter the system during the isochoric part

 

[orthovector]

you guys are all missing the point. you must trust that the 3 equations I've listed are absolutely correct.

what is ambiguous is the derivation on the $$PV^{\gamma} = constant_2$$
because we must either assume that the work done in the isobaric pathway is equal to the work done in the adiabatic pathway OR
we must blindly accept that if you differentiate the ideal gas law, you obtain a dT that is equal to the dT of the adiabatic pathway $$PdV + VdP = nRdT$$

PLEASE TRY THE DERIVATION ON YOUR OWN. NO MORE "WORK IS NOT DONE IN ISOCHORIC PATHWAY" RESPONSES...DUH! WE ALL KNOW THAT.

 

[Dadface]

you guys are all missing the point. you must trust that the 3 equations I've listed are absolutely correct.

what is ambiguous is the derivation on the $$PV^{\gamma} = constant_2$$
because we must either assume that the work done in the isobaric pathway is equal to the work done in the adiabatic pathway OR
we must blindly accept that if you differentiate the ideal gas law, you obtain a dT that is equal to the dT of the adiabatic pathway $$PdV + VdP = nRdT$$

PLEASE TRY THE DERIVATION ON YOUR OWN. NO MORE "WORK IS NOT DONE IN ISOCHORIC PATHWAY" RESPONSES...DUH! WE ALL KNOW THAT.

I think you will see it more clearly if you sketch out the PV graph, the adiabatic change rises fairly steeply to the final state and you need to integrate to find the work done.In effect you are finding the area under the graph.With the isobaric/isochoric change you have a section parallel to the V axis where work is done and a section parallel to the V axis where no work is done.The work done in the isobaric section is P*change in V.Clearly the areas are different so you need to recheck your maths and the assumptions you made.

 

[Mapes]

you guys are all missing the point. you must trust that the 3 equations I've listed are absolutely correct.

what is ambiguous is the derivation on the $$PV^{\gamma} = constant_2$$
because we must either assume that the work done in the isobaric pathway is equal to the work done in the adiabatic pathway OR
we must blindly accept that if you differentiate the ideal gas law, you obtain a dT that is equal to the dT of the adiabatic pathway $$PdV + VdP = nRdT$$

PLEASE TRY THE DERIVATION ON YOUR OWN. NO MORE "WORK IS NOT DONE IN ISOCHORIC PATHWAY" RESPONSES...DUH! WE ALL KNOW THAT.

The equation $P\,dV+V\,dP=nR\,dT$ is correct, but it's not correct that it's the "total energy," which you said twice above. This mistake may be connected to the contradiction you're finding.

It should be clear that this isn't the total energy because the energy change for an ideal gas is always the integral of $nc_V\,dT$, which is different from $nR\,dT$.

The work done in the isobaric pathway is not equal to the work done in the adiabatic pathway; one can see this by comparing the P-V curves, as Dadface says.

Here's how I would derive the fact that $PV^\gamma=\mathrm{constant}$:

$$\Delta U=W$$

$$nc_V\,dT=-p\,dV$$

$$nc_V\,dT=-\frac{nRT}{V}\,dV$$

$$\frac{c_V}{R}\int^{T_2}_{T_1}\frac{dT}{T}=-\int^{V_2}_{V_1}\frac{dV}{V}$$

$$\frac{c_V}{R}\ln\left(\frac{T_2}{T_1}\right)=-\ln\left(\frac{V_2}{V_1}\right)$$

$$\left(\frac{T_2}{T_1}\right)^{c_V/R}=\left(\frac{V_1}{V_2}\right)$$

$$\left(\frac{P_2 V_2}{P_1 V_1}\right)^{c_V/R}=\left(\frac{V_1}{V_2}\right)$$

$$P_1 V_1^{c_P/c_V}=P_2 V_2^{c_P/c_V}$$

Does this resolve the issue?

 

[orthovector]

What is wrong with my method? It should work! Three ways exist to expand this ideal gas to obtain same energy state

 

[Mapes]

I've read your original post at least 10 times, and I'm with Redbelly98; I don't know what contradiction you're finding with approach #3. Can you show the steps in detail?

 

[orthovector]

how do you go from

$$\left(\frac{P_2 V_2}{P_1 V_1}\right)^{c_V/R}=\left(\frac{V_1}{V_2}\right)$$

to
$$P_1 V_1^{c_P/c_V}=P_2 V_2^{c_P/c_V}$$

 

[orthovector]

The equation $P\,dV+V\,dP=nR\,dT$ is correct, but it's not correct that it's the "total energy," which you said twice above. This mistake may be connected to the contradiction you're finding.

It should be clear that this isn't the total energy because the energy change for an ideal gas is always the integral of $nc_V\,dT$, which is different from $nR\,dT$.

2 STEP PROCESS TO GET TO SAME ENERGY STATE AS DIRECT ADIABATIC EXPANSION
PdV + VdP = nRdT ought to be the total energy change for the isochoric then the isobaric pathway to expand the gas.

VdP is the total energy change in the isochoric pathway from T1 to T2.

PdV is the total energy change in the isobaric pathway from T2 to T3.

the total energy change in this 2 step process would be nRdT where integral of dT = (T3 - T1)

and we know this for a fact:
Adiabatically, the temperature change would be T3 - T1. The change in Energy state from T1 to T3 would equal the work of the system to expand the gas adiabatically where T1 > T3.

are these statements correct? if there are any flaws with these statements, then the contradiction is valid. However, if all these statements are correct so far, then I will tell you where the contradiction lies again.

THANK YOU.

 

[Mapes]

how do you go from

$$\left(\frac{P_2 V_2}{P_1 V_1}\right)^{c_V/R}=\left(\frac{V_1}{V_2}\right)$$

to
$$P_1 V_1^{c_P/c_V}=P_2 V_2^{c_P/c_V}$$

$$\left(\frac{P_2 V_2}{P_1 V_1}\right)^{c_V/R}=\left(\frac{P_2 V_2}{P_1 V_1}\right)^{c_V/(c_P-c_V)}=\left(\frac{V_1}{V_2}\right)$$

$$\left(\frac{P_2 V_2}{P_1 V_1}\right)=\left(\frac{V_1}{V_2}\right)^{(c_P-c_V)/c_V}=\left(\frac{V_1}{V_2}\right)^{c_P/c_V-1}$$

$$\left(\frac{P_2}{P_1}\right)=\left(\frac{V_1}{V_2}\right)^{c_P/c_V}$$

$$P_2 V_2^{c_P/c_V}=P_1 V_1^{c_P/c_V}$$

 

[Mapes]

PdV + VdP = nRdT ought to be the total energy change for the isochoric then the isobaric pathway to expand the gas.

This isn't true, for the reason I stated in post #15. The equation is perfectly correct, but your interpretation of it signifying the total energy change is incorrect. (Of course, I encourage you to prove me wrong with a textbook reference.)

EDIT: We can also see that the assignment of $V\,dP=\mathrm{isochoric}$ and $P\,dV=\mathrm{isobaric}$ energies is totally wrong by constructing a P-V diagram. During the isobaric expansion, the gas does work ($-P\,dV$), but it is also heated (we know this because it moves to a higher isotherm). So the energy change during this step does not simply equal $-P\,dV$.

 

[orthovector]

This isn't true, for the reason I stated in post #15. The equation is perfectly correct, but your interpretation of it signifying the total energy change is incorrect.

right... VdP = energy change for isochoric pathway

nCpdT - PdV = energy change for the isobaric pathway. I think I've just corrected my mistaken statement.

HOWEVER, this is what GIANCOLI BOOK HAS DONE:

$$dT = \frac{VdP + PdV}{nR}$$

and this dT is plugged into the equation
$$C_VndT + P_{systA}dV = 0$$

the result is

$$C_V VdP + C_V PdV + RP_{systA}dV = 0$$

$$C_V VdP + PdV (C_V + R) = 0$$

$$C_V VdP + PdV (C_p) = 0$$

I'm not sure why this book did these steps because I don't think that

$$PdV = P_{systA}dV$$


maybe i do NOT understand how you differentiate the ideal gas law.

how does
PV = nRT turn into PdV + VdP = nRdT ?

 

[Mapes]

right... VdP = energy change for isochoric pathway

nCpdT - PdV = energy change for the isobaric pathway. I think I've just corrected my mistaken statement.

No, this is still wrong (EDIT: the second equation is correct, see next post). The equations of state for an ideal gas are $dU=nc_V\,dT$ and $PV=nRT$. You should be able to derive everything from these (and the definitions of heat, $T\,dS$, and work, $-P\,dV$).

From these equations, I get that the energy change during an isochoric process is

$$dU=\frac{c_V V}{R}\,dP$$

and during an isobaric process is

$$dU=\frac{c_V P}{R}\,dV$$

I think the problem you're running into is assuming that $d(PV)=P\,dV+V\,dP$ means that $P\,dV$ corresponds to one process and $V\,dP$ to another. That's not the case. They're just mathematical terms; the only thing you can really say is that the work done by the gas is always $-P\,dV$, but this isn't the same as the total energy change because there is heat flow during an isobaric process.

So if you drop this faulty assumption, I think your problem with $-P\,dV$ (the origin of this thread) will go away, yes? Giancoli is not trying to mix Ps from different processes; it's the same P.

 

[Mapes]

nCpdT - PdV = energy change for the isobaric pathway.

I apologize, this equation is correct (and equal to $(c_V P/R)dV$, as above).

 

[orthovector]

right...I'll get back to you.

 

[orthovector]

No, this is still wrong (EDIT: the second equation is correct, see next post). The equations of state for an ideal gas are $dU=nc_V\,dT$ and $PV=nRT$. You should be able to derive everything from these (and the definitions of heat, $T\,dS$, and work, $-P\,dV$).

From these equations, I get that the energy change during an isochoric process is

$$dU=\frac{c_V V}{R}\,dP$$

and during an isobaric process is

$$dU=\frac{c_V P}{R}\,dV$$

yes, those are correct statements above. However, I STILL believe that

$$VdP + PdV = nRdT$$ tells us that the isochoric and isobaric energy changes equal the nRdT from T1 to T3 for the adiabatic expansion of the ideal gas.

I can conclude the same statement when I expand the gas ISOTHERMALLY, and it works out perfectly. I can also say the above statement for the adiabatic expansion.

here are the energy changes.

isochoric:
$$dU_{ic} = 3/2 nRdT_{ic}$$ where $$dT_{ic} = \frac{V}{nR}dP_{ic}$$

$$\frac{3}{2}VdP_{ic} = dU_{ic}$$

isobaric:
$$dU_{ib} = \frac{3}{2}nRdT_{ib}$$ where $$dT_{ib} = \frac{P}{nR} dV_{ib}$$

$$dU_{ib} = \frac{3}{2}PdV_{ib}$$

finally, total internal energy change of the gas is

$$\frac{3}{2}nRdT_{adiabatic}$$

$$dU_{ic} + dU_{ib} = dU_{adiabatic}$$

$$VdP_{ic} + PdV_{ib} = nRdT_{adiabatic}$$

these statements are ALL CORRECT! ARE THEY NOT?

 

[Mapes]

We're just going in circles. Three times you've said that $P\,dV+V\,dP=nR\,dT$ is the energy charge during adiabatic expansion. I've argued that that that's not the energy change, which is actually $(c_V/R)(P\,dV+V\,dP)=nc_V\,dT$. If you ignore the $c_V/R$ term, the math won't work out and you'll find an apparent paradox.

All your other equations look fine, so I wonder if it's the $c_V/R$ term that's causing all the trouble.

 

[orthovector]

We're just going in circles. Three times you've said that $P\,dV+V\,dP=nR\,dT$ is the energy charge during adiabatic expansion. I've argued that that that's not the energy change, which is actually $(c_V/R)(P\,dV+V\,dP)=nc_V\,dT$. If you ignore the $c_V/R$ term, the math won't work out and you'll find an apparent paradox.

All your other equations look fine, so I wonder if it's the $c_V/R$ term that's causing all the trouble.

The $$C_V/R$$ term is perfectly clear. What is NOT CLEAR is why you must use $$\frac{C_V}{R} VdP = dU_v$$ when $$dU_V = \frac{3}{2}nRdT_v$$ and $$dU_V = \frac{C_V}{R} VdP$$ because $$dU_V = dQ_{in} - 0$$

the same question applies to the isobaric pathway; why choose one form of dU over the other?

 

[Mapes]

I'm not sure what you mean; the dU terms are equivalent. Giancoli only uses the $P\,dV$ and $V\,dP$ forms so he can collect the dV terms, as you wrote in post #22. Note that Giancoli is not dividing the process up into isobaric and isochoric parts, though. This is very important to realize. His $P\,dV$ and $V\,dP$ have different values from the isobaric and isochoric divisions, because he's considering the adiabatic process. It's like the equations

$$a+a=2a$$

$$1.1a+0.9a=2a$$

Just as the final energy change is the same no matter what process(es) you take, the equations add up to the same total; however, the components are different. Again, this is very important to realize, and I think maybe this is the source of all the confusion.

Why have you switched to using 3/2? It's only for a monatomic gas that $c_V=\frac{3}{2}R$. Can we stay general? Then

$$dU_\mathrm{isochoric}=\frac{c_V V_\mathrm{isochoric}}{R}\,dP_\mathrm{isochoric}=nc_V\,dT_\mathrm{isochoric}$$

for the isochoric process. And

$$dU_\mathrm{isobaric}=\frac{c_V P_\mathrm{isobaric}}{R}\,dV_\mathrm{isobaric}=nc_V\,dT_\mathrm{isobaric}$$

for the isobaric process. And

$$dU_\mathrm{adiabatic}=\frac{c_V}{R}( V_\mathrm{adiabatic}\,dP_\mathrm{adiabatic}+P_\mathrm{adiabatic}\,dV_\mathrm{adiabatic})=nc_V\,dT_\mathrm{adiabatic}$$

for the adiabatic process. But

$$V_\mathrm{isochoric}\,dP_\mathrm{isochoric}\neq V_\mathrm{adiabatic}\,dP_\mathrm{adiabatic}$$

$$P_\mathrm{isobaric}\,dV_\mathrm{isobaric}\neq P_\mathrm{adiabatic}\,dV_\mathrm{adiabatic}$$

even though

$$nc_V(dT_\mathrm{isochoric}+dT_\mathrm{isobaric})=nc_V\,dT_\mathrm{adiabatic}$$

If you apply subscripts to each of Giancoli's terms, I think things will be a lot clearer.

 

[orthovector]

i figured it out a few weeks ago...

if you did it in the 2 step way...it would have been an isochoric process then an isobaric, isothermal process that tells us nothing new.

thanks anyway...