Adiabatic or Isobaric process?

  • #1
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Main Question or Discussion Point

Consider the following problem:
Gaseous helium (assumed ideal) filled in a horizontal cylindrical vessel is separated from its surroundings by a massless piston. Both piston and cylinder are thermally insulating. The ambient pressure is suddenly tripled without changing the ambient temperature. How many times of its initial value will the volume of the helium become, when the piston finally stops?

Initial pressure and volume be P1 and V1. Now the ambient pressure is suddenly made P2 = 3 P1. The gas compresses under constant pressure P2, so the process must be isobaric. However the walls are also thermally insulating, so the process must also be Adiabatic. I find this confusing.
If I use the expression for adiabatic process i.e [tex] PV^\gamma =constant [/tex], then I don't get the answer at the back.
However, I can use 1st law to write[tex] W=-P_2\Delta V=3/2 nR\Delta T=3/2\Delta (PV)=3/2(P_2V_2-P_1V_1) [/tex] to get the correct answer.
I think the second way is fine. But why doesn't the first method work?
 

Answers and Replies

  • #2
mjc123
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Because PVγ = constant only applies to an equilibrium adiabatic process.
 
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  • #3
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Because PVγ = constant only applies to an equilibrium adiabatic process.
Ok. So if I understand correctly, the process is adiabatic, but PVγ=constant is not valid because it isn't an eqbm process. Thanks.
SO is the formula valid only if the pressure is increased gradually? If I have an equilibrium process which is adiabatic and also constant pressure, then T and V should also remain constant, (i.e there is no process at all). Is this correct?
 
  • #4
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Your second method is correct. Even though you refer to this as an isobaric process, the final pressure of the gas in the cylinder is not the same as its initial pressure. The correct equation to use for P-V work (done by the system on the surroundings) is always ##\int{P_{ext}dV}##, where ##P_{ext}## is the force per unit area exerted by the surroundings (in this case, the inside piston face) on the gas. In the rapid deformation that you describe, the gas pressure is not uniform spatially within the cylinder, and viscous forces in the gas are contributing to the force on the inside piston face (because of Newton's 3rd law, the force the gas exerts on the piston face is equal and opposite to the force that the piston face exerts on the gas). The force per unit area imposed by the surroundings on the gas will be consistent with the rate at which the gas is deforming so that Newton's 3rd law is always satisfied at the piston face.
 
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  • #5
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Your second method is correct. Even though you refer to this as an isobaric process, the final pressure of the gas in the cylinder is not the same as its initial pressure. The correct equation to use for P-V work (done by the system on the surroundings) is always ##\int{P_{ext}dV}##, where ##P_{ext}## is the force per unit area exerted by the surroundings (in this case, the inside piston face) on the gas. In the rapid deformation that you describe, the gas pressure is not uniform spatially within the cylinder
Thanks a lot. That cleared a lot of my confusion. So the process is not really isobaric, even though external pressure is constant during the compression. We can't even talk about the 'pressure of the gas' during the process of the compression because it doesnt take a single value. So basically it is a non-quasistatic adiabatic process.
My first thought was maybe there is something wrong with the massless piston assumption. Since pressure on both sides of massless piston should be equal, i thought the gas pressure is forced to change suddenly from p1 to p2 along with the external pressure. As a result , I thought the volume should also suddenly change along with the pressure. In this case the 2nd method would be wrong because both pressure and volume are changing together.
But one question i still have is, how do we know the process is as you described, i.e non- equilibrium process, with viscous forces contributing to the pressure etc . Can we know this from the problem statement itself or is it general knowledge?
 
  • #6
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Thanks a lot. That cleared a lot of my confusion. So the process is not really isobaric, even though external pressure is constant during the compression. We can't even talk about the 'pressure of the gas' during the process of the compression because it doesnt take a single value. So basically it is a non-quasistatic adiabatic process.
Yes.
My first thought was maybe there is something wrong with the massless piston assumption. Since pressure on both sides of massless piston should be equal, i thought the gas pressure is forced to change suddenly from p1 to p2 along with the external pressure. As a result , I thought the volume should also suddenly change along with the pressure. In this case the 2nd method would be wrong because both pressure and volume are changing together.
The force per unit area exerted on the inner face of the piston by the gas also changes suddenly, primarily due to viscous forces.
But one question i still have is, how do we know the process is as you described, i.e non- equilibrium process, with viscous forces contributing to the pressure etc . Can we know this from the problem statement itself or is it general knowledge?
When the external pressure is suddenly changed, the gas experiences a very rapid rate of deformation. Viscous stresses are proportional to, not the amount of deformation, but to the rate of deformation. Any process where the rate of deformation of a fluid is very rapid is going to have viscous stresses contributing to the overall stress. Viscous stresses cause dissipation of mechanical energy to internal energy, which is a key contributor to irreversibility. Another key contributor to irreversibility are temperature gradients, which give rise to high heat fluxes.

What's happening here is very similar to what happens in a system consisting of a spring and damper in parallel, where the spring is analogous to the PV behavior of the gas and the damper is analogous to the viscous behavior of the gas. Here is a reference to an Physics Forums Insights article I wrote that discusses this analogy in more detail: https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/
 
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  • #7
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Yes.

The force per unit area exerted on the inner face of the piston by the gas also changes suddenly, primarily due to viscous forces.
When the external pressure is suddenly changed, the gas experiences a very rapid rate of deformation. Viscous stresses are proportional to, not the amount of deformation, but to the rate of deformation. Any process where the rate of deformation of a fluid is very rapid is going to have viscous stresses contributing to the overall stress. Viscous stresses cause dissipation of mechanical energy to internal energy, which is a key contributor to irreversibility. Another key contributor to irreversibility are temperature gradients, which give rise to high heat fluxes.

What's happening here is very similar to what happens in a system consisting of a spring and damper in parallel, where the spring is analogous to the PV behavior of the gas and the damper is analogous to the viscous behavior of the gas. Here is a reference to an Physics Forums Insights article I wrote that discusses this analogy in more detail: https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/
Thank you! That was helpful.
 
  • #8
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So what is the solution? What is the ratio of the final volumes at the end? I want to see if I can work it out.

Thanks!
 
  • #9
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So what is the solution? What is the ratio of the final volumes at the end? I want to see if I can work it out.

Thanks!
Let’s see what you come up with.
 
  • #10
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I take the ratios of the equation of state for the He after and before, using the fact that the pressure afterwards is 3 times the pressure before. This gives the volume ratio as 1/3 the temperature ratio. But is there more?
 
  • #11
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I take the ratios of the equation of state for the He after and before, using the fact that the pressure afterwards is 3 times the pressure before. This gives the volume ratio as 1/3 the temperature ratio. But is there more?
Are you saying that the volume ratio is 1/3?
 
  • #12
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Are you saying that the volume ratio is 1/3?
No. I am saying that the volume ratio is 1/3 the temperature ratio. I.e.,

##\frac{V_2}{V_1} = \frac{T_2}{3T_1}##
 
  • #13
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No. I am saying that the volume ratio is 1/3 the temperature ratio. I.e.,

##\frac{V_2}{V_1} = \frac{T_2}{3T_1}##
What is the temperature ratio equal to (an actual number)?
 
  • #14
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I don't know how to answer that without assuming something about the process.
 
  • #15
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I don't know how to answer that without assuming something about the process.
The OP correctly determined the work done by the surroundings on the system (method 2) in his first post. Do you know how to combine that with the first law of thermodynamics and the ideal gas law to determine the temperature ratio?
 
  • #16
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I can't isolate the temperature ratio or the volume ratio alone, independent of temperature and pressure. Anyone else can help?

##c_v \Delta T = -P_2 \Delta V##
##c_v \frac{T_1}{V_1} (\frac{T_2}{T_1}-1) = -P_2 (\frac{V_2}{V_1}-1)##
 
  • #17
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I can't isolate the temperature ratio or the volume ratio alone, independent of temperature and pressure. Anyone else can help?

##c_v \Delta T = -P_2 \Delta V##
##c_v \frac{T_1}{V_1} (\frac{T_2}{T_1}-1) = -P_2 (\frac{V_2}{V_1}-1)##
It should be $$nC_v(T_2-T_1)=-P_2(V_2-V_1)=-P_2\left[\frac{nRT_2}{P_2}-\frac{nRT_1}{P_1}\right]$$or, equivalently,
$$C_v(T_2-T_1)=-RT_2+\frac{P_2}{P_1}RT_1$$
 
  • #18
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Here's how I worked it out.

For each state k we have ## P_kV_k = mRT_k##. So the ratio of two states of the same mass of gas gives ## \frac{P_2V_2}{P_1V_1} = \frac{T_2}{T_1}##. We also have ##P_2=3P_1##, and substituting this into the state ratio yields

## \frac{V_2}{V_1} = \frac{T_2}{3T_1}...\big(1\big)##

The work done on the mass during the compression can be calculated from

##W=\int_{1}^{2} p dV = P_2\Delta V = P_2\Big(V_2-V_1 \Big)##

With no heat transfer and the assumption of constant specific heat, the 1st law for the gas is

##mc_v\Big(T_2-T_1 \Big) = -P_2\Big(V_2-V_1 \Big)##

or

##mc_v \frac{T_1}{V_1} \Big(\frac{T_2}{T_1}-1\Big) = -P_2 \Big(\frac{V_2}{V_1}-1\Big)##

Substitute into the above expression the given ##P_2=3P_1## to yield

##mc_v \frac{T_1}{V_1} \Big(\frac{T_2}{T_1}-1\Big) = -3P_1 \Big(\frac{V_2}{V_1}-1\Big)##

##mc_v \frac{T_1}{3P_1V_1} \Big(\frac{T_2}{T_1}-1\Big) = \Big(1-\frac{V_2}{V_1}\Big)##

## \frac{c_v}{3R} \Big(\frac{T_2}{T_1}-1\Big) = \Big(1-\frac{V_2}{V_1}\Big)##

Now use (1) to get rid of the temperature ratios

## \frac{c_v}{3R} \Big(\frac{3V_2}{V_1}-1\Big) = \Big(1-\frac{V_2}{V_1}\Big)##

This can be solved for the volume ratios to get

##\frac{V_2}{V_1} = \Big(\frac{1+\frac{c_v}{3R}}{1+\frac{c_v}{R}}\Big)##

Now for an ideal monoatomic gas, ##c_v = \frac{3R}{2}##, and so the volume ratio is ##\frac{V_2}{V_1} = \frac{3}{5}##.
 
  • #19
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Here's how I worked it out.

For each state k we have ## P_kV_k = mRT_k##. So the ratio of two states of the same mass of gas gives ## \frac{P_2V_2}{P_1V_1} = \frac{T_2}{T_1}##. We also have ##P_2=3P_1##, and substituting this into the state ratio yields

## \frac{V_2}{V_1} = \frac{T_2}{3T_1}...\big(1\big)##

The work done on the mass during the compression can be calculated from

##W=\int_{1}^{2} p dV = P_2\Delta V = P_2\Big(V_2-V_1 \Big)##

With no heat transfer and the assumption of constant specific heat, the 1st law for the gas is

##mc_v\Big(T_2-T_1 \Big) = -P_2\Big(V_2-V_1 \Big)##

or

##mc_v \frac{T_1}{V_1} \Big(\frac{T_2}{T_1}-1\Big) = -P_2 \Big(\frac{V_2}{V_1}-1\Big)##

Substitute into the above expression the given ##P_2=3P_1## to yield

##mc_v \frac{T_1}{V_1} \Big(\frac{T_2}{T_1}-1\Big) = -3P_1 \Big(\frac{V_2}{V_1}-1\Big)##

##mc_v \frac{T_1}{3P_1V_1} \Big(\frac{T_2}{T_1}-1\Big) = \Big(1-\frac{V_2}{V_1}\Big)##

## \frac{c_v}{3R} \Big(\frac{T_2}{T_1}-1\Big) = \Big(1-\frac{V_2}{V_1}\Big)##

Now use (1) to get rid of the temperature ratios

## \frac{c_v}{3R} \Big(\frac{3V_2}{V_1}-1\Big) = \Big(1-\frac{V_2}{V_1}\Big)##

This can be solved for the volume ratios to get

##\frac{V_2}{V_1} = \Big(\frac{1+\frac{c_v}{3R}}{1+\frac{c_v}{R}}\Big)##

Now for an ideal monoatomic gas, ##c_v = \frac{3R}{2}##, and so the volume ratio is ##\frac{V_2}{V_1} = \frac{3}{5}##.
A still simpler method is:

$$\frac{C_v}{R}(P_2V_2-P_1V_1)=-P_2(V_2-V_1)$$or
$$\frac{3}{2}(3V_2-V_1)=-3(V_2-V_1)$$or$$\frac{15}{2}V_2=\frac{9}{2}V_1$$
 
  • #20
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It is impossible to "suddenly (instantaneously?)" change the pressure to 3P1.
 
  • #22
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It is impossible to "suddenly (instantaneously?)" change the pressure to 3P1.
Well what if you slowly change the external pressure from P1 to P2 while holding the piston fixed and then release the piston? That is effectively the same problem.
 
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  • #23
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I have a really hard time with calculating work for non-quasistatic processes. I know that chemistry text books teach that W = Pext ΔV.
Many physics text books, however, are very careful about this. See Young and Freedman for example. They simply only calculate work for quasistatic processes. Obviously that means that Pext is always the same as Pint in those processes. The work for isothermal and adiabatic processes then becomes an integral over a pressure that depends on V and T and changes.
So my questions: Is there a "formal" way of showing that W = Pext ΔV for every process?
If there's no gas in the cylinder being compressed, is W = Pext ΔV still correct then? I think I know the answer to that is no.
The obvious follow up question to that is: How many atoms do you need in the cylinder to be allowed to use W = Pext ΔV?
 
  • #24
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I have a really hard time with calculating work for non-quasistatic processes. I know that chemistry text books teach that W = Pext ΔV.
Many physics text books, however, are very careful about this. See Young and Freedman for example. They simply only calculate work for quasistatic processes. Obviously that means that Pext is always the same as Pint in those processes. The work for isothermal and adiabatic processes then becomes an integral over a pressure that depends on V and T and changes.
So my questions: Is there a "formal" way of showing that W = Pext ΔV for every process?
This equation applies only to the "PV" work done by the system on its surroundings. It can be shown in a formal way:

If we believe Newton's 3rd law, then we know that, at the interface between the system and its surroundings, the force per unit area exerted by the gas on its surroundings (typically at the piston face) is equal in magnitude and opposite in direction to the force per unit area exerted by the surroundings on the gas ##P_{ext}##. So the force exerted by the gas on its surroundings must be ##F=P_{ext}A##, where A is the area of the piston. From this, it follows that the work done by the gas on its surroundings is $$dW=Fdx=P_{ext}Adx=P_{ext}dV$$where dV=Adx.

It is important to note that, in a rapid irreversible expansion, (a) the pressure of the gas within the cylinder is not uniform spatially (so there is no one single value of the pressure that we can point to to call the gas pressure), (b) the ideal gas law (which applies only to thermodynamic equilibrium situations) does not describe the force per unit area on the piston face, and (c) the force per unit area exerted by the gas on the piston face is composed of two parts: the local pressure (based on the local temperature and specific volume of the gas at the piston face) and viscous stresses (which depend not on the amount the gas has been deformed, but on the rate at which the gas is deforming). So, in the absence of detailed knowledge of the gas dynamics within the cylinder, we must resort to working entirely with the force we apply or specify externally ##P_{ext}A## in determining the amount of work done for a irreversible process.

If there's no gas in the cylinder being compressed, is W = Pext ΔV still correct then? I think I know the answer to that is no.
The obvious follow up question to that is: How many atoms do you need in the cylinder to be allowed to use W = Pext ΔV?
If there is no gas in the cylinder, then the force per unit area exerted by the gas on its surroundings at the piston face ##P_{ext}## is zero.
 
  • #25
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Makes sense! Thanks.
I've also been wondering about the reverse process. If you have a gas at high pressure and you suddenly release it, for example the CO2 in a bottle of champagne when you let the cork fly. This is often given as an example of adiabatic expansion because it happens so quickly that practically no heat can be transferred. Shouldn't one instead treat it as an isobaric expansion with the outside pressure as the constant pressure and calculate W and the temperature change for an isobaric process?
 

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