Another Thermodynamics Problem

  • #1
Klaus von Faust
37
10
Member warned that some effort must be shown, and the template is required
1.
In vacuum and weightlessness, at the bottom of a
cylindrical vessel (a cup), there is a layer of solid substance
of molar mass µ. This substance sublimes slowly (evaporates
from the solid phase into gaseous phase) and pushes thereby
the vessel to the opposite direction. Estimate the terminal
speed of the vessel. The mass of the vessel M, and the initial
mass of the substance m ≪ M; the temperature of the ves-
sel is T; the process can be assumed to be isothermal (cooling
due to evaporation and heat radiation remains negligible). The
cross-sectional area of the vessel is A.

I think a molecular kinetic approach is required to solve this problem, but I am quite unsure about what to start with.
 
Last edited by a moderator:

Answers and Replies

  • #3
Klaus von Faust
37
10
What is the exact statement of the problem?
This is the exact statement. No more information is provided in the problem
 
  • #4
22,071
5,031
Hint (I think): If they are talking about a terminal velocity of the cup, then the cup must have gained kinetic energy up to the point where the solid totally evaporated. This must have required work.
 
  • Like
Likes Klaus von Faust
  • #5
Lord Jestocost
Gold Member
871
699
Maybe, one should think of the conservation of total momentum after the substance has completely evaporated in a "directional" way into vacuum.
 
  • Like
Likes Klaus von Faust
  • #6
22,071
5,031
Here's another hint: What is the pressure at the very surface of the solid when it is evaporating? What is the density of the gas at the very surface of the solid as it is evaporating?
 
  • Like
Likes Klaus von Faust
  • #7
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,475
The problem doesn't give any number for heat of sublimation (or enthalpy of sublimation) of substance m?
 
  • Like
Likes Klaus von Faust
  • #8
Klaus von Faust
37
10
The problem doesn't give any number for heat of sublimation (or enthalpy of sublimation) of substance m?
No. This is all data given in the problem. But I guess I'll come up with a solution, or, at least, an attempt, soon.
 
  • #9
22,071
5,031
You need to obtain an equation for the rate at which work is being done on the expelled vapor. This is equal to the pressure times the volume rate of expulsion of vapor from the surface. This will be equal to rate at which the vapor does work on the cup.
 
  • Like
Likes Klaus von Faust and Delta2
  • #10
Klaus von Faust
37
10
I have an attempt of a solution.(a clumsy one, I guess). I suppose in this case a molecular kinetic approach is indeed required. First, it is stated in the problem that m<<M. That means that we can consider the impact between molecules and the cup wall as a perfectly elastic collision, and write the equation for conservation of momentum.
Let us assume that the initial velocity of one molecule is u, and it's terminal velocity is also u. Now we can write the equation for conservation of momentum
μ*u=M*Δv-μ*u
Where Δv denotes velocity increment due to collision with one molecule. So 2*μ*u=M*Δv
In order to find terminal velocity of the cup, we have to take the summ of all impacts(but not to forget that in real case not all the molecules have speed u, so we have to take the average speed of all molecules)
Σ(2*μ*u)=M*ΣΔv
2*<u>*m*0.5=M*v (0.5 because according to statistical isothropy one half of the molecules will go in the positive direction of the axis, and the other half will go in the negative one)
The next step is to find the average speed of one molecule, according to Maxwell's distribution,
the average kinetic energy related to the motion along the x axes is
<0.5*m*u^2>=0.5*k*T then
<μ*n*u^2>=k*T Where n is number of moles

Our next step is to find the velocity of one molecule, so when substituting N/N_A instead of n and giving N the value 1, we get
u=sqrt{RT/μ}
The final answer is
v=m/M*sqrt{RT/μ}


 
  • #11
Klaus von Faust
37
10
You need to obtain an equation for the rate at which work is being done on the expelled vapor. This is equal to the pressure times the volume rate of expulsion of vapor from the surface. This will be equal to rate at which the vapor does work on the cup.
I tried to solve it differently. Is my solution correct?
 
  • #12
22,071
5,031
The mechanical work that the evaporating vapor does on the cup is ##W=P_{vapor}(V_{vapor}-V_{solid})\approx P_{vapor}V_{vapor}=n_{vapor}RT##, where ##n_{vapor}## are the number of moles that evaporate: $$n_{vapor}=\frac{m}{\mu}$$The work W must equal the final kinetic energy of the cup:$$\frac{1}{2}Mv_{terminal}^2=\frac{m}{\mu}RT$$So, $$v_{terminal}=\sqrt{\frac{2mRT}{M\mu}}$$
 
  • Like
Likes Klaus von Faust and EVERYTHINGISNOTHING
  • #13
Klaus von Faust
37
10
The mechanical work that the evaporating vapor does on the cup is ##W=P_{vapor}(V_{vapor}-V_{solid})\approx P_{vapor}V_{vapor}=n_{vapor}RT##, where ##n_{vapor}## are the number of moles that evaporate: $$n_{vapor}=\frac{m}{\mu}$$The work W must equal the final kinetic energy of the cup:$$\frac{1}{2}Mv_{terminal}^2=\frac{m}{\mu}RT$$So, $$v_{terminal}=\sqrt{\frac{2mRT}{M\mu}}$$
The mechanical work that the evaporating vapor does on the cup is ##W=P_{vapor}(V_{vapor}-V_{solid})\approx P_{vapor}V_{vapor}=n_{vapor}RT##, where ##n_{vapor}## are the number of moles that evaporate: $$n_{vapor}=\frac{m}{\mu}$$The work W must equal the final kinetic energy of the cup:$$\frac{1}{2}Mv_{terminal}^2=\frac{m}{\mu}RT$$So, $$v_{terminal}=\sqrt{\frac{2mRT}{M\mu}}$$
Why is the volume of substance in solid state much smaller than the volume of substance in gaseous state?
I checked their official answer and I will give the link of the problem, there is only answer and no solving process provided. https://www.ioc.ee/~kalda/ipho/Thermodyn.pdf
Their answer matches with mine, but I am sure that my process of solving is wrong, and it is a matter of coincidence that I got the same results. It is 9th problem
 
  • #14
22,071
5,031
Why is the volume of substance in solid state much smaller than the volume of substance in gaseous state?
What is the molar volume of ice at 0 C?
What is the molar volume of water vapor at 0 C and the equilibrium vapor pressure?
What is the ratio?

I checked their official answer and I will give the link of the problem, there is only answer and no solving process provided. https://www.ioc.ee/~kalda/ipho/Thermodyn.pdf
Their answer matches with mine, but I am sure that my process of solving is wrong, and it is a matter of coincidence that I got the same results. It is 9th problem
I don't understand the molecular dynamics derivation, but I'm confident of the continuum thermodynamic approach I employed. So, I stand by my answer.
 
  • Like
Likes Klaus von Faust
  • #15
Klaus von Faust
37
10
What is the molar volume of ice at 0 C?
What is the molar volume of water vapor at 0 C and the equilibrium vapor pressure?
What is the ratio?


I don't understand the molecular dynamics derivation, but I'm confident of the continuum thermodynamic approach I employed. So, I stand by my answer.
Thank you
 
  • #16
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,475
@Chestermiller The text of the problem clearly states to view the process as isothermal, while your approach at post #12 assumes (correct me if I am wrong) that the process is isobaric.
 
  • #17
22,071
5,031
@Chestermiller The text of the problem clearly states to view the process as isothermal, while your approach at post #12 assumes (correct me if I am wrong) that the process is isobaric.
No. The pressure in the equation I wrote is just the equilibrium vapor pressure of the material at the interface with the solid. This interface is where the specific volume of the material changes from that of the solid to that of the saturated vapor. So the pressure in the equation is just the force per unit area at the surface of the evaporating solid. As the vapor recedes in the wake of the evaporating solid, it drops in pressure rapidly. But, the work to accelerate the cup takes place at the solid surface, driven by the equilibrium vapor pressure. This work is also equal in magnitude to the work done by the solid in forcing the vapor backward. The rate at which this work is carried out is ##P_{sat}(v_{vapor}-v_{solid})(-\frac{dm}{dt})## where v's are the specific volumes of the saturated solid and vapor, and -dm/dt is the rate of evaporation. The integral of this with respect to time is just ##p_{sat}v_{vapor}m=\frac{m}{\mu} RT##
 
  • #18
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,475
No. The pressure in the equation I wrote is just the equilibrium vapor pressure of the material at the interface with the solid. This interface is where the specific volume of the material changes from that of the solid to that of the saturated vapor. So the pressure in the equation is just the force per unit area at the surface of the evaporating solid. As the vapor recedes in the wake of the evaporating solid, it drops in pressure rapidly. But, the work to accelerate the cup takes place at the solid surface, driven by the equilibrium vapor pressure. This work is also equal in magnitude to the work done by the solid in forcing the vapor backward. The rate at which this work is carried out is ##P_{sat}(v_{vapor}-v_{solid})(-\frac{dm}{dt})## where v's are the specific volumes of the saturated solid and vapor, and -dm/dt is the rate of evaporation. The integral of this with respect to time is just ##p_{sat}v_{vapor}m=\frac{m}{\mu} RT##
So you assume that the vapor equilibrium pressure ##P_{sat}## remains constant throughout the process. Is that a realistic assumption (maybe it is) but is it compatible with the assumption that the sublimation process has to be at constant temperature?
 
  • #19
22,071
5,031
So you assume that the vapor equilibrium pressure ##P_{sat}## remains constant throughout the process. Is that a realistic assumption (maybe it is) but is it compatible with the assumption that the sublimation process has to be at constant temperature?

The equilibrium vapor pressure is a uniquely determined by the temperature. So, if the temperature is constant, the equilibrium vapor pressure is constant.
 
  • #20
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,475
@Chestermiller , because the volume of the vapor is increasing (so we have some sort of expansion) and the temperature of the vapor is assumed to be constant , what I have in mind is an isothermal expansion of some sort. (I understand is not completely the same). But ##PV=nRT## and if T remains constant and V keeps increasing , P cant be constant, so seems to me that pressure cant be constant at the interface of the vapor with the solid.
 
  • Like
Likes Klaus von Faust
  • #21
Klaus von Faust
37
10
@Chestermiller , because the volume of the vapor is increasing (so we have some sort of expansion) and the temperature of the vapor is assumed to be constant , what I have in mind is an isothermal expansion of some sort. (I understand is not completely the same). But ##PV=nRT## and if T remains constant and V keeps increasing , P cant be constant, so seems to me that pressure cant be constant at the interface of the vapor with the solid.
@Delta², do you see any mistakes in my solution? I don't know where did I commit mistakes.
 
  • #22
22,071
5,031
@Chestermiller , because the volume of the vapor is increasing (so we have some sort of expansion) and the temperature of the vapor is assumed to be constant , what I have in mind is an isothermal expansion of some sort. (I understand is not completely the same). But ##PV=nRT## and if T remains constant and V keeps increasing , P cant be constant, so seems to me that pressure cant be constant at the interface of the vapor with the solid.
We're talking about a change of phase here, from a solid to a vapor. You are aware that the ideal gas law does not apply to a change of phase, right?
 
  • #23
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,475
We're talking about a change of phase here, from a solid to a vapor. You are aware that the ideal gas law does not apply to a change of phase, right?

But you use the ideal gas law in your post #12 you write ##P_{vapor}V_{vapor}=n_{vapor}RT##...
 
  • #24
22,071
5,031
But you use the ideal gas law in your post #12 you write ##P_{vapor}V_{vapor}=n_{vapor}RT##...
Yes. That all applies to the vapor.

I have an alternative derivation which I think you guys will find interesting and which will please you.

Let the rate of mass flow of vapor at the solid-vapor boundary be given by ##\phi##. This can be expressed as $$\phi=\rho v A$$ where ##\rho## is the mass density of the vapor at the boundary, v is the velocity of the vapor leaving the boundary, and A is the cross sectional area of the boundary. The change in velocity of the evaporating material at the boundary is just v, since the (entering) velocity of the solid is essentially zero. So the rate of change in momentum of the material ejected at the boundary is
$$rate\ of\ change\ of\ momentum=\phi v=\phi \left(\frac{\phi}{\rho A}\right)=\frac{\phi^2}{\rho A}$$But, from the ideal gas law, $$\rho = \frac{P\mu}{RT}$$ where P is the pressure of the vapor at the boundary. If we combine these equations, we obtain:$$rate\ of\ change\ of\ momentum=\frac{\phi^2 RT}{\mu PA}$$But the force PA at the boundary is equal to the rate of change of momentum of the material ejected at the boundary: $$F=PA=\frac{\phi^2 RT}{\mu PA}$$From this it follows that the force at the boundary is given by $$F=\phi\sqrt{\frac{RT}{\mu}}$$But, $$\phi=-\frac{dm}{dt}$$ So, $$F=\left(-\frac{dm}{dt}\right)\sqrt{\frac{RT}{\mu}}$$where m is the remaining mass of solid at time t. In addition, we know from Newton's 3rd law that $$F=M\frac{dV}{dt}$$ where V is the velocity of the cup. So,
$$M\frac{dV}{dt}=\left(-\frac{dm}{dt}\right)\sqrt{\frac{RT}{\mu}}$$If we integrate from time 0 to time t, we obtain the following for the terminal velocity:
$$V=\frac{m_0}{M}\sqrt{\frac{RT}{\mu}}$$
So, this matches the result in your book and I stand corrected.

Chet
 
  • Like
Likes Klaus von Faust, Delta2 and Charles Link
  • #25
Charles Link
Homework Helper
Insights Author
Gold Member
5,352
2,632
I'm joining this discussion after it is already perhaps nearly completed, but I see something in the analysis that I'm not completely convinced of. The mass ## M ## can be considered to be much more than ## m_o ##, so that the terminal velocity ## V ## will be quite small. The gas molecules will leave the solid state and enter the vapor state with some average x-velocity ## \bar{v}_x ##. The total momentum delivered to the container will then be ## m_o \bar{v}_x ##. The problem is then to calculate and/or estimate ## \bar{v}_x ##. ## \\ ## It appears the assumption is being made above that the cylindrical container is essentially long and narrow enough that it is taking on some of the properties of a closed container, where there is actually a pressure that gets built up at the solid-vapor interface. Without this assumption, it would appear the problem may have a somewhat different answer. ## \\ ## The above analysis seems to make use of and/or derives the result that using ## \frac{1}{2} \mu \, \bar{v_x^2}=\frac{1}{2} k_B T ## , that ## P=\frac{N}{V} k_B T=\frac{N}{V} \mu \, \bar{v_x^2} ##. A good estimate for ## \bar{v}_x ## may be ## \bar{v}_x \approx \sqrt{\frac{k_BT}{\mu} } ##, but I don't believe the result is exact. (## \bar{v_x^2}=\frac{k_B T}{\mu} ## could mean that ## \bar{v}_x \approx \sqrt{\frac{k_B T}{\mu}} ## for this sublimation case.)## \\ ## With this approximation, I get a result that agrees with the above. The above problem is one of a non-equilibrium nature, and it appears the best one can do with computations would be to use results that are from equilibrium cases for best estimates, and perhaps even to assume a quasi-equilibrium, as in @Chestermiller 's calculation. ## \\ ## Edit: It then has occurred to me, with a long tube, it could be made into an effusion type problem with a small aperture. I do think that perhaps then ## \bar{v}_x=\sqrt{\frac{k_B T}{\mu}} ##, for molecules leaving the aperture, but I need to do some additional calculations to see if that is the case. A quick analysis comparing the average x-velocity during the effusion process to one involving the pressure that occurs on the wall of a container suggests that perhaps ## \bar{v}_x=\sqrt{\frac{k_B T}{\mu}} ##. A more careful analysis, however, is showing a slightly different result: ## \\ ## Considering x-momentum ## p_x ## that gets outside of the container: ## \frac{dp_x}{dt}=(\frac{dm}{dt})\bar{v}_x=[\mu \frac{N \bar{v}}{4 V}A] \bar{v}_x ## (where ## \bar{v}_x ## is the average x-velocity of the escaping molecules.) ## \\ ## Now ## F=PA=2 \frac{dp_x}{dt}=\frac{N}{V} k_B T \, A =2 \mu [\frac{N \bar{v}}{4 V}A]\bar{v}_x ## where ## \bar{v}=\sqrt{\frac{8 k_BT}{\pi \mu}} ## is the average speed in a Maxwell-Boltzmann distribution. Solving for ## \bar{v}_x ##, we get ## \\ ## ## \bar{v}_x=\sqrt{\frac{\pi k_BT}{2 \mu}} ##. ## \\ ## Additional edit: I think I see why the effusion result is slightly different: The effusion process will artificially raise the mean velocity because the distribution of the velocities of the emerging particles is such that their average speed will be faster than the average speed of the particles in the container. The emerging particles with higher velocities has the effect of cooling the remaining gas, which gets re-energized by the walls of the container that keep it at temperature ## T ##. In that sense, this result with the effusion process can not be expected to give the same answer as @Chestermiller 's result.
 
Last edited:
  • Like
Likes Lord Jestocost
  • #26
22,071
5,031
I'm joining this discussion after it is already perhaps nearly completed, but I see something in the analysis that I'm not completely convinced of. The mass ## M ## can be considered to be much more than ## m_o ##, so that the terminal velocity ## V ## will be quite small. The gas molecules will leave the solid state and enter the vapor state with some average x-velocity ## \bar{v}_x ##. The total momentum delivered to the container will then be ## m_o \bar{v}_x ##. The problem is then to calculate and/or estimate ## \bar{v}_x ##. ## \\ ## It appears the assumption is being made above that the cylindrical container is essentially long and narrow enough that it is taking on some of the properties of a closed container, where there is actually a pressure that gets built up at the solid-vapor interface. Without this assumption, it would appear the problem may have a somewhat different answer.
I don't think this assumption is made at all. The only assumption (if you want to even call it an assumption) is that the local gas pressure at the phase boundary is equal to the equilibrium vapor pressure of the solid. It, of course, rapidly decreases with distance from the phase boundary.

## \\ ## The above analysis seems to make use of and/or derives the result that using ## \frac{1}{2} \mu \, \bar{v_x^2}=\frac{1}{2} k_B T ## , that ## P=\frac{N}{V} k_B T=\frac{N}{V} \mu \, \bar{v_x^2} ##. A good estimate for ## \bar{v}_x ## may be ## \bar{v}_x \approx \sqrt{\frac{k_BT}{\mu} } ##, but I don't believe the result is exact. (## \bar{v_x^2}=\frac{k_B T}{\mu} ## could mean that ## \bar{v}_x \approx \sqrt{\frac{k_B T}{\mu}} ## for this sublimation case.)
That's not how I (at least in my analysis) arrived at the vapor bulk velocity of the gas at the phase boundary at all. I arrived at the result by setting the rate of change of momentum of the evaporating vapor equal to the force on the phase boundary.

I think that this result is totally amazing. It shows that the bulk velocity of the vapor at the phase boundary is independent of the mass flux and the density of the vapor at the boundary, and is a function only of the temperature. It also shows that the density of the vapor at the boundary is directly proportional to the mass flux at the boundary, with the constant of proportionality being a function only of the temperature. I find all these results quite astonishing and unexpected.
 
  • #27
Charles Link
Homework Helper
Insights Author
Gold Member
5,352
2,632
I don't think this assumption is made at all. The only assumption (if you want to even call it an assumption) is that the local gas pressure at the phase boundary is equal to the equilibrium vapor pressure of the solid. It, of course, rapidly decreases with distance from the phase boundary.


That's not how I (at least in my analysis) arrived at the vapor bulk velocity of the gas at the phase boundary at all. I arrived at the result by setting the rate of change of momentum of the evaporating vapor equal to the force on the phase boundary.

I think that this result is totally amazing. It shows that the bulk velocity of the vapor at the phase boundary is independent of the mass flux and the density of the vapor at the boundary, and is a function only of the temperature. It also shows that the density of the vapor at the boundary is directly proportional to the mass flux at the boundary, with the constant of proportionality being a function only of the temperature. I find all these results quite astonishing and unexpected.
@Chestermiller I was quite impressed with the results of your derivation as well. I need to study this whole scenario in more detail=I first looked at it tonight, but it is quite an interesting problem.
 
  • Like
Likes Chestermiller
  • #28
22,071
5,031
@Chestermiller I was quite impressed with the results of your derivation as well. I need to study this whole scenario in more detail=I first looked at it tonight, but it is quite an interesting problem.
I totally agree. Fascinating.
 
  • Like
Likes Charles Link
  • #30
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,475
@Chestermiller it is obvious from this

##F=PA=\frac{\phi^2 RT}{\mu PA}##

that the pressure ##P## in the boundary cannot be considered constant , since in the most general case ##\phi=\phi(t)=-\frac{dm}{dt}## will be time varying.
 
  • #31
Charles Link
Homework Helper
Insights Author
Gold Member
5,352
2,632
@Delta² and @Chestermiller Please see my Edited last part of post 25, where I have once again made a couple additional changes and inputs. I don't know that Chestermiller is taking ## P ## to be constant, but there are a couple of items that I still think need to be looked at very carefully. I think there are a couple of underlying assumptions in @Chestermiller 's analysis, but I am quite impressed by it, in any case.
 
Last edited:
  • #32
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,475
@Delta² and @Chestermiller Please see my Edited last part of post 26, where I have once again made a couple additional changes and inputs. I don't know that Chestermiller is taking ## P ## to be constant, but there are a couple of items that I still think need to be looked at very carefully. I think there are a couple of underlying assumptions in @Chestermiller 's analysis, but I am quite impressed by it, in any case.

He was taking ##P## to be constant in his approach in post #12. In his new approach , he doesn't make such an assumption.
 
  • Like
Likes Charles Link
  • #33
Charles Link
Homework Helper
Insights Author
Gold Member
5,352
2,632
He was taking ##P## to be constant in his approach in post #12. In his new approach , he doesn't make such an assumption.
In his latest approach, the one difficulty I am having with the introduction of pressure ## P ## is that it seems there needs to be some obstruction either by multiple collisions, or introducing a closed container with a small hole in it like I am doing in the last part of post 25. Otherwise, there doesn't seem to be a mechanism to cause any pressure to occur. If the walls of the container are supplying this reverse pressure by slowing some of the particles, then there could be some momentum loss in the x-direction that needs to be computed there.## \\ ## Pressure would imply a force from particles moving in the minus-x direction, and there is for the most part, no particle motion in the minus x-direction, unless some additional assumptions and/or additional boundaries are applied. ## \\ ## Edit: I do think the assumption of many collisions between the particles (in order to create a pressure ## P ##) may be inherent in @Chestermiller 's calculations. In any case, if this is assumed, the equations of @Chestermiller 's derivation look to be correct.
 
Last edited:
  • #34
22,071
5,031
@Chestermiller it is obvious from this

##F=PA=\frac{\phi^2 RT}{\mu PA}##

that the pressure ##P## in the boundary cannot be considered constant , since in the most general case ##\phi=\phi(t)=-\frac{dm}{dt}## will be time varying.
How do we know that dm/dt won't be constant. Apparently it will if P is constant. And there is nothing in the other equation coupling the force F with dV/dt to prevent this.
 
  • #35
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,475
How do we know that dm/dt won't be constant. Apparently it will if P is constant. And there is nothing in the other equation coupling the force F with dV/dt to prevent this.

Well to be honest I am not an expert in sublimation process, but I think that at start dm/dt gonna be large, since the mass that is sublimating is larger in volume and surface boundary, and afterwards , dm/dt will get smaller and smaller as the volume and surface boundary shrinks due to loss of mass...What do you think?

EDIT: OK, if the sublimation is done in such a way that the layer of the sublimating substance has always surface A, then dm/dt probably will be constant.

But where exactly your approach at post #12 goes wrong if it isn't because of the non constant pressure?
 
Last edited:

Suggested for: Another Thermodynamics Problem

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
Top