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Another Thermodynamics Problem

  • #1
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.
 
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Answers and Replies

  • #2
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What is the exact statement of the problem?
 
  • #3
What is the exact statement of the problem?
This is the exact statement. No more information is provided in the problem
 
  • #4
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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.
 
  • #5
Lord Jestocost
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Maybe, one should think of the conservation of total momentum after the substance has completely evaporated in a "directional" way into vacuum.
 
  • #6
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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?
 
  • #7
Delta2
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The problem doesn't give any number for heat of sublimation (or enthalpy of sublimation) of substance m?
 
  • #8
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
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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.
 
  • #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
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
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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, $$v_{terminal}=\sqrt{\frac{2mRT}{M\mu}}$$
 
  • #13
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
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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.
 
  • #15
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
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@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
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@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
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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
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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
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@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.
 
  • #21
@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
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@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
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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
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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
 
  • #25
Charles Link
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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.
 
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