How does stopping a moving gas affect its temperature?

[vinter]

OK, here is a question I found in a book some time ago :-

An insulated box containing a monatomic gas of molar mass M moving with a velocity v is suddenly stopped. Find the increment in the gas temperature as a result of stopping the box.

I thought on this and what came to my mind was this - I should first know the the exact definition of temperature to tackle this question. As I know, temperature is defined very precisely by the zeroth law of thermodynamics. But here, I need a definition based on the energy of the molecules. Of course, temperature is directly proportional to the total internal energy of the gas. And what is internal energy? It is the sum of several forms of mechanical energies one of them being the kinetic energy. Now, we know that the KE depends on the frame of reference, it comes that the temperature should also depend on the frame of reference. But does it? This is one question that comes from the main question.

The problem is not as simple as it seems.

I asked several people the same question but all did some baseless energy conservation calculations and actually reached the answer. No one seems to be realising that there is something serious here.

If you reach the answer, then you may like to think on the validity of your arguments by applying the same method to a slightly modified question - instead of the container stopping to zero velocity, its velocity suddenly changes to -v. Now find the change in temperature.

Help please!

 

[Clausius2]

OK, here is a question I found in a book some time ago :-

An insulated box containing a monatomic gas of molar mass M moving with a velocity v is suddenly stopped. Find the increment in the gas temperature as a result of stopping the box.

I thought on this and what came to my mind was this - I should first know the the exact definition of temperature to tackle this question. As I know, temperature is defined very precisely by the zeroth law of thermodynamics. But here, I need a definition based on the energy of the molecules. Of course, temperature is directly proportional to the total internal energy of the gas. And what is internal energy? It is the sum of several forms of mechanical energies one of them being the kinetic energy. Now, we know that the KE depends on the frame of reference, it comes that the temperature should also depend on the frame of reference. But does it? This is one question that comes from the main question.

The problem is not as simple as it seems.

I asked several people the same question but all did some baseless energy conservation calculations and actually reached the answer. No one seems to be realising that there is something serious here.

If you reach the answer, then you may like to think on the validity of your arguments by applying the same method to a slightly modified question - instead of the container stopping to zero velocity, its velocity suddenly changes to -v. Now find the change in temperature.

Help please!

An interesting question. Thermodynamic temperature doesn't depend on the reference frame. That temperature is based on the <mean> temperature of the molecules and is measured somehow by definition in the reference frame attached to the center of mass of the group of molecules taken into account. It has no sense asking oneself about the thermodynamic temperature viewed from another reference frame, because that temperature won't represent the real thermal state of the substance.

If the container is moving at velocity $$v$$ towards the right, it will have two components of energy:

-One measured in the reference frame I said above, and it represents the local thermodynamic state: $$U_1=mc_vT_1$$ is the internal energy which contains the microscopic kinetic energy of the molecules. Under the assumption of local thermodynamic equilibrium there exists a mean velocity of fluctuation of the molecules, which is given by the Kinetic Theory.

-Another measured in the reference frame you want, always if you are coherent with the rest of the calculation. I have chosen the reference frame attached to ground. It represents the macroscopic kinetic energy: $$E_1=1/2 m v^2$$

Once the container have been stopped, its total energy is only the thermal energy based on the local thermodynamic properties: $$U_2=mc_vT_2$$

Under the assumption of adiabatic walls, the container doesn't exchange any heat with the external environment. And suppossing the effects of internal (viscous dissipation) are negligible, the desacceleration process can be approximated as an isentropic one:

$$U_1+E_1=U_2$$

which gives us an important result:

$$T_1+\frac{v^2}{2c_v}=T_2$$

This can be reshaped into a similar form:

$$\frac{T_2}{T_1}=1+\frac{v^2}{2c_vT_1}$$

This last result is similar to one found in Gas Dynamics Theory. The last term is of the order of the Mach Number. If you study Gas Dynamics some time, you will realize that the behavior of a compressible fluid particle inside a flow when being accelerated/desaccelerated, can be described with the whole approximation of this container problem. The temperature $$T_2$$ represents the stagnation temperature of the flow, and it depends by this proper definition on the reference frame chosen. On the other hand, the temperature $$T_1$$ is the so called Static temperature, and it is only a function of the local thermodynamic state as I said.

I hope this helps you. Otherwise post your doubts.

 

[Timbuqtu]

OK, here is a question I found in a book some time ago :-

An insulated box containing a monatomic gas of molar mass M moving with a velocity v is suddenly stopped. Find the increment in the gas temperature as a result of stopping the box.

I thought on this and what came to my mind was this - I should first know the the exact definition of temperature to tackle this question. As I know, temperature is defined very precisely by the zeroth law of thermodynamics. But here, I need a definition based on the energy of the molecules. Of course, temperature is directly proportional to the total internal energy of the gas. And what is internal energy? It is the sum of several forms of mechanical energies one of them being the kinetic energy. Now, we know that the KE depends on the frame of reference, it comes that the temperature should also depend on the frame of reference. But does it? This is one question that comes from the main question.

First of all temperature is NOT defined by the average kinetic energy of a gas. It is defined as the proportionality between change of entropy and change of energy.

But yes, in case of an ideal gas the temperature is proportional to the average kinetic energy of the gas. But in an ideal gas the particle-motion is isotropic, the same number of particles moving to the left as moving to the right. So only is the frame of reference in which the average velocity is zero the temperature is proportional to the average kinetic energy. So the answer is no: temperature doesn't depend on the frame of reference.

 

[vinter]

Thanks Clausius for your reply. But does your equation apply to my second problem? Suppose the container, instead of stopping completely, changes its velocity to -v. In that case, your $$E_{1}$$ term does not change. Will the temperature not change in that case?

Secondly, I would prefer a much rigorous treatment taking care of the definition of temperature. First, what's the definition? Is it -
The avg. KE of the molecules as seen from the centre of mass reference frame?

If yes, then I would prefer to think like :-
The initial temperature = avg. of half * m * (v-v0)^2
where v0 is the velocity of the centre of mass from the ground frame
Similarly, find the expression for the final temperature and then see what types of relations they exhibit.

 

[vinter]

First of all temperature is NOT defined by the average kinetic energy of a gas. It is defined as the proportionality between change of entropy and change of energy.

So do you mean, that (in the classical sense) if you have been told the KE's of all the molecules of a gas, and their masses also, then you can't find the temperature?

 

[Clausius2]

Thanks Clausius for your reply. But does your equation apply to my second problem? Suppose the container, instead of stopping completely, changes its velocity to -v. In that case, your $$E_{1}$$ term does not change. Will the temperature not change in that case?

Of course not if you assume an isentropic acceleration/decceleration. What happens in real world?. Although you could suppose adiabatic walls, there will be internal irreversibilities such as viscous dissipation which enhances a greater temperature at the final of the process.

Secondly, I would prefer a much rigorous treatment taking care of the definition of temperature. First, what's the definition? Is it -
The avg. KE of the molecules as seen from the centre of mass reference frame?

If yes, then I would prefer to think like :-
The initial temperature = avg. of half * m * (v-v0)^2
where v0 is the velocity of the centre of mass from the ground frame
Similarly, find the expression for the final temperature and then see what types of relations they exhibit.

As Tumbuctú has said, the definition of temperature is the partial of the internal energy respect to the entropy remaining volume constant. In some way it will have to match with the result of the kinetic theory in which the root mean square of the molecular velocity is proportional to the square root of the temperature. In other words, temperature is proportional to the square of the rms velocity and so to the average kinetic energy.

Leave me some time to see if I am able to explain this mathematically.

 

[toocool_sashi]

This is a JEE 2 marks question hehe :D and yes i did conservation of energy without thinking too much. I always had this small niggling doubt that Kinetic Energy of a system depends on your frame of reference. I cannot give an alternate definition of temperature or anything of that sort but i can say that if the velocity changes from v to -v there will be no temperature change. Firstly let this be clear...we are talking abt temperature change of box. now we have decreased the velocity from v to 0 and the temperature of box can be calculated from conservation of energy(friction, sound dissipation absent). i do not know anything abt temperature's relevance to frame...i just know that if conservation of energy holds...i can calculate temperature using conservation principle. immediately heat transfers from the box to the surrounding to maintain thermal equilibrium. now when box is again given a velocity of -v ... it gains KE and logically...and of course mathematically...box shud attain same temp as it had when it had velocity of v.

 

[Gokul43201]

You're forgetting that work needs to be done by some external body to bring the box to rest. How do you use energy conservation without including this term ?

 

[toocool_sashi]

You're forgetting that work needs to be done by some external body to bring the box to rest. How do you use energy conservation without including this term ?

Here conservation of energy implies the 7th class stuff that we learn, "Energy in an isolated system must remain conserved". Initially, only energy is KE of the box. Finally, only energy wud be the heat energy. we equate both of them.

 

[Gokul43201]

How is the system isolated if an external body is interacting with it ? And since when did JEE start asking "7th class stuff" ?

 

[vinter]

Hey are you all Indians here like me? If not, how do you know about JEE. In case you want to tell that it's a very prestigious exam and is famous throughout the world then do tell me because I just qualified in the exam one month ago ;-)
And yes, applying energy conservation is not that simple here, toocoolsashi. Indeed, the system is isolated and you can use conservation of energy, i.e., the total energy = constant. Now, some part of this total energy is related to the temperature and some to miscelleneous stuff like KE.
So, you have,
energy related to temperature + miscelleneous stuff = constant.
To get anything out of this equation, you need to know too much about the first term. You need to be aware of the relation between temp. and energy, for example. And this is not simple, since you may say that temperature is just a proportionality constant times the avg. KE of the molecules, but, KE from which reference frame? Ground reference frame? Or the frame of the container? Or that of the Zem mattresses of the planet Squornshellous Beta?

 

[Gokul43201]

Hey are you all Indians here like me? If not, how do you know about JEE. In case you want to tell that it's a very prestigious exam and is famous throughout the world then do tell me because I just qualified in the exam one month ago ;-)

Not all. And the JEE is virtually unknown outside India.

Indeed, the system is isolated and you can use conservation of energy, i.e., the total energy = constant.

Please explain to me how this system is isolated when there is an external agent exchanging momentum (and hence, energy) with the body. You can not bring something to rest without doing work on it.

And this is not simple, since you may say that temperature is just a proportionality constant times the avg. KE of the molecules, but, KE from which reference frame? Ground reference frame? Or the frame of the container? Or that of the Zem mattresses of the planet Squornshellous Beta?

Even jolly ol' Adams can tell you that temperature is defined (approximately, or for an ideal gas) as the average KE in the CoM frame of the gas (or in this case, the rest frame of the box). If it weren't defined this way, there would be all kinds of double-counting and make a mess of energy conservation.

It's not hard to calculate the work done by the external body - just use momentum conservation. Of course, this may lead to surprising results, but such is physics.

 

[Gokul43201]

An insulated box containing a monatomic gas of molar mass M moving with a velocity v is suddenly stopped.

I believe there is no increment in the temperature, but let me find some time to write up a proof.

 

[krab]

An insulated box containing a monatomic gas of molar mass M moving with a velocity v is suddenly stopped.

The word "suddenly" is very important here. Consider that the box is stopped in a time that is short compared with the time it takes a sound wave to travel the length of the box. Then stopping requires no work (except for the work on the mass of the box itself), and all the kinetic energy of the uniform motion of the gas is translated first into a sound wave, then gradually as the sound dissipates, into a temperature rise.

OTOH, if the deceleration takes a long time, the kinetic energy of the gas is allowed to do work on the decelerator, and in the limit, there is no temperature rise.

 

[Gokul43201]

The word "suddenly" is very important here. Consider that the box is stopped in a time that is short compared with the time it takes a sound wave to travel the length of the box. Then stopping requires no work (except for the work on the mass of the box itself), and all the kinetic energy of the uniform motion of the gas is translated first into a sound wave, then gradually as the sound dissipates, into a temperature rise.

Krab, I'm not sure I follow this argument entirely. Just after the box is stopped "instantaneously", the molecules of the gas will have a net momentum (=Nmv_0, for N molecules, each of mass m) and hence a sound wave propagates through the box. How is this net momentum erased without the addition of an external impulse ?

 

[krab]

It's like throwing a ball of putty against a wall. The momentum is transferred to the earth. But the energy heats the putty.

 

[Gokul43201]

But the energy heats the putty.

Does it really ? How do you explain this physically ? Why does the wall not get heated as well ? I can't see how you can transfer momentum without transferring energy.

 

[krab]

I can't see how you can transfer momentum without transferring energy.

Old question. Often answered.

https://www.physicsforums.com/showthread.php?t=74337
https://www.physicsforums.com/showthread.php?t=51321

Momentum transferred is $M\delta v=mv$ (M is mass of Earth and m is mass of gas.) Energy transferred is $M(\delta v)^2/2$, which is utterly negligible compared with $mv^2/2$; it is in fact smaller by a factor m/M (do the math). So all the energy except about 1 part in $10^{26}$ stays locally.

 

[Gokul43201]

Old question. Often answered.

https://www.physicsforums.com/showthread.php?t=74337
https://www.physicsforums.com/showthread.php?t=51321

Only, not in either of those threads.

Momentum transferred is $M\delta v=mv$ (M is mass of Earth and m is mass of gas.) Energy transferred is $M(\delta v)^2/2$, which is utterly negligible compared with $mv^2/2$; it is in fact smaller by a factor m/M

I completely agree. The KE transferred to the Earth is negligible (By the way, where do we have the Earth in the OP's problem ? Could the box not be stopped by another box?)

So all the energy except about 1 part in $10^{26}$ stays locally.

In an elastic collision. The putty getting smashed into a wall is hardly elastic. What about the strain and heat energies ?

Anyway, I don't think this is pertinent to this problem because I don't see why the box must be stopped by an object that is rigidly coupled to the earth.

(do the math)

I guess I shall. I may end up proving myself wrong after all.

 

[Doc Al]

In an elastic collision. The putty getting smashed into a wall is hardly elastic. What about the strain and heat energies ?

An elastic collision is hardly necessary for the energy to remain in the putty. The translational kinetic energy of the putty is transformed into internal energy. Since the point of application of the wall's force on the putty does not move, that force does no real work on the putty and thus does change its total energy (to the degree that krab estimated).

A similar situation involving deformable bodies is that of person vertically jumping off the ground. The ground exerts a force on the person, but again that force does no real work. In this case the internal (chemical) energy of the person is transformed into KE of the center of mass.

 

[Gokul43201]

Let the initial RMS velocity of the molecules of the gas molecules (in the rest frame of the box) be $\hat {v}$ and the final RMS velocity be $\hat {v'}$. Let there be N molecules in the box, the box having negligible mass (whatever the mass of the box, it doesn't affect the calculation, so just call it zero). N/6 molecules of the gas can be treated as moving along each of the 6 cartesian directions at any instant of time, with a speed, $\hat {v}$ in the box-frame. Their velocities in the ground-frame are given by $\mathbf {\hat {v} + v_0}$, where $v_0$ is the speed of the box in the ground-frame.

The initial momentum of the gas in the ground-frame is

$$P_{initial} = \frac {Nm}{6} \sum_{i=1}^6 \mathbf {\hat {v_i} + v_0} = \frac {Nm}{6} \cdot 6v_0 = Nm \mathbf{v_0}$$

Now, let some constant force $\mathbf{F}$ act on the box for a time $\tau$, at the end of which the molecules of the gas have no net momentum (ie : the box has come to rest). In other words :

$$P_{final} = \frac {mN}{6} \sum_{i=1}^6 \mathbf {\hat {v'_i}} = 0$$

$$\Delta P = -Nmv_0 = -F \tau$$

Now, the work done by this force is given by :

$$W = \int \mathbf{F \cdot dx} = - \int_0^{ \tau } Fvdt = - \int_0^{ \tau } Fatdt$$
$$= - \frac {F^2}{Nm} \frac{\tau ^2}{2} = - \frac {(F \tau)^2}{2Nm}$$

Plugging in for $F \tau$ from above, the work done by the box is :

$$- W = \frac {N^2m^2v_0^2}{2Nm} = \frac {1}{2} Nmv_0^2$$

[contd. after dinner...]

 

[Gokul43201]

Next we find the initial KE of the gas molecules in the ground frame. Two-thirds of the molecules are moving (in the box-frame) along directions perpendicular to the motion of the box; one-sixth are moving in the same direction and one-sixth in the opposite direction.

$$KE_{initial} = \frac {1}{2} \frac {Nm}{6} \sum_{i=1}^6 |\mathbf {\hat {v_i} + v_0}|^2$$
$$= \frac {Nm}{12} \sum_{i=1}^6 (4[ \sqrt {\hat {v}^2 + v_0^2}~]^2 + [\hat {v} - v_0]^2 + [\hat {v} + v_0]^2 )$$
$$= \frac {Nm}{12} (4[ \hat {v}^2 + v_0^2] + 2[\hat{v}^2 + v_0^2]) = \frac {Nm}{2} (\hat {v}^2 + v_0^2)$$

The final KE, after the box has come to rest is much simpler to calculate. It is simply :

$$KE_{final} = \frac {1}{2}Nm \hat{v'}^2$$

So, by the work-energy theorem :

$$\frac {Nm}{2} (\hat {v}^2 + v_0^2) - \frac {1}{2}Nm \hat{v'}^2 = \frac {1}{2} Nmv_0^2$$
$$\implies \hat{v'}^2 = \hat {v}^2$$

$$\implies T' = T$$

 

[Gokul43201]

An elastic collision is hardly necessary for the energy to remain in the putty. The translational kinetic energy of the putty is transformed into internal energy. Since the point of application of the wall's force on the putty does not move, that force does no real work on the putty and thus does change its total energy (to the degree that krab estimated).

Okay, I guess I need to think about inelastic collisions a lot more. There's so much I just hadn't thought about.

Nevertheless, I don't think that's pertinent to the present problem. Have I made a mistake somewhere in my calculation ?

 

[vinter]

Next we find the initial KE of the gas molecules in the ground frame. Two-thirds of the molecules are moving (in the box-frame) along directions perpendicular to the motion of the box; one-sixth are moving in the same direction and one-sixth in the opposite direction.

$$KE_{initial} = \frac {1}{2} \frac {Nm}{6} \sum_{i=1}^6 |\mathbf {\hat {v_i} + v_0}|^2$$
$$= \frac {Nm}{12} \sum_{i=1}^6 (4[ \sqrt {\hat {v}^2 + v_0^2}~]^2 + [\hat {v} - v_0]^2 + [\hat {v} + v_0]^2 )$$
$$= \frac {Nm}{12} (4[ \hat {v}^2 + v_0^2] + 2[\hat{v}^2 + v_0^2]) = \frac {Nm}{2} (\hat {v}^2 + v_0^2)$$

The final KE, after the box has come to rest is much simpler to calculate. It is simply :

$$KE_{final} = \frac {1}{2}Nm \hat{v'}^2$$

So, by the work-energy theorem :

$$\frac {Nm}{2} (\hat {v}^2 + v_0^2) - \frac {1}{2}Nm \hat{v'}^2 = \frac {1}{2} Nmv_0^2$$
$$\implies \hat{v'}^2 = \hat {v}^2$$

$$\implies T' = T$$

I could not quite keep track of the algebra you have done, but I don't agree with the final result you got.

Well, I am now myself confused about why we are conserving energy. But assuming that the total KE of the molecules DOES remain constant in the full process, we can do the following :-
First recall the theorem that the KE of a system of particles as seen from a frame S can be broken into two components viz, the KE of the particles as seen from the centre of mass reference frame and the KE that would be observed from S if the full mass was concentrated at the centre of mass and moving with the velocity of the centre of mass as seen from S. (If you doubt, you can verify it, it's got a simple proof). To make things simple, we will call the first component "first component" and the second...
Now consider the molecules as a system.
Initial KE = first component + second component
= N*k*T + half*N*M*v^2
where T = initial temerature and k is the proportionality constant between T and avg. KE
Final KE = N*k*T(final)
equating the two, we get an expression for T(final). And we can see that T(final) is not equal to T.

 

[Doc Al]

Now, let some constant force $\mathbf{F}$ act on the box for a time $\tau$, at the end of which the molecules of the gas have no net momentum (ie : the box has come to rest). In other words :

The force acts on the box, not directly on the particles in the box. If the box is stopped suddenly (which is the key assumption), time is still required for the center of mass of the particles (COM) to be stopped.

$$P_{final} = \frac {mN}{6} \sum_{i=1}^6 \mathbf {\hat {v'_i}} = 0$$

$$\Delta P = -Nmv_0 = -F \tau$$

OK.

Now, the work done by this force is given by :

$$W = \int \mathbf{F \cdot dx} = - \int_0^{ \tau } Fvdt = - \int_0^{ \tau } Fatdt$$
$$= - \frac {F^2}{Nm} \frac{\tau ^2}{2} = - \frac {(F \tau)^2}{2Nm}$$

This is where the problem arises (if I understand what you're doing correctly). What I believe you are calculating is better called pseudo-work, than real work. If the box hits an "immovable" object, the force on the box does no real work (thus does not transfer energy to or from the box). But you can still apply the "pseudo-work energy" theorem (also called the "center of mass" equation): $F_{net} \Delta x_{com} = 1/2 m v_{com}^2$. This is just a result of Newton's 2nd law.

Work, in the conservation of energy sense, is not done on the box by the force F, so the total energy of the box remains the same. But the macroscopic translational KE of the particles is transformed to random KE of the particles (that is, internal energy).

Of course if the box does move while being stopped, you can calculate the real work done by applied force on the box: $W = F_{applied} \Delta x_{box}$, where by $\Delta x_{box}$ I mean the distance that the point of application of the force on the box moves. Only in the special case of a perfectly rigid body would this negative work exactly equal the change in macroscopic translational KE of the particles.

 

[Gokul43201]

So, you're saying that my calculation will work only if

(i) the box is rigid, and

(ii) the box is brought to rest, say, by compressing a spring ?

The only assumption I thought I made was that the box is rigid (and when I said "the box is stopped" I was referring the the CoM of the gas as coming to rest)...need to think more (later perhaps).

Thanks, krab and Doc !

 

[toocool_sashi]

heavy duty stuff going on in there, maybe this is purely exam oriented, but this was a 2 mark question and i think time alloted wud be around 3 minutes. I am not in2 too much depth of physics, i just know what JEE physics requires me 2 know, and maybe that's way 2 little compared 2 every1 here but what's wrong in equating KE = heat energy? obviously velocity of KE is measured from the CoM of the box.

And oh yes gokul I've found many JEE questions related 2 just high school level topics, if ur interested, there's a 1997 or a 1998 maths problem, co-geo,(centroid-circle prob) which can be done using pure geometry of 9th level and it was for 10 marks ;) cheers!