-1

Suppose in a gaseous container we have two movable pistons , consider its moving with speeds u and v and between them is the monoatomic gas , now at later time its temperature might increase , when we write first law of thermodynamics between two stages why dont we consider energy change due to change in momentum of gas particles inside also in the linear momentum equation of whole system(gas + piston)? My Sir said usually the olympiad problems like this one the change in momentum is very negligible and neglected, does there a proof of this? Like this pf problem too:

enter image description here

(in case not readable: In a long fixed horizontal pipe two identical pistons $A$ and $B$ each mass $m=415 \mathrm{~g}$ can slide without friction. The pipe and the pistons are made of perfect heat insulating materials. The space between the pistons is filled with $n=0.1$ mole of gaseous helium. The piston $\mathrm{A}$ is given an initial velocity $u=12 \mathrm{~m} / \mathrm{s}$ towards the piston B. Find maximum temperature change of the gas in ensuing motion of pistons.)

Qmechanic
  • 201,751
Orion_Pax
  • 512
  • My Sir said something along lines that change in momentum is very negligible thats why is not considered , i didnt know why so i asked for proof of this . In the problems like above – Orion_Pax Dec 10 '21 at 06:36
  • 2
    Maybe your teacher was talking about momentum of the bulk. The temperature in a box does not change if the box gets a momentum. The initial impact is distributed among $10^{23}$ particles per mole, and then there is a steady state. – anna v Dec 10 '21 at 07:23
  • Are you assuming that, once the system reaches equilibrium, everything is moving at a constant speed or that the system comes to a stop at long times? – Chet Miller Dec 10 '21 at 12:35
  • Here for example as u all can see in the second problem they asked for max temperature change which happens when both moves with same speed (although i dont have a proof as to why bc i dont know how we can neglect change in momentum from intial to same speed time ) @Chet_Miller – Orion_Pax Dec 10 '21 at 12:38
  • So you are neglecting viscous dissipation at the tube wall, and, in the end, everything is moving at the same speed (as determined by conservation of momentum). But the bulk kinetic energy has changed, right? You are basically treating this as an inelastic collision. – Chet Miller Dec 10 '21 at 12:58
  • Sir by bulk kinetic energy can u explain a bit what u meant to say , do u meant gas particles KE? And how its inelastic collision Sir energy is conserved here isnt ? (Ignoring viscous drag) and Sir can u give a detail microscopically whats going on so as therefore we r "ignoring momentum of gas particles" at any point of time when writing momentum equation for whole system(gas+piston)? @Chet_Miller – Orion_Pax Dec 11 '21 at 12:10
  • And can u also describe the full motion of this gas-piston setup , will the gas after reaching max temp with same speed of pistons , will now they remain like this and continue to go on forever ? – Orion_Pax Dec 11 '21 at 12:15
  • @Orion_Pax It's hard to read the text in your image, but the image only shows a velocity $v$ for piston A. Does the text in the image say the velocity of piston B is $u$? – Bob D Dec 11 '21 at 15:45
  • I gave two examples Sir one having both pistons some speed , other one in pic which has only one having speed u . Wait i will use mathpix to latex that one for better reading the text. – Orion_Pax Dec 11 '21 at 15:50
  • Also @anna_v Mam by" momentum of bulk" do u mean " momentum of gas particles"? And how u r sure that temp will not change when the momentum from box gets transferred to gas part , isnt gas particles kinetic energy may fall or increase ? So that leads to temp change? Pls give a proof in the answer or just explain here if possible in detail? – Orion_Pax Dec 11 '21 at 21:07
  • @Bob_D Sir now is it understandable? – Orion_Pax Dec 13 '21 at 08:38

1 Answers1

3

"Why doesn't the First Law (e.g., $\Delta U=W+Q$, where $U$ is internal energy, $W$ is work, and $Q$ is heat) incorporate bulk motion or momentum?"

For simplicity and convenience. We often assume that the system is motionless in our frame of reference, but this is not essential. I quote at length from Callen's Thermodynamics and an Introduction to Thermostatics:

In accepting the existence of a conserved macroscopic energy function as the first postulate of thermodynamics, we anchor that postulate directly in Noether's theorem and in the time-translation symmetry of physical laws.

An astute reader will perhaps turn the symmetry argument around. There are seven "first integrals of the motion" (as the conserved quantities are known in mechanics). These seven conserved quantities are the energy, the three components of linear momentum, and the three components of the angular momentum; and they follow in parallel fashion from the translation in "space-time" and from rotation. Why, then, does energy appear to play a unique role in thermostatistics? Should not momentum and angular momentum play parallel roles with the energy? In fact, the energy is not unique in thermostatistics. The linear momentum and angular momentum play precisely parallel roles. The asymmetry in our account of thermostatistics is a purely conventional one that obscures the true nature of the subject.

We have followed the standard convention of restricting attention to systems that are macroscopically stationary, in which case the momentum and angular momentum arbitrarily are required to be zero and do not appear in the analysis. But astrophysicists, who apply thermostatistics to rotating galaxies, are quite familiar with a more complete form of thermostatistics. In that formulation the energy, linear momentum, and angular momentum play fully analogous roles. [emph. added]

Callen then gives an example involving a stellar atmosphere in motion.

"Classically, why don't we see a difference in temperature for two identical systems of gas moving at different speeds?"

Because the temperature is defined to eliminate bulk motion; it measures the activity of the molecules relative to their center of mass, not the movement of the center of mass.

"How is this relevant to the piston–gas problem?"

Since we're free to assume any inertial frame regarding changes in temperature, let's assume for convenience that we're moving to the right at speed $u/2$, half the initial speed of the left piston. Now the problem is symmetric, with both pistons moving inward at the same speed. (The gas is now moving to the left at speed $u/2$, but this doesn't matter because we're going to focus on the gas temperature; as noted above, this is defined to be independent of the gas speed.)

Because the speeds are far less than the speed of sound in the gas, the pressure is approximately the same everywhere in the gas; this pressure decelerates the pistons to a stop, which occurs for both pistons at approximately the same time because of symmetry. At this moment, the pistons have zero kinetic energy in our frame, and this is the point where the gas achieves its highest temperature.

"So is the momentum of a moving region of gas ever relevant?"

Yes, if the bulk speed changes, which isn't the case in the frictionless piston–gas problem. This is the origin of the stagnation temperature, which measures the increase in the bulk temperature if the gas is brought to a halt. Here, the bulk kinetic energy of the gas has been converted entirely into thermal energy.

  • Sir in that frame of u/2 i didnt get few things in this frame what will be the comment on the momentum of gas particles ? And in this how does speed of sound of gas comes in effect and how from there concluded that they both need to stop at approximately same time ? And why does in this frame only we r considering zero kinetic energy effect why cant we chose any other frame? And how does zero kinetic energy proofs that all energy get passed into that gas molecules inside and causing there translational kinetic energy to increase and hence the rise in temperature ? – Orion_Pax Dec 11 '21 at 12:01
  • And why its true that any piston will come to stop , i want to know whats going on inside microscopically which causes the pistons to come to stop ? And momentum in case of gas-piston problem why is it ignored (momentum of gas ) when considering whole system (piston+ gas) ? And in this piston-gas problem isnt the region of gas inside moving ,How its different from the last example u shared Sir (bulk one )? And by "momentum of a moving region of gas" can u explain in brief what u meant ? – Orion_Pax Dec 11 '21 at 12:06
  • 1
    Thank you; I've edited my answer. – Chemomechanics Dec 11 '21 at 20:05
  • Sir somethings which r still not cleared to me : what u meant by "bulk speed " is it that of gas particles speed u meant ? Or piston + gas combined speed ? Or piston speed ? And secondly does in"any inertial frame only "temp is invariant the max will be achieved same as if just considered ground frame? Thirdly whats the difference between my problem setup (piston+ gas) and yours one gas in container ,as such in both cases gas temp rises isnt? Fourthly i dont get why are u including" speed of sound of gas" here in the explanation i mean how is it connected to my question ? – Orion_Pax Dec 13 '21 at 08:01
  • We r talking about speed of gas particles, not speed of sound of gas here isnt ? And fifthly when at any time the pistons goes a bit inside in the frame of u/2 , will the pressure change in the gas inside be changed to some other value than previously which will be constant in the whole gas chamber ? And whats the aftermotion looks like will after the max temperature it gained will the pistons speeds be remaining same as u/2 and it will go on forever or pressure inside gas will cause some affect which will change the speed of the piston and temperature will fall of from max in aftermotion ? – Orion_Pax Dec 13 '21 at 08:08
  • "We often assume that the system is motionless in our frame of reference, but this is not essential." Here" by system "are we referring to "gas molecules" to be motionless and thats why zero momentum of this ? And sir consider the post @anna_v made above and the reply i gave to her , what u think will be answer to my queries in that post i made? – Orion_Pax Dec 13 '21 at 08:37
  • Part of gaining intuition and familiarity in physics is learning how to answer questions yourself. Your new set of questions asks about the meaning of "bulk speed" and "motionless." These are definitions you can look up yourself. The nuances are even discussed on this site, as can be found with a simple search. – Chemomechanics Dec 13 '21 at 19:41
  • For choices and assumptions like a certain moving frame or a very high speed of sound, try solving the problem by making different choices or without making those assumptions; does the problem become easier or harder? – Chemomechanics Dec 13 '21 at 19:42