Gas in a piston
Helium gas is enclosed in a gas-tight piston which is held in position by the clamping force of an elastic spring. In mechanical equilibrium, the gas pressure p = − A F corresponds to the tensioning force F of the spring divided by the area A . The gas is then brought to a temperature of T = 6 0 0 K by rapid heating from room temperature T = 3 0 0 K .
What is the final temperature of the gas (in units of Kelvin) after setting a new mechanical balance with the spring?
Details and Assumptions:
- The volume of the gas is V = A x , where A is the cross-sectional area and x is the piston position.
- The mechanical force of the spring equals F = − k x , where k is the spring constant.
- The piston does not move during the heating from 3 0 0 K to 6 0 0 K .
- Helium can be treated as an ideal gas.
- Energy is transferred only by mechanical work. The piston is perfectly thermally insulated.
The answer is 525.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
I don't quite understand the second to last line- why is it 2nRT'?
I use the same method as before
pV=-(F/A)(Ax')=2W'
so W'=(1/2)pV=(1/2)nRT'
so shouldn't it be
U'+W'=nRT'+(1/2)nRT'?
Where have I gone wrong? Thanks!
Log in to reply
kx^2 = nRT Then : (kx^2)/2 = nRT/2 Therefore : U + W = 3nRT/2 + nRT/2 = 4nRT/2 = 2nRT (Sorry for my bad english)
I believe this solution might be flawed. It assumes that the spring is solely responsible for receiving energy during the expansion. But, it is the case that the pressure and spring force are not balanced when the piston is released. This would lead to a quick compression and the spring would end with some energy that manifests as harmonic oscillation about some length.
To be a process which does expand reversibly, an additional force must be supplied to balance the pressure while the piston relaxes. The gas does work against this force, removing more energy from the system. This means the final temperature is lower than suggested. The nature of the force is not necessary to analyze the problem; one could look at the constants of an adiabatic process (see link below) to obtain the final temperature. https://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_(reversible_process)
Using this method, I obtained the answer T f = 2 3 / 4 T 0 → T f ≈ 5 0 4 K .
Log in to reply
You are right, that an adiabatic expansion results in different temperature, that is lower then the proposed solution. But the thermal energy cannot vanish, since the system is perfectly thermally insulated. Therefore, the missing energy is stored as kinetic energy of the piston, so that we would get a harmonic oscillation of the piston. (In the same way also the temperature of the gas is oscillating.) However, the ideal gas has a finite viscosity, that causes a friction force acting on the piston. Therefore, the oscillation would be attenuated over time, so that the kinetic energy is again converted into heat. In the end we get the equilibrium with a final temperature T = 525 K.
The problem assumes, that the system reaches an equilibrium state in the end, but does not tell how we actually get there. If this equilibrium exists, the final temperature has to be T = 525 K, because of energy conversion. The temperature you proposed is only an intermediate temperature on the way to the equilibrium.
Log in to reply
Thank you, I arrived at a similar conclusion later. I think the solution fails to acknowledge this and such things are interesting and should be mentioned.
This resolution method looks alright to me except for one thing, we're told that the gas is Helium, which is a diatomic particle. Therefore U=5/2nRT because the molecules have two more degrees of liberty than hydrogen molecules. If you do the calculations again with this in mind, you find T' = 550 K
Log in to reply
Helium is a noble gas, that never bonds with ather atoms. Therefore, Helium is a monoatomic particle!
Relevant wiki: Ideal Gas Laws
Let the initial temperature be T 0 . Then the initial equilibrium yields n R T 0 = p 0 V 0 = A k x 0 ⋅ A x 0 = k x 0 2 . Similarly, the final equilibrium at temperature T f yields n R T f = k x f 2 . The thermodynamic work done by expansion of the gas is W t d = U f − U i = 2 d n R ( 2 T 0 − T f ) , where d = 3 is the degrees of freedom for a mono-atomic gas like helium. This work is equal to the mechanical work done in compressing the spring, W m e c h = 2 1 k ( x f 2 − x 0 2 ) = 2 1 n R ( T f − T 0 ) . Equating the amounts of work gives 2 d n R ( 2 T 0 − T f ) = 2 1 n R ( T f − T 0 ) ; ( 2 d + 1 ) T 0 = ( d + 1 ) T f ; T f = d + 1 2 d + 1 T 0 . Intuitive interpretation: The spring provides one additional degree of freedom per gas molecule. As the system approaches equilibrium, the increased energy in the molecular degrees of freedom becomes evenly shared with the additional, mechanical degree of freedom.
With the given values, T f = 4 7 ⋅ 3 0 0 = 5 2 5 K .
I followed same line of approach that the difference in spring potential energy (1/2kXf^2 - 1/2kXo^2) is equal to work done by gas expansion. Then I realized that overall heat energy added to the system is equal Q=nCv(600-300) where Cv=3/2R is heat capacity at constant voilume (Assumption that piston does not move during heating!!). However, only a portion of added heat can be converted into work that equals n3/2R(600 - Tf). Thus, using state equation PV=nRT for initial and final gas state and derived expressions kXo^2=nR300 as well as kXf^2=nRTf, one can write n3R(600 - Tf)=nRTf - nR300. This equals 1800 - 3Tf=Tf - 300 or Tf=525K.
A straightforward solution leads to a different answer.
Initial stage : stage 0,
Isochoric Heating to 600 K = stage 1,
Isentropic expansion to stage 2,
If x is the displacement of spring then from mechanical equilibrium:
Initial : P0.A = k xo,
Final : P2. A = k x2,
Therefore P2 = P0 (x2/x0) : equation (1),
Process 0 -1 : isochoric heating : PV = n RT,
i.e P1/P0 = T1/T0 = 600/300 = 2 : equation (2),
Isentropic expansion PV^γ = const.,
Then P2 (A.x2)^γ = P1 (A.x1)^γ --> P2 = P1 (x1/x2)^γ --> P2 = 2P0 (x0/x2)^γ equation (3),
From equations 1 and 3 and with γ = 1.4 = 7/5 (diatomic gas):
(x2/x0) = 2^1/(1+γ) = 1,3348 equation (4) ,
From equation 4, 2 and 1 : P2/P1 = 1,3348/2 = 0,6674,
Using the relation for isentropic expansion : T2/T1 = (P2/P1)^(γ-1)/γ get T2 = 534,5
For γ= 1,6667 = 5/3 (monoatomic gas) T2 = 504,5. This is the correct answer. The other proofs presented above violate the 1st Law.
Log in to reply
525 K solution from internal energy is crystal clear and seems the right one. But I learn more from mistake than from solution. I've used 2 other paths to solution with different result and it would be very helpful if someone could point the mistake. First I used the same path as above which lead to the same result, also given by Bradley Treece: 504,5°C. And I don't undestand what is wrong. I don't agree with Markus Michelmann. Even if there are oscillations, T° should tend to 504°C. The expansion from 600K to final T° is adiabatic as there is no exchange with the outside of the spring and gaz box. And I assume the expansion to be also isentropic. So PV^y should be correct. I also used the work path. Gas work during expansion is fully transfered to the spring. Then 1/2.k.(x2^2 - x0^2) = nR.(T2-T1)/(y-1) which gives a 431K, also wrong. In a conclusion I feel that expansion is not isentropic. Right?
Note: I don't find how to edit my post in order to modify my formula into LaTeX. How do you manage to use so cute LaTeX formula in comments? No preview even if reload the page...
Relevant wiki: Ideal Gas Laws
For an one-atomic ideal gas applies U p V = 2 3 n R T = n R T with the internal energy U , the amount of substance n and the gas constant R . In equilibrium, pressure and volume are determined by the piston position n R T = p V = − A F A x = k x 2 = 2 W with the mechanical energy W = 2 1 k x 2 of the spring. The total energy is then given by E = U + W = 2 3 n R T + 2 1 k x 2 = 2 n R T By supplying the amount of heat Q , the internal energy of the gas is doubled for a short time, resulting in a new total energy E ′ and a final temperature T ′ : E ′ = E + Q = 2 U + W ⇒ 2 7 n R T ⇒ T ′ = U ′ + W ′ = 2 n R T ′ = 4 7 T = 4 7 3 0 0 K = 5 2 5 K