Filling Flat Tire - Adiabatic Reversible Process

Click For Summary
SUMMARY

The discussion centers on calculating the final temperature of air in a bicycle tire after a reversible adiabatic compression, starting from an initial pressure of 1.00 bar and temperature of 298.0 K, and ending at a pressure of 5.00 bar with a final volume of 1 L. Participants highlight the importance of understanding the assumptions regarding volume and pressure during the process, particularly the constant volume of the tire and the ambiguity in measuring final pressure. The key equations referenced include the adiabatic condition P1-γ Tγ = constant and the relationship PV = NkT, which are essential for solving the problem.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with the ideal gas law (PV = NkT)
  • Knowledge of the adiabatic condition (P1-γ Tγ = constant)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation and application of the adiabatic process equations
  • Learn how to calculate final temperature using the ideal gas law
  • Explore the concept of reversible processes in thermodynamics
  • Investigate the implications of constant volume in gas compression scenarios
USEFUL FOR

Students studying thermodynamics, particularly those tackling problems involving gas laws and adiabatic processes, as well as educators looking for examples to illustrate these concepts.

HLxDrummer
Messages
2
Reaction score
0

Homework Statement


A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Take the initial P and T of the air (before it is put into the tire) to be 1.00 bar and 298.0K. The final volume of air (after it is in the tire), is 1 L and the final pressure is 5.00 bar. Calculate the final temperature (be sure to state your assumptions).

P initial = 1.00 bar
V initial = ?
T initial = 298K

P final = 5.00 bar
V final = 1 L
T final = ?

moles = ?



Homework Equations



Cval dT = P dV

P 1-γ T γ = constant

P internal = P external (since it is reversible)


The Attempt at a Solution



The biggest issue I am having is figuring out the volumes. My professor gave us a hint saying that we can assume constant volume, but what does he mean - I know we can assume the tire volume is constant, but the volume of air used to fill the tire can't be the same as the volume of the tire can it? Is that volume even relevant.

The first equation I derived but if volume is constant like my professor was saying, dV will be zero which ruins that equation. I found the PV=constant online but it seems way too easy. Is this assuming everything else is constant?

I don't care if someone comes out and gives me a direct answer I just need a little guidance. The homework is due Wens so I have a bit of time. Thank you!

There are just a ton of variables unaccounted for that could affected pressure which makes it really confusing to me - temp, moles, volume, etc.
 
Physics news on Phys.org
Volume of the tire is constant, volume of the air before and after compression is not.

Question is ambiguous, as it doesn't state whether the final pressure is measured at 298 K or immediately after pumping (when the air is hotter). That's where you have to assume something - just state it in the final result.
 
Thanks for the reply!

I would assume since the gas is at lower pressure when it is 298k that it is before it is pumped. Does this mean I can use the equations above?

Also, does the gas have to be pumped to fill the tire if it is a reversible process Pinternal=Pexternal? I was thinking in this fake model that atmospheric pressure may go to 5 bar to fill the tire without any work?
 
you've got to use PV=NkT, which holds both before and after compression. You then have two equations you can combine (eliminate Nk) to something like which can be solved for Tf.
that means Pi*Vi=Pf*Vf*Ti/Tf.

If course, first you must find the value for Vi first by using the adiabatic identity P^γ V =const.
 

Similar threads

Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K
Replies
1
Views
2K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K