Filling Flat Tire - Adiabatic Reversible Process

[HLxDrummer]

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

C_val 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.

 

[Borek]

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.

 

[HLxDrummer]

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?

 

[Cipherflak]

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.

 

[DeeNos]

First of all, great job on recognizing the need for guidance and seeking help. As a scientist, it is important to always ask questions and seek clarification when you are unsure about something.

To solve this problem, we can use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. We can also use the specific heat at constant volume, Cv, which is the amount of heat required to raise the temperature of a substance by 1 degree at constant volume.

Since the tire is nearly flat, we can assume that the volume of the tire is negligible compared to the volume of the air being pumped into it. Therefore, we can assume that the volume of the air before and after it is pumped into the tire is the same, and we can use the given final volume of 1 L.

Now, let's set up the equation using the ideal gas law and the specific heat at constant volume:

P1V1/T1 = P2V2/T2

Since the process is adiabatic, there is no exchange of heat between the air and its surroundings. This means that Q = 0, and therefore, ΔU = W (the change in internal energy is equal to the work done).

We can express the work done on the air as:

W = -PΔV

Since the process is reversible, we can use the equation Pinternal = Pexternal. Therefore, we can rewrite the work equation as:

W = -PexternalΔV

Since the process is adiabatic, we can also use the equation PVγ = constant, where γ is the ratio of specific heats. In this case, since we are dealing with air, we can use γ = 1.4.

We can rewrite the work equation as:

W = -Pexternal(Vinitial1-γ - Vfinal1-γ)

Now, we can plug in the given values and solve for the final temperature:

W = -5 bar(1 L^1.4 - 1 L^1.4) = 0

ΔU = 0 (since Q = 0)

ΔU = nCvΔT

0 = nCv(Tfinal - 298 K)

Tfinal = 298 K