Feynman problem 39-2: Calculations for an adiabatic process

Click For Summary
SUMMARY

The discussion centers on the calculations for an adiabatic process in the context of Feynman problem 39-2, specifically regarding the behavior of gases when air is pumped into a tire. The participants clarify that an adiabatic process requires no heat exchange, and the introduction of air at a non-zero temperature does add energy. The correct application of the ideal gas law, PV = nRT, is emphasized, with the final temperature calculation yielding T1 = 174.5 K, which aligns closely with the book's answer of 173 K after accounting for gauge pressure. The importance of understanding gauge pressure and the constancy of the number of molecules in the system is highlighted.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with the ideal gas law (PV = nRT)
  • Knowledge of gauge pressure and its implications
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the principles of adiabatic processes in thermodynamics
  • Learn about the implications of gauge pressure in fluid mechanics
  • Explore the ideal gas law and its applications in real-world scenarios
  • Investigate the concept of isothermal versus adiabatic processes
USEFUL FOR

Students and professionals in physics and engineering, particularly those focused on thermodynamics, fluid mechanics, and gas laws.

aa_o
Messages
24
Reaction score
4
Homework Statement
See attached
Relevant Equations
P*V / T= constant
P*V^(y) = constant (if process is adiabatic)
I'm not sure that this is an adiabatic process. As far as i can read, it is adiabatic if no HEAT or ENERGY is added. But pumping in molecules that are a non-zero temperature is an addition of energy, no?
Anyway - my solution with the assumption of an adiabatic process.
(skipping units for brevity):
P0 = 14.7
P1 = 50.0
T0 = 293
y = 1.40

We have:
P0 * V0^y = P1 *V1^y
So:
V1 / V0 = (P0 / P1)^(1/y)

And
P0*V0 / T0 = P1*V1 / T1
So:
P1 / P0 * V1 / V0 = T1 / T0
Inserting:
(P0 / P1)^(-1) * (P0 / P1)^(1/y) = T1 / T0
T1 = (P0 / P1)^(1/y - 1) * T0

Inserting the values we get

T1 = 415.7 K = 142.55 C

The answer in the book says 173 C.
Are my assumptions about adiabatic wrong? Or am i using the wrong equations?
Can i really use P*V / T = constant if we are adding air molecules with the pump?

Any suggestions would be appreciated.
 

Attachments

  • Untitled.png
    Untitled.png
    8.2 KB · Views: 229
Last edited by a moderator:
Physics news on Phys.org
aa_o said:
Can i really use P*V / T = constant if we are adding air molecules with the pump?

No. PV = nRT, and n is changing.
 
Okay. So the pump has a start state of:
P0 * V0 = n0 * R * T0
And end state of:
P1 * V0 = n1 * R * T1

I thus have 2 unknowns i need to solve the problem T1, which is what i need in the end, and the ratio of the number of molecules between the 2 states. I just can't find that extra connection that gives me the information about the ratio!

I still haven't used the fact that y = 1.40. I just can't see where that fits in.

Any suggestions?
 
Thinking about it again, if a constant amount of air is being compressed within the pump to 50 psig before being injected into the tyre (is this how the pump works?), then I think your original method is correct. But read carefully. The exit pressure is 50 psi gauge - what does that mean?
 
Thanks, mjc123.

So n is constant for the whole system (tire + pump), but the volume then changes (compresses).

Ahh, i think that gauge was the missing piece. I didn't know about the meaning of gauge and simply skipped over it without paying much attention.

With that information (P1 = P0 + 50) i get an answer of 174.5 - close enough to the one in the book.
 
aa_o said:
So n is constant for the whole system (tire + pump), but the volume then changes (compresses).
No, n is constant for the air in the pump. As I read it, the pump compresses an amount of air from atmospheric to 50 psig, then injects it into the tyre (presumably the valve opens at that pressure).
 
  • Like
Likes aa_o
mjc123 said:
No, n is constant for the air in the pump. As I read it, the pump compresses an amount of air from atmospheric to 50 psig, then injects it into the tyre (presumably the valve opens at that pressure).
Yeah, that makes sense. I think there was a lot of assumptions that had to be made that wasn't explicitly stated in the problem.

But thanks a lot for the help!
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
9K
Replies
5
Views
2K
  • · Replies 60 ·
3
Replies
60
Views
10K
  • · Replies 22 ·
Replies
22
Views
6K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K