Adiabatic compressed air and energy calculations

  • #1
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Homework Statement


. Consider a pump that is required to compress air in a factory. The cylinder in the pump has an inner diameter of 2.00 cm and length 60.0 cm. Air is drawn into the pump at atmospheric pressure and 18 degrees celcius and the pump adiabatically compresses the air to a pressure of 17 atmospheres.

(a) Calculate the volume and temperature of the compressed air.

(b) What is the increase in internal energy of the gas during the compression?

(c) Suppose that the air in the pump is compressed once and the compressed air and the steel of the cylinder on the pump the come into thermal equilibrium. If the cylinder is 1.00 mm thick, what will be the increase in temperature of the steel after one compression?

How should i go about solving this question? I don't fully understand thermodynamics. I would love it if you just gave me a worked example. I find it easier to understand since my English is not very good.


Homework Equations


P1V1gamma =P2v2gamma


The Attempt at a Solution


I calculated the volume to be 3.16 me but that's wrong according to the answer and I'm quite stuck on how to calculate energy
 

Answers and Replies

  • #2
haruspex
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I calculated the volume to be 3.16 me but that's wrong
I'm trying to see your working so that I can find your mistake, but my crystal ball doesn't seem to have the necessary range.
 
  • #3
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I am sorry?.
 
  • #4
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Do you know how to solve this problem?
 
  • #5
haruspex
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I am sorry?.
You wrote that you got the wrong answer for the volume. My task, as a homework helper, is to find where you went wrong. But if you do not post your working I cannot possibly find your error.
Please post your working (as typed text, not as an image).
 
  • #6
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I'm sorry. Here is my working.
I used PiVi^gamma = PiVi^gamma.
In this case gamma = 7/5 as air is a diatomic molecule therefore Cp/Cv =7/2R divided by 5/2R = 1.4
Vi = Pi*(0.01)^2* 0.6 = 1.885x10(-4) m^3
I reformulated it to be
Vf = Vi*(Pi/Pf)^1.4
Vf 3.57x10(-6) m^3
 
  • #7
haruspex
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Vi = Pi*(0.01)^2* 0.6
Risk of confusion between π and Pi. Use the buttons above the text area, X2 for subscript, Σ for special characters.
I reformulated it to be
Vf = Vi*(Pi/Pf)^1.4
There's your mistake. Try that reformulation again.
 
  • #8
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I calculated the temperature using TiVi^(gamma -1) = TfVf^(gamma -1)
Basically it's the same method as before.
Tf = Ti*(Vi/Vf)^2/5
Tf = (18 +273.15)(1.885x10(-4)/3.57x10(-6))^2/5
Tf = 291.15(4.887)
Tf = 1422.89 K
This doesn't seem right. I think it's because of turning 18 Degrees to Kelvin?
If i didn't it becomes 52.8 K which still doesn't sound right
 
  • #11
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I am sorry. They are all wrong you cannot swap the position of the power therefore
Vf = square root by gamma(Pi*Vi/Pf)
This gives me an answer 2.88x10-4
 
  • #12
haruspex
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Vf = square root by gamma(Pi*Vi/Pf)
Still no.
Starting with PfVfγ=PiViγ, take it one step at a time and post all the steps.
 
  • #13
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Vfγ = Pi*Viγ/Pi
Vf = square root by γ (Pi*Viγ/Pi)
 
  • #14
haruspex
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Vfγ = Pi*Viγ/Pi
Vf = square root by γ (Pi*Viγ/Pi)
Right, but "square root" is the wrong terminology. You could write it as Vf=Vi(Pf/Pi)1/γ, or in LaTeX, ##V_f=V_i\left(\frac{P_i}{P_f}\right)^{\frac 1{\gamma}}##.
What numerical answer do you get now? If you still get 2.88x10-4, please post your working.
 
  • #17
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oh yes sorry that is a typo. I get -5 also.
 
  • #18
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With this i calculated the energy to be 654 K. Which i think is right.
I calculated this using Tf = Ti(1/17)2/5
 
  • #19
haruspex
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With this i calculated the energy to be 654 K. Which i think is right.
I calculated this using Tf = Ti(1/17)2/5
I agree with the number but not the algebra. I think you would have used Tf = Ti(17)2/7.
 
  • #20
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I'm sorry I do not understand how you derived it to that from:
TiViγ-1 = TfVfγ-1
 
  • #21
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It's fine I understand now.
For part b). I think you have to calculate the number of moles initially using the idea gas law
PV = nRT
where n = PV/RT
n = 1.013x10(5)*1.885x10(-4)/8.314*291.15
n = 19.1/2421
n =7.89x10(-3) moles
This is used in the equation ΔE = 2.5 nR ΔT:
ΔE = 2.5*7.89x10(-3)*8.314*(654 - 291.15)
ΔE = 59.5 Joules of Energy.
Therefore, there is an increase of 59.5 Joules of energy.
However, I do not understand the third question at all. I know I need to make some assumptions(This was a hint the teacher gave me.)
 
  • #22
haruspex
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I'm sorry I do not understand how you derived it to that from:
TiViγ-1 = TfVfγ-1
I didn't derive it from that. The 17 is the ratio of pressures, not volumes. I derived it from TγP1-γ is constant.
How did you get
Tf = Ti(1/17)2/5
?
That would give a final temperature below the starting temperature.
 
  • #23
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yes sorry i typed it wrong what i meant was:
TiViγ-1 = TfVfγ-1
Tf = Tf(Vi/Vf)
to find Tf
 
  • #24
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I do not understand how to do part c). Could you please help me with that?
 
  • #25
haruspex
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Tf = Tf(Vi/Vf)
It is hard to follow your work if you keep making typos. Please check it before posting.
Did you mean Tf=Ti(Vf/Vi)1-γ?
 

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