Thermodynamics: cylindrical figures contains a gas

  • Thread starter mtayab1994
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  • #1
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Homework Statement



A cylindrical figures contains a gas and we have a piston that is capable of changing the gases volume.
Initial state: V0=1l P0=10^5p T=300°K

We pressurize on the piston that has a mass m=300g and a surface S=20cm^2.
The new temperature T1=540°K

Homework Equations



Calculate the change in internal energy.

The Attempt at a Solution



Well I know that the change in internal energy is: ΔU=W+Q or ΔU=-P1(V1-V0)

and that p=F/S. but since my teacher didn't explain the lesson well enough I have no clue on how to answer it. Any help would be very well appreciated.
 
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Answers and Replies

  • #3
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i believe i would have to use this one right?

265d7d7b21674406fdcb29c850301fe0.png


I would get: ΔU=p1(V0-V2)+Q right?
 
  • #4
Simon Bridge
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OK, I'll let you off the hook:
Internal energy for an ideal gas depends only on it's temperature.
[tex]\Delta U = \frac{3}{2}nR\Delta T[/tex]

You know the change in temperature - it's given to you.
You need to find nR.
 
  • #5
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OK, I'll let you off the hook:
Internal energy for an ideal gas depends only on it's temperature.
[tex]\Delta U = \frac{3}{2}nR\Delta T[/tex]

You know the change in temperature - it's given to you.
You need to find nR.

alright i'll give it a try right now.
 
  • #6
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I got n=.0401 by using the equation of ideal gases and i got a change of internal energy as 55J correct?
 
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  • #7
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I got n=.0401 by using the equation of ideal gases and i got a change of internal energy as 24.5J.

Is this correct?
 
  • #8
Simon Bridge
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I've not done the math so I cannot tell you.
But it's numbers-in-numbers-out, hard to go wrong.
 

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