Thermodynamics: cylindrical figures contains a gas

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Homework Help Overview

The discussion revolves around a thermodynamics problem involving a gas contained within a cylindrical figure with a piston. The initial conditions include specific values for volume, pressure, and temperature, with a change in temperature introduced after applying pressure through the piston. The task is to calculate the change in internal energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between internal energy and temperature, with some suggesting the use of specific equations for ideal gases. Questions arise regarding the validity of different internal energy equations and the implications of the given conditions.

Discussion Status

There is an ongoing exploration of the internal energy calculations, with some participants attempting to derive values for the change in internal energy based on the ideal gas law. Multiple interpretations of the equations and approaches are being considered, but no consensus has been reached regarding the correctness of the calculations.

Contextual Notes

Participants express uncertainty about the clarity of the lesson provided by the instructor, which may impact their understanding of the problem. There is also mention of needing to find specific values such as nR for the calculations.

mtayab1994
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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.
 
Last edited by a moderator:
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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?
 
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.
 
Simon Bridge said:
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.
 
I got n=.0401 by using the equation of ideal gases and i got a change of internal energy as 55J correct?
 
Last edited:
mtayab1994 said:
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?
 
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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