Isobaric Process: Finding Work Done by Ideal Gas

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Homework Help Overview

The discussion revolves around an isobaric process involving an ideal gas, where heat is added and the volume changes while pressure remains constant. Participants are exploring how to determine the change in internal energy, the specific heat at constant pressure (Cp), and the specific heat at constant volume (Cv) based on the provided data.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply the first law of thermodynamics and relevant equations to find the change in internal energy and specific heats. Questions are raised about the nature of the process and how to calculate temperature from energy values.

Discussion Status

The discussion is active, with participants confirming the nature of the process as constant pressure. Guidance has been provided regarding the application of the first law of thermodynamics, and there is an ongoing exploration of how to calculate work done by the system in this context.

Contextual Notes

Participants are working with specific values and equations but have noted uncertainty regarding the calculation of temperature in Kelvin from Joules, indicating a potential gap in information or understanding.

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Homework Statement


When 23.3 J was added as heat to a particular ideal gas, the volume of the gas changed from 54.7 cm3 to 100 cm3 while the pressure remained at 1.00 atm.
(a) By how much did the internal energy of the gas increase?
J

(b) If the quantity of gas present was 2.00 10-3 mol, find Cp.
J/mol·K

(c) Find CV.
J/mol·K


Homework Equations


PV=nRT
Cp=Cv+R
Eint=CvRT



The Attempt at a Solution


I have tried to use the equations above but am not sure how to calculate the temperature in kelvin from Joules
 
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What kind of process is this, constant volume, constant pressure, isothermal or adiabatic?
 
constant pressure
 
Correct. What does the first law of thermodynamics say?
 
change in internal energy U=Q-W
change in internal energy=heat added-work done by system
 
tigers4 said:
change in internal energy U=Q-W
change in internal energy=heat added-work done by system
Correct. You know that 23.3 J of heat was added to the gas. If you can find W, the work done by the system, then you have the change in internal energy. How does one find the work done by the system (gas) when one has a constant pressure process?
 

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