Isobaric Process: Finding Work Done by Ideal Gas

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The discussion focuses on calculating the change in internal energy for an ideal gas undergoing an isobaric process, where 23.3 J of heat is added and the volume changes from 54.7 cm³ to 100 cm³ at a constant pressure of 1.00 atm. The first law of thermodynamics is applied, stating that the change in internal energy equals the heat added minus the work done by the system. Participants are encouraged to determine the work done using the formula for work in a constant pressure process. Additionally, there are inquiries about calculating specific heat capacities, Cp and Cv, based on the given data. The conversation emphasizes understanding the relationships between heat, work, and internal energy in thermodynamic processes.
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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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