Internal energy and pv diagrams

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SUMMARY

The discussion focuses on the application of the first law of thermodynamics to calculate internal energy and work done in a piston system. The user correctly identifies the formula E = Q + W, where Q is the heat added (200 J in this case) and W is the work done, calculated as W = -PdV. The internal energy (U) at the specified point is determined to be 185 J, indicating a proper understanding of the thermodynamic principles involved.

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  • Familiarity with pressure-volume (PV) diagrams
  • Knowledge of isobaric processes
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coffeem
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[PLAIN]http://img689.imageshack.us/img689/511/pistonquerstion.jpg

This is my part attempt...

[PLAIN]http://img830.imageshack.us/img830/7342/internalenergyanswer.jpg

And ideas on how I can do the rest of it?
 
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Actuall I think I have managed to work outish a). I used the first law of thermodynamics: E = Q + W

then giving that in the question it says: Q(isobarbic)= 200j

So since W = -PdV = -P(Vf-Vi)

I for internal energy of the syster at this point: U = 185j

Am I doing it right?
 

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