Thermal Kinetic Physics Question involving PV diagram

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SUMMARY

The discussion centers on a thermal kinetic physics (TKP) problem involving the first law of thermodynamics, specifically the equation dU=dQ-PdV. The participant is preparing for a resit exam and is analyzing an isochoric process where no work is done (PdV=0), leading to the conclusion that the change in internal energy (dU) equals the heat added to the system (dQ), which is 20J. The conversation emphasizes the importance of precise language in conveying thermodynamic principles.

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Familiarity with isochoric processes in thermodynamics
  • Knowledge of internal energy and heat flow concepts
  • Ability to interpret PV diagrams
NEXT STEPS
  • Study the implications of isochoric processes in thermodynamics
  • Learn about the relationship between heat transfer and internal energy changes
  • Explore detailed examples of PV diagrams and their interpretations
  • Review the first law of thermodynamics with practical applications
USEFUL FOR

This discussion is beneficial for students preparing for thermodynamics exams, educators teaching thermal kinetic physics, and anyone looking to deepen their understanding of the first law of thermodynamics and isochoric processes.

KaseyKC
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Homework Statement


I'm stuck on question A4 that is linked below. University won't provide me with any solutions. I have a TKP resit exam next month.

Homework Equations


dU=dQ-PdV

The Attempt at a Solution


Internal energy is equal to heat flow plus the work done. Work done is 0 because it's isochoric. So Internal energy is equal to the heat flow right? 20J.
 

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Hello, KaseyKC. Welcome to PF!

Be careful with how you word your answer. The heat added to the system plus the work done on the system equals the change in internal energy of the system.

Otherwise, your answer looks good.
 
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