PV Diagram Isothermas and Isochloric Processes

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SUMMARY

The discussion focuses on the analysis of isothermal and isochoric processes in thermodynamics, specifically using the PV diagram. The participant correctly identifies that for process DA, the work done (W) is zero, leading to the conclusion that the change in internal energy (ΔU) equals the heat added (Q). For process AB, the work done is calculated using the formula W = nRT(ln(Vf/Vi)), confirming that both processes result in positive heat flow into the gas. The relationship between positive work and positive heat flow in isothermal processes is clarified, emphasizing that ΔU equals zero in such cases.

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  • Understanding of thermodynamic processes, specifically isothermal and isochoric processes.
  • Familiarity with the ideal gas law and its applications.
  • Knowledge of the first law of thermodynamics, particularly the relationship between work, heat, and internal energy.
  • Ability to interpret and analyze PV diagrams.
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  • Study the first law of thermodynamics in detail, focusing on internal energy, work, and heat transfer.
  • Learn how to derive and apply the equations for isothermal processes, including W = nRT(ln(Vf/Vi)).
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lc99
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Homework Statement



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Homework Equations

The Attempt at a Solution


These questions are kinda tricky for me even though i might get the right answer. I just want to know if my reasonings are correct.

The answer i chose is C. For process DA, W = 0 , so delta U = Q. Q = 3/2nR(deltaT). Temp increases along DA while temp decreases along BC. So DA has a positive Q.

For process AB, W = nrt(ln(vf/vi)) . since Vf>Vi, W is positive so Q is positive.

For both processes, Q is positive which means heat flows into the gas.
 

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lc99 said:
For process AB, W = nrt(ln(vf/vi)) . since Vf>Vi, W is positive so Q is positive.
Why does W > 0 imply that Q > 0 for process AB?

Otherwise, everything looks OK.
 
TSny said:
Why does W > 0 imply that Q > 0 for process AB?

Otherwise, everything looks OK.
Oh. Cause deltaU = 0 for a isothermic process. Since W is positive, Q is positive.
 
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lc99 said:
Oh. Cause deltaU = 0 for a isothermic process. Since W is positive, Q is positive.
OK. Good work.
 
TSny said:
OK. Good work.
Thanks for the help!
 

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