Work done BY the gas in a cyclic thermodynamic process

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SUMMARY

The work done by the gas in a cyclic thermodynamic process is calculated using the formula P1*(V2-V1), which represents the area under the pressure-volume curve for the segment from V1 to V2. To determine the total work done by the gas throughout the entire cycle, one must also account for the work done during the other segments, necessitating the use of P2. The question implies a need for the net work done by the gas, which encompasses all segments of the cycle, rather than just the work from V1 to V2.

PREREQUISITES
  • Understanding of thermodynamic processes
  • Familiarity with pressure-volume (P-V) diagrams
  • Knowledge of the first law of thermodynamics
  • Basic calculus for area calculations
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  • Research the calculation of net work in cyclic thermodynamic processes
  • Learn about the significance of P-V diagrams in thermodynamics
  • Explore the implications of the first law of thermodynamics on work and heat
  • Study different types of thermodynamic cycles, such as Carnot and Otto cycles
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Homework Statement
What is the work done by the gas in the cycle shown below?
Relevant Equations
V1=3L, V2=5L, P1=1atm, =P2=2atm
image8.png
Since the assignment asks the work done by the gas, that should be equal to P1*(V2-V1) aka the area under the P1 line. Do I have to subtract the work done to the system or is this the solution already? If so, why do I need P2?
 
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P1*(V2-V1) is the work done by the gas only for the leg of the cycle from V1 to V2. For the complete cycle you must add to this the work done by the gas for the other three legs. That's why you need P2.
 
I get what you are saying and I see the emphasis in the title of the post on the word “BY”, but I suspect you need to read the question as what NET work is done by the gas. Sure, maybe they could have specified explicitly, but that is not an unreasonable way to ask the question particularly when they say “in the cycle”. That is going to mean “net” (i.e. all the way around the cycle) every time.
 

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