Question on expansion and Compression of gas?

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SUMMARY

The discussion focuses on the thermodynamic processes involving the expansion and compression of gas under isothermal conditions. The key equations utilized include the ideal gas law (PV=nRT), the first law of thermodynamics (ΔU=q+w), and the work done formula (w=p*ΔV). The participant correctly identifies that during isothermal processes, the change in internal energy (ΔU) is zero, leading to the conclusion that heat (q) is equal to the negative work done (q=-p*ΔV). The calculations for final volume after expansion and the implications of returning to initial pressure are also addressed.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Familiarity with the first law of thermodynamics (ΔU=q+w)
  • Knowledge of Boyle's Law for isothermal processes
  • Basic concepts of thermodynamic work (w=p*ΔV)
NEXT STEPS
  • Study isothermal processes in greater detail, focusing on real-world applications.
  • Explore the implications of the first law of thermodynamics in various thermodynamic cycles.
  • Learn about the differences between isothermal and adiabatic processes.
  • Investigate the effects of temperature changes on gas behavior using the ideal gas law.
USEFUL FOR

Students studying thermodynamics, physics enthusiasts, and professionals in engineering fields who require a solid understanding of gas behavior under varying pressure and volume conditions.

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Homework Statement



If a pressure of gas is suddenly reduced and we let the gas to expand again until it reaches equilibrium. What is final volume, work done and heat in this process? and what would be change in internal energy?
Again the pressure is suddenly increased back to initial pressure and gas is compressed to initial volume. What would be work done heat and change in internal energy?
Temperature is same throughout all process

Homework Equations



PV=nRT
ΔU=q+w
w=p*ΔV

The Attempt at a Solution



Here All the process is done at isothermal condition so Boyles law apply
P1V1=P2V2

where V2 is volume after the gas comes to equilibrium, so,
V2=P1V1/P2

since the process is carried in isothermal condition
ΔU=0

w=p*ΔV

0=q+p*ΔV -------> q=-p*ΔV

Now when the pressure is increased back to initial condition

ΔU=0 ( since process is still isothermal)
ΔU=q+p*ΔV ---------> q=-p*ΔV, here change in volume is volume after decreasing pressure - initial volume

I am not sure if I am correct or not? Help
 
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