An ideal gas closed system reversible process

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SUMMARY

The discussion focuses on the analysis of a reversible process involving an ideal gas at an initial temperature of 70°C and pressure of 1 bar. The process includes three stages: adiabatic compression to 150°C, cooling at constant pressure from 150°C to 70°C, and isothermal expansion back to the original state. Key equations utilized include dU = dQ + dW, dU = CvdT, and the relationship between temperature and pressure during the process. The main challenge identified is determining the pressure to use during the isothermal expansion phase.

PREREQUISITES
  • Understanding of the first law of thermodynamics (dU = dQ + dW)
  • Knowledge of specific heat capacities (Cv and Cp)
  • Familiarity with the ideal gas law and its applications
  • Concept of reversible processes in thermodynamics
NEXT STEPS
  • Study the derivation and application of the ideal gas law in various thermodynamic processes
  • Learn about the implications of adiabatic and isothermal processes on internal energy and enthalpy
  • Explore the concept of partial derivatives in thermodynamic equations
  • Investigate the calculation of work done during isothermal expansion of gases
USEFUL FOR

Students and professionals in thermodynamics, particularly those studying ideal gas behavior and reversible processes, will benefit from this discussion.

HethensEnd25
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Homework Statement


an Ideal gas at T = 70 C and 1 bar undergoes following reversible
processes:
a: Adiabatically compressed to 150 C
b: then, cooled from 150 to 70 C at constant pressure
c: finally, expanded isothermally to the original state (T=70 C and P = 1 bar)

Homework Equations


1-dU=dQ+dW
2-dU=CvdT
3-(T2/T1)=(P2/P1)(R/Cp) from TP(1-ϒ)/ϒ=constant
4-CpdT=dH

The Attempt at a Solution


So far for part a) I have done the equation dU=dQ+dW, since it is an adiabatic process I have dQ=0, thus leaving dU=dW. I know that dW=-PdV so I can calculate dU=CvdT to get my change in internal energy. From there I calculated P2 using equation 3.

For b) knowing it is an Cp problem I used equation 4 having CpdT=dH where my change in temperature is 150 to 70 C

I am stuck on part c)

I know it is an isotherm that means that there is no change in temperature thus dU and dH are equal to zero since they are temperature dependent. So dQ=dW. Do I make it so that T is a constant and solve with partial derivatives giving me dT=(V/R)dP+(P/R)dV. With dT=0 we get -PdV=VdP. Using the ideal gas equation to give V=RT/P ==> -(RT/P) dP. Integrating gives me -RTln(P2/P1)

My problem is I don't know what pressure to use. Do i just use the pressures calculated in my part b going from there back to the original state?
 
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HethensEnd25 said:

Homework Statement


an Ideal gas at T = 70 C and 1 bar undergoes following reversible
processes:
a: Adiabatically compressed to 150 C
b: then, cooled from 150 to 70 C at constant pressure
c: finally, expanded isothermally to the original state (T=70 C and P = 1 bar)

Homework Equations


1-dU=dQ+dW
2-dU=CvdT
3-(T2/T1)=(P2/P1)(R/Cp) from TP(1-ϒ)/ϒ=constant
4-CpdT=dH

The Attempt at a Solution


So far for part a) I have done the equation dU=dQ+dW, since it is an adiabatic process I have dQ=0, thus leaving dU=dW. I know that dW=-PdV so I can calculate dU=CvdT to get my change in internal energy. From there I calculated P2 using equation 3.

For b) knowing it is an Cp problem I used equation 4 having CpdT=dH where my change in temperature is 150 to 70 C

I am stuck on part c)

I know it is an isotherm that means that there is no change in temperature thus dU and dH are equal to zero since they are temperature dependent. So dQ=dW. Do I make it so that T is a constant and solve with partial derivatives giving me dT=(V/R)dP+(P/R)dV. With dT=0 we get -PdV=VdP. Using the ideal gas equation to give V=RT/P ==> -(RT/P) dP. Integrating gives me -RTln(P2/P1)

My problem is I don't know what pressure to use. Do i just use the pressures calculated in my part b going from there back to the original state?
Yes.
 

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