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An ideal gas closed system reversible process

  • #1

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?
 

Answers and Replies

  • #2
mjc123
Science Advisor
Homework Helper
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You haven't told us the questions you need to answer!
 
  • #3
19,690
3,993

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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