# An ideal gas closed system reversible process

## 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?

Related Biology and Chemistry Homework Help News on Phys.org
mjc123
Homework Helper
You haven't told us the questions you need to answer!

Chestermiller
Mentor

## 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.