- #1

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?