Entropy of gas at constant pressure and volume

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Homework Statement



1m^3 of air is heated reversibly at constant pressure from 288K to 573K. Then it is cooled reversibly at constant volume back to the initial T. Initial P is 103kPa Calculate overall change in entropy.
Cp=1.02
Cv=0.702

Homework Equations



dS=Cp x ln(T2/T1)-R x ln(P2/P1)


The Attempt at a Solution



I have found P2 to be 911kPa but when I put all the data into the above equation I end up with the wrong answer (0.076kJ/K). Please help!

I have found
 
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tell me how u got 911kPa..
its wrong from thr only...
 
for 1st proces which is isobaric...
V1/T1=V3/T3... V1= 1m^3 so one can find V3...3 is the intermediate stage
and then for isochoric process...
P3/T3=P2/T2...
n bcoz T2= T1=288 and P3=P1...
n T3=573
so jus find value n den see...
u r goin in wrong direction...dats actually for adiabatic processes