Entropy of gas at constant pressure and volume

AI Thread Summary
The discussion focuses on calculating the overall change in entropy for a gas undergoing two processes: heating at constant pressure and cooling at constant volume. The initial conditions include 1m³ of air heated from 288K to 573K at 103kPa. The user initially calculated the final pressure after heating to be 911kPa but received an incorrect entropy change result. Other participants pointed out the misuse of equations applicable to adiabatic processes rather than the correct approach for isobaric and isochoric processes. Clarification on the correct application of thermodynamic principles is emphasized to resolve the calculation errors.
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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
 
Physics news on Phys.org
tell me how u got 911kPa..
its wrong from thr only...
 
I used T2/T1 = (P2/P2)^(n-1/n)

and n = Cp/Cv
 
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
 

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