Question

2. One mole of an ideal gas, CP - (7/2)R and CV - (5/2)R, is compressed adiabatically in a piston/cylinder device from 2 bar and 25°C to 7 bar. The process is irreversible and requires 35% more work than a reversible, adiabatic compression from the same initial state to the same final pressure. What is the entropy change of the gas?

Answer 1

Solution- write the equation for the entoropy change as follows: asen[Comch)- ]. - Here, - as is entropy change, cp is heat capacity at constant pressure, n is the number of moles of vas, P, is initial pressure, fe is final temperature initial poressure , LP, ) is a bor initial temperature iti zis 25% Final pressione, (.P2>is y barn write the equation for the work done in adiabatic reversible process as follows:

Woney - n RT ** 7 (495)*-*] -6 Y-1 Herre, Woren is work done for reversible process. whese, Y = Cp P osts 7 78 79 SR12 - 1.4 1 Y=1-4) substitute the corresponding values in eluation (2) to calculate the work done as Ғoldows : ولا

= imol x 8. 314 Il molky 298.15k 14-1 7 baos 1-4 -1 1.4 2 bass -1 =[26675 2 667 J write the fiors z law of thermodynamics as : والدم su = @tw Here, wis w0912 done four imoreversible and is equal to 1.35 times of work done process foor reversible process ...poor adiabatic process, heat transfer is elual - to zero So, the above equation is written as follours!

به = به 1:35 Wypr = nav (T2-,2 Tz = Tit 135Wgter now substitute the connes ponding value sin the above ecuation to calculate the final temperature TE = T, + 135 Wger ncy = 298.156 + 1.35 x 26675 1. Imal x 52 x 8-314Jl Molk = 298.156+173-22 K - 1477.37K] Hence, the final temperature is 471-37K substitute the corresponding values in ecuation (1)

to calculate the entropy change as follows: OS - NR -19 ***** [2m(7) -(52] = imod C8313 / maiK> [ 33 km/ 41137* )- 2981K = (8.314JIK2 C1-603-1-253) = [2-91J/K IJIK si The sequined enthropy change i-32.915/ K -X Thank you