Question

2.70 moles of an ideal gas with CV,m=5R/2are transformed from an initial state T = 660. Kand P = 1.65 bar bar to a final state T = 298 Kand P = 4.75 bar.

Calculate ΔU for this process.

Calculate ΔH for this process.

Calculate ΔS for this process.

Please show work

Answer 1

Given that

No of moles of gas , n = 2.70 mol

Initial temperature T1 = 660 K

Initial pressure P1 = 1.65 bar

Final temperature T2 = 298 K

Final pressure P2 = 4.75 bar

Cv,m = 5R/2          

Cp,m = Cv,m + R = 5R/2 + R = 7R/2

1) ΔU

ΔU = n Cv,m dT

      = n (5R/2) (T2-T1)

      = (2.70 mol) ( 5 x 8.314 J/K/mol )/2 ( 298 K-660 K)

      = -20315 J

Therefore, ΔU = -20315 J

2) ΔH

ΔH = nCp,m dT

     = n (7R/2) (T2-T1)

      = (2.70 mol) ( 7 x 8.314 J/K/mol )/2 ( 298 K-660 K)

      = -28441 J

Therefore, ΔH = -28441 J

3) ΔS

When T and P are variables ,

ΔS = [nCp,m In (T2/T1)] - [nR In (P2/P1)]

     = [n (7R/2) In (T2/T1)] - [nR In (P2/P1)]

      = nR {[ (7/2) In (T2/T1) ]- [ In (P2/P1)]}

       = (2.70 mol) (8.314 J/K/mol ){ [ (7/2) In(298/660)] - [In (4.75/1.65)]

      = -86.21 J/K

ΔS = -86.21 J/K

Therefore, ΔU = -20315 J

                 ΔH = -28441 J

               ΔS = -86.21 J/K