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An ideal gas, initially at 30°C and 100 kPa, undergoes the following cyclic processes in a closed system: (a) In mechanically reversible processes, it is first compressed adiabatically to 500 kPa, then cooled at a constant pressure of 500 kPa to 30°C, and finally expanded isothermally to its original state.

Respuesta :

Answer:

a) compressed adiabatically:

T2 = 576.8 K

ΔU = 3422 J/mol

b) cooled at P contant to 30°C:

Q = - 5965.04 J/mol

c) expanded isotermally to P=100 KPa:

W = - 4054.403 J

Explanation:

ideal gas in a mechanically reversible process:

∴ T1 = 30°C = 303 K

∴ P1 = 100 KPa

a) compressed adiabatically to 500 KPa:

  • ΔU = Q + W      ∴ Q = 0

⇒ ΔU = W = CvΔT.....(1)

∴ Cv = (3/2)R = 12.5 J/K.mol

∴ W = - PδV........(2)

(1) = (2):

⇒ [(R+Cv)/R] Ln (T2/T1) = Ln (P2/P1)

∴ R+Cv/R = 5/2

⇒ (5/2) Ln(T2/T1) = 1.6094

⇒ LnT2 - LnT1 = 0.64376

⇒ LnT2 = 0.64376 + 5.7137 = 6.3575

⇒ T2 = 576.8 K

⇒ ΔU = W = (12.5 J/K.mol)(576.8 - 303 ) = 3422 J/mol

b)n cooled at constant P = 500KPa to 30°C:

∴ T2 = 303 K

∴ T1 = 576.8 K

∴ ΔU = Q + W

⇒ Q = CpΔT

∴ Cp = (5/2)R = 20.8 J/K.mol

⇒ Q = (20.8 J/K.mol)(303 - 576.8)

⇒ Q = - 5695.04 J/mol

c) expanded isothermally a P=100 KPa

∴ ΔU = 0

∴ T = 303 K

∴ P1 = 500 KPa

∴ P2 = 100 KPa

∴ W = nRT Ln(P2/P1)......assuming n = 1 mol

⇒ W = (1 mol)(8.314 J/K,mol)(303 K) Ln(100/500)

⇒ W = - 4054.403 J

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