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Suppose 2.00 mol of an ideal gas is contained in a heat- insulated cylinder with a moveable frictionless piston. Initially, the gas is at 1.00 atm and 0°C. The gas is compressed reversibly to 2.00 atm. The molar heat capacity at constant pressure, cP, equals 29.3 J K21 mol 21. Calculate the final temperature of the gas, the change in its internal energy, DU, and the work done on the gas.

Respuesta :

Answer:

a) T2 = 332.342 K

b) ΔU = 1245.355 J/mol

c) W = - 986.738 J

Explanation:

ideal gas:

∴ n = 2 mol

∴ P1 = 1.00 atm

∴ T1 = 0 °C = 273 K

∴ P2 = 2.00 atm

∴ Cp,m = 29.3 J/mol.K

  • Cp,m - Cv,m = R........ideal gas

⇒ Cv,m = Cp,m - R = 29.3 - 8.314 = 20.986 J/mol.K

  • Q + W = ΔU....first law

∴ Q = 0.........heat-insulated cylinder

⇒ W = ΔU

∴ ΔU = CvΔT

∴ W = - PδV......compressed reversibly

ideal gas:

  •  PV = RT

⇒ PδV + VδP = RδT

⇒ PδV = RδT - VδP

⇒ RδT - VδP = - Cv,mδT,,,,,first law

⇒ RδT - (RT/P)δP = - Cv,mδT

⇒ RδT + Cv,mδT = (RT/P)δP

⇒ (R+Cv,m)δT/T = RδP/P

⇒ ((R+Cv,m)/R)∫δT/T = ∫δP/P

⇒ ((R+Cv,m)/R) Ln(T2/T1) = Ln (P2/P1)

∴ (R+Cv,m)/R = 3.524

⇒ Ln (T2/T1) = Ln (P2/P1) / 3.524

⇒ Ln (T2/T1) = Ln (2.00/1.00) / 3.524

⇒ Ln (T2/T1) = 0.693 / 3.524 = 0.1967

⇒ T2/T1 = 1.2174

⇒ T2 = (273K)×(1.2174) = 332.342 K

b) ΔU = Cv,mΔT

⇒ ΔU = (20.986 J/mol.K)(332.342 K - 273 K)

⇒ ΔU = 1245.355 J/mol

c) W = - PδV

⇒ W = - P2V2 + P1V1

∴ P1V1 = nRT1 = (2mol)(8.314 J/mol.K)(273 K) = 4539.444 J

∴ P2V2 = nRT2 = (2mol)(8.314 J/mol.K)(332.342 K) = 5526.183 J

⇒ W = - 5526.183 J + 4539.444 J = - 986.738 J

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