Answer:
a) Q = 0; W = 0; ΔU = 0; ΔH = 0; ΔS = 0.09 atm.L/K
b) Q = 1250 J; W = 0; ΔU = 1250 J; ΔH = 1250 J; ΔS = -0.0235 atm.L/K
c) Q = 3653.545 J; W = - 3653.545 J; ΔU = 0; ΔH = 0; ΔS = - 3653.545 J
d) Q = - 2080 J; W = 830 J; ΔU = - 1250 J; ΔH = - 2080 J; ΔS = - 5.984 J/K
Explanation:
a) If there is a vacuum, the work is zero, as it is a free expansion, the volume increases, the pressure decreases, the temperature is constant and the internal energy is constant.
∴ n = 1 mole
∴ PV = RTn....ideal gas
∴ P1 = 10 atm
∴ R = 0.082 atm.L/K.mol
∴ T = 300 K = T2
∴ V2 = 3*V1
⇒ W = 0.....expands freely into vacuum
⇒ ΔU = Q = 0....first law
⇒ ΔS = - nR Ln(P2/P1).....ideal gas
∴ V1*P1/T1 = V2*P2/T2
∴ T1 = T2 = 300 K
⇒ P2 = V1*P1 / V2 = V1*P1 / 3V1 = 10 atm/3 = 3.33 atm
⇒ ΔS = - (1mol)*(0.082 atm.L/K.mol) Ln ( 3.33/10)
⇒ ΔS = 0.09 atm.L/K
∴ ΔH = ΔU + (P2V2 - P1V1) = 0 + 0 = 0
b) heated reversibly at constant volume:
⇒ W = 0 ...at constant volume
∴ T2 = 400 K; T1 = 300 K
∴ V1 = V2
⇒ Q = ΔU = CvΔT....first law
∴ Cv = 12.5 J/K.mol.....monoatomic ideal gas
∴ ΔT = 400 - 300 = 100 K
⇒ Q = ΔU = 12.5 J/mol.K * 100K = 1250 J/mol * 1 mol = 1250 J
∴ ΔH = ΔU + PΔV = ΔU + 0 = 1250 J
∴ ΔS = - nR Ln (P2/P1)
∴ P2/T2 = P1/T1...constant volume
∴ P1 = 3.33 atm
⇒ P2 = P1*T2 / T1 = (3.33 atm)*(400K) / (300K) = 4.44 atm
⇒ ΔS = - (1mol)*(0.082atm.L/K.mol) Ln (4.44/3.33)
⇒ ΔS = - 0.0235 atm.L/K
c) reversibly expanded at constant temperature:
∴ T1 = T2 = 400K
∴ V2 = 3*V1
∴ ΔU = 0...constant temperature
⇒ Q = - W....fisrt law
∴ W = - ∫ PdV..... reversibly expansion
∴ P = nRT/V... ideal gas
⇒ W = - nRT ∫ dV/V
⇒ W = - nRT Ln (V2/V1)
⇒ W = - (1mol)*(8.314 J/K.mol) Ln (3)
⇒ W = - 9.134 J/K *400K = - 3653.545 J
⇒ Q = - W = 3653.545 J
⇒ ΔH = ΔU + P1V1 - P2V2 = 0 + nRT1 - nRT2 = 0 + 0 = 0
∴ ΔS = - nR Ln(P2/P1)
∴ P1 = 4.44 atm
⇒ P2 = V1*P1*T2/ V2*T1 = V1*(4.44atm)*(400K) / (3.V1)*(400K)
⇒ P2 = 4.44atm/3 = 1.48 atm
⇒ ΔS = - (1mol)*(8.314 J/mol.K) Ln (1.48/4.44)
⇒ ΔS = -9.134J/K * 400K = - 3653.545 J
d) reversibly cooled at constant pressure:
∴ T2 = 300 K; T1 = 400 K
∴ P2 = P1
⇒ Q = ΔH = CpΔT
∴ Cp = 20.8 J/K.mol
∴ ΔT = 300 - 400 = - 100 K
⇒ Q = ΔH = 20.8 J/mol.K * ( -100K) = - 2080 J/mol * 1mol = - 2080 J
⇒ ΔU = nCvΔT = (1mol)*(12.5 J/mol.K)*( - 100K) = -1250 J
⇒ W = ΔU - Q = ΔU - ΔH = -1250 J - ( - 2080 J ) = 830 J
∴ ΔS = ∫ δQ/T = ∫ nCpdT/T
⇒ ΔS = nCp Ln (T2/T1)
⇒ ΔS = (1mol)*(20.8 J/mol.K) Ln (300/400) = - 5.984 J/K
