17 April, 14:39

# One mole of a monatomic ideal gas is subjected to the following sequence of steps: a. Starting at 300 K and 10 atm, the gas expands freely into a vacuum to triple its volume. b. The gas is next heated reversibly to 400 K at constant volume. c. The gas is reversibly expanded at constant temperature until its volume is again tripled. d. The gas is finally reversibly cooled to 300 K at constant pressure. Calculate the values of q and w and the changes in U, H, and S.

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1. 17 April, 15:06
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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