An monatomic ideal gas at 300 K has a volume of 15 liters at a pressure of 15 atm. Calculate a. The final volume of the system b. The work done by the system c. The heat entering or leaving the system d. The change in the internal energy e. The change in the enthalpy when the gas undergoes i. A reversible isothermal expansion to a pressure of 10 atm ii. A reversible adiabatic expansion to a pressure of 10 atm The constant-volume molar heat capacity of the gas, Cy, has the value 1.5 R.

Question

Ayaan Mayer

Chemistry

25 July, 00:40

An monatomic ideal gas at 300 K has a volume of 15 liters at a pressure of 15 atm. Calculate a. The final volume of the system b. The work done by the system c. The heat entering or leaving the system d. The change in the internal energy e. The change in the enthalpy when the gas undergoes i. A reversible isothermal expansion to a pressure of 10 atm ii. A reversible adiabatic expansion to a pressure of 10 atm The constant-volume molar heat capacity of the gas, Cy, has the value 1.5 R.

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Answers (1)

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Juan Bass · Answer 1

i) a) V2 = 22.5 L

b) W = 9249.451 J

c) Q = - 9249.451 J

d) ΔU = 0 J/mol

e) ΔH = 0 J/mol

ii) a) V2 = 19.154 L

b) W = 3390.346 J

c) Q = 0 J

d) ΔU = - 562.124 J/mol

e) ΔH = - 936.884 J/mol

Explanation:

an monoatomic ideal gas:

∴ PV = RTn

contants molar heat capacity:

∴ Cp, n = 2.5 R

∴ Cv, n = 1.5 R

i) a reversible isothermal expansion:

∴ T1 = T2 = 300 K

∴ P1 = 15 atm

∴ V1 = 15 L

∴ P2 = 10 atm

⇒ n = P1V1 / RT1 = ((15) * (15)) / ((0.082) * (300)) = 9.146 mol

a) V2 = RT2n/P2 = ((0.082) * (300) * (9.146)) / 10 = 22.5 L

b) W = nRT Ln (P1/P2)

⇒ W = (9.146) * (8.314) * (300) Ln (15/10)

⇒ W = 9249.451 J

c) Q = - W = - 9249.451 J

d) ΔU = CvΔT

∴ ΔT = 0

⇒ ΔU = 0

e) H = U + PV ... ideal gas

⇒ ΔH = ΔU + ΔPV = 0 + (P1V1 - P2V2) = 0 + 0 = 0

ii) a reversible adiabatic expansion:

∴ P2 = 10 atm

a) P1 (V1) ∧γ = P2 (V2) ∧γ ... reversible adiabatic

∴ γ = Cp, n / Cv, n = 2.5R / 1.5R = 1.666

⇒ (V2) ∧γ = ((15) (15) ∧γ) / 10

⇒ (V2) ∧γ = 136.849

⇒ V2 = (136.849) ∧ (1/1.666)

⇒ V2 = 19.154 L

b) W = P1V1 - P2V2

⇒ W = ((15) * (15)) - ((10) * (19.154)) = 225 - 191.54 = 33.46 atm. L

⇒ W = 33.46 atm. L * (Pa/9.8692 E-6atm) * (m³/1000L) = 3390.346 Pa. m³

⇒ W = 3390.346 J

c) Q = 0 J ... adiabatic

d) ΔU = Cv, n*ΔT

∴ T2/T1 = (V1/V2) ∧ (R/Cv, n)

⇒ T2 = T1 * [ (V1/V2) ∧ (R/Cv, n) ]

∴R/Cv, n = R/1.5R = 1/1.5 = 0.666

⇒ T2 = 300K * [ (15/19.154) ∧ (0.666) ]

⇒ T2 = 254.925 K

⇒ ΔU = Cv, n*ΔT = 1.5 * (8.314) * (254.925 - 300) = - 562.124 J/mol

e) ΔH = ∫Cp, n*δT = Cp, n*ΔT ... constant molar heat capacity

⇒ ΔH = 2.5 * (8.314) * (254.925 - 300) = - 936.884 J/mol