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.

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 + PVideal gas

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

ii) a reversible adiabatic expansion:

∴ P2 = 10 atm

a) P1(V1)∧γ = P2(V2)∧γreversible 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*ΔTconstant  molar heat capacity

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

2b2t

explanation:

2b2t