Answer:
The total heat flow is Q = -24.46 J.
Explanation:
Transformation 1:
V₁ = 17.0 L
V₂ = 64.0 L
P = 1.00 atm
T₁ → T₂
T₁ < T₂
ΔU₁ = Q₁ + W₁
Transformation 2:
V = V₂ = 64.0 L
T₂ → T₁
T₁ < T₂
ΔU₂ = Q₂ + W₂, W₂ = 0 ∴ ΔU₂ = Q₂
Transformation 3:
T = T₁
V₂ → V₁
ΔU₃ = Q₃ + W₃, ΔU₃ = 0 ∴ Q₃ = - W₃
ΔU is a state function, therefore the total ΔU = 0.
ΔU = ΔU₁ + ΔU₂ + ΔU₃ = 0
Q₁ + W₁ + Q₂ + 0 = 0
Q₁ + Q₂ = W₁
Q₁ + Q₂ = - PΔV
Q₁ + Q₂ = -1.00 atm x (64.0 - 17.0)L
Q₁ + Q₂ = - 47.0 J
Q₃ = -W₃
Q₃ = - nRTln(V₁/V₂)
PV = nRT
nRT = 1.00 atm x 17.0L
Q₃ = - 17 x ln (17/64)
Q₃ = 22.54 J
Q = Q₁ + Q₂ + Q₃
Q = -47.0 + 22.54
Q = -24.46 J
