An ideal gas expands from 17.0 l to 64.0 l at a constant pressure of 1.00 atm. then, the gas is cooled at a constant volume of 64.0 l back to its original temperature. it then contracts back to its original volume without changing temperature. find the total heat flow, in joules, for the entire process.

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

diaria maybe go see a doctor about that