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14 April, 05:53

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.

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  1. 14 April, 06:13
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    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
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