Air contained in a piston–cylinder assembly, initially at 2 bar, 200 K, and a volume of 1 L, undergoes a process to a final state where the

Question

Air contained in a piston–cylinder assembly, initially at 2 bar, 200 K, and a volume of 1 L, undergoes a process to a final state where the pressure is 8 bar and the volume is 2 L. During the process, the pressure–volume relationship is linear. Assuming the ideal gas model for the air, determine the work and heat transfer, each in kJ.

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niczorrrr 2 weeks 2021-09-01T11:22:27+00:00 1 Answers 0 views 0

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    2021-09-01T11:23:45+00:00

    Answer:

    W = 0.5 KJ

    Q = 6.97 KJ

    Explanation:

    Given

    P₁ = 2 bar = 200000 Pa

    V₁ = 1 L = 0.001 m³

    T₁ = 200 K

    P₂ = 8 bar = 800000 Pa

    V₂ = 2 L = 0.002 m³

    Workdone = PdV

    If the pressure volume relationship is linear, then

    W = P(avg) (V₂ – V₁)

    P(avg) = (P₁ + P₂)/2 = (2+8)/2 = 5 bar = 500000 Pa

    W = 500000 (0.002 – 0.001) = 500 J = 0.5 KJ

    b) To calculate the heat transfer, we will apply the mathematical statement of the first law of thermodynamics.

    Q – W = ΔU

    Q = ΔU + W

    ΔU = m (C₂T₂ – C₁T₁)

    We need to obtain T₂ first

    Using the PVT relation for ideal gases

    P₁V₁/T₁ = P₂V₂/T₂

    T₂ = (P₂V₂T₁)/(P₁V₁) = (800000×0.002×200)/(200000×0.001)

    T₂ = 2400 K

    C₁ = specific heat capacity of air at the temperature T₁ = 200 K is 1.07 KJ/K.kg (from literature)

    C₂ = specific heat capacity of air at the temperature T₂ = 2400 K is 0.8625 KJ/K.kg (from literature)

    Then we calculate mass from the ideal gas relation, P₁V₁ = mRT₁

    R = gas constant for air = 287 KJ/kg.K

    (200000 × 0.001) = m × 287 × 200

    m = 0.003484 kg

    ΔU = m (C₂T₂ – C₁T₁)

    ΔU = 0.003484 [(0.8625)(2400) – (1.07)(200)]

    ΔU = 6.47 KJ

    Q = ΔU + W = 6.47 + 0.5 = 6.97 KJ

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