A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3 . The propane undergoes a process to a final p

Question

A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3 . The propane undergoes a process to a final pressure of 4 bar, during which the pressure–volume relationship is pV1.1 = constant. For the propane, evaluate the work and heat transfer, each in kJ. Kinetic and potential energy effects can be ignored.

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Philomena 4 years 2021-08-24T16:58:02+00:00 1 Answers 280 views 1

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  1. Nick
    0
    2021-08-24T16:59:33+00:00
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    Answer:

    Heat transfer in the process is -19.6 kJ

    Work done in the process is – 28 kJ

    Explanation:

    As we know that the process equation is given as

    P_1V_1^{1.1} = P_2V_2^{1.1}

    now we know that

    P_1 = 1 bar

    V_1 = 0.2 m^3

    P_2 = 4 bar

    now from above equation we have

    1(0.2)^{1.1} = 4(V)^{1.1}

    V = 0.057 m^3

    now work done in this process is given as

    W = \frac{P_1V_1 - P_2V_2}{N - 1}

    here we know N = 1.1

    so we have

    W = \frac{10^5 (0.2) - 4 \times 10^5(0.057)}{1.1 - 1}

    W = -2.8 \times 10^4 J

    W = -28 KJ

    Now we have

    \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

    T_2 = \frac{P_2V_2 T_1}{P_1V_1}

    T_2 = \frac{4\times 10^5 (0.057) (300)}{10^5 (0.2)}

    T_2 = 342 K

    now we have

    \Delta U = \frac{f}{2} n R(\Delta T)

    \Delta U = 3(P_2 V_2 - P_1 V_1)

    \Delta U = 3(4 \times 10^5(0.057) - 10^5(0.2))

    \Delta U = 8400

    so we have

    Q = \Delta U + W

    Q = -28 kJ + 8.4 kJ

    Q = -19.6 kJ

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