Helium gas contained in a piston cylinder assembly undergoes an isentropic polytropic process from the given initial state with P1 = 0.02 ba

Question

Helium gas contained in a piston cylinder assembly undergoes an isentropic polytropic process from the given initial state with P1 = 0.02 bar, T1 = 200 K to the final state with P2 = 0.14 bar. Determine the work and heat transfer (per unit mass) involved in this process.

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Ben Gia 5 years 2021-08-22T23:39:03+00:00 1 Answers 12 views 0

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  1. thienthi
    0
    2021-08-22T23:40:16+00:00
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    Answer:

    Heat transfer during the process = 0

    Work done during the process = – 371.87 KJ

    Explanation:

    Initial pressure P_{1} = 0.02 bar

    Initial temperature T_{1} = 200 K

    Final pressure P_{2} = 0.14 bar

    Gas constant for helium R = 2.077 \frac{KJ}{kg k}

    This is an isentropic polytropic process so temperature – pressure relationship is given by the following formula,

    \frac{T_{2} }{T_{1} } = [\frac{P_{2} }{P_{1} } ]^{\frac{\gamma - 1}{\gamma} }

    Put all the values in above formula we get,

    ⇒ \frac{T_{2} }{200} = [\frac{0.14 }{0.02 } ]^{\frac{1.4 - 1}{1.4} }

    ⇒  \frac{T_{2} }{200} = 1.74

    ⇒ T_{2} = 348.72 K

    This is the final temperature of helium.

    For isentropic polytropic process heat transfer to the system is zero.

    ⇒ ΔQ = 0

    Work done W = m × ( T_{1} – T_{2} ) × \frac{R}{\gamma - 1}

    ⇒ W = 1 × ( 200 – 348.72 ) × \frac{2.077}{1.4 - 1}

    ⇒ W = 371.87 KJ

    This is the work done in this process. here negative sign shows that work is done on the gas in the compression of gas.

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