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Discussion :: Chemical Engineering Thermodynamics

  1. On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line (∂P/∂V)S and the slope of the reversible isothermal line (∂P/∂V)T are related as (where, y = Cp/Cv) )

    2. A.

     (∂P/∂V)S = (∂P/∂V)T
    B.

     (∂P/∂V)S = [(∂P/∂V)T]Y
    C.

     (∂P/∂V)S = y(∂P/∂V)T
    D.

     (∂P/∂V)S = 1/y(∂P/∂V)T

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    Answer : Option C

    Explanation :

    For an adiabatic process → PVy = constant
    For an isothermal process → PV = constant
    So, slope for adiabatic process in PV plane is dpdv=−ypv
    Slope for isothermal process is dpdv=−pv
    Hence, (∂p∂v)S=−y(∂p∂v)T

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