Anonymous
Anonymous asked in Science & MathematicsPhysics · 6 years ago

A sample of an ideal gas goes through the process shown below. From A to B, the process is adiabatic; from B t?

A sample of an ideal gas goes through the process shown below. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 kJ of energy entering the system by heat. From C to D, the process is isothermal; and from D to A, it is isobaric with 371 kJ of energy leaving the system by heat. Determine the difference in internal energy, Eint,B − Eint,A.

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  • 6 years ago
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    Because the process from B to C is isothermal the internal energy in state B and state C is the same.

    E_int,B =E_int;C

    The reason for this is, that the internal for an ideal ideal gas is a function of the temperature alone:

    E_int = n∙Cv∙T

    where

    n = number moles

    Cv = molar heat capacity at constant volume

    T = thermodynamic temperature

    That makes it possible you can compute the change internal energy for process A→B from the changes for the processes B→C and D→A.

    ∆E_int_B→C = E_int,C - E_int;B

    ∆E_int_D→A = E_int,A - E_int;D

    =>

    E_int;B = E_int,C - ∆E_int_B→C

    E_int,A = E_int,D + ∆E_int_D→A

    Hence,

    ∆E_int_A→B = E_int;B - E_int,A

    = E_int,C - ∆E_int_B→C - (E_int,D + ∆E_int_D→A )

    = - ∆E_int_B→C - ∆E_int_D→A

    You have enough information to compute the change in internal energy from B to C as well as from D to A.

    The change internal energy equals heat added to the gas minus work done by it:

    ∆E_int = Q - W

    The heats are given, The work done by the gas in these isobaric processes is given by:

    W = P∙∆V

    Pressure and volumes can be found in the diagram.

    Hence

    W_B→C = P∙(V_C - V_B)

    = 3∙101.3 kPa ∙ ( 0.4 m² - 0.09 m³)

    = 94.2 kPa∙m³

    = 94.2 kJ

    =>

    ∆E_int_B→C = Q_B→C - W_B→C

    = 345 kJ - 94.2 kJ

    = 250.8 kJ

    W_D→A = P∙(V_A - V_D)

    = 1∙101.3 kPa ∙ ( 0.2 m² - 1.2 m³)

    = - 101.3 kPa∙m³

    =>

    ∆E_int_D→A = Q_D→A - W_D→A

    = - 371 kJ - (-101.3 kJ )

    = - 269.7 kJ

    =>

    ∆E_int_A→B = - ∆E_int_B→C - ∆E_int_D→A

    = - (250.8 kJ ) - (-268.7 kJ)

    = 18.9 kJ

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