A sample of an ideal gas goes through the process shown in the figure below. Fro

Question

A sample of an ideal gas goes through the process shown in the figure 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 E_int, B - E_int, A. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully, kJ

Explanation / Answer

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 = nCvT
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 AB from the changes for the processes BC and DA.
E_int_BC = E_int,C - E_int;B
E_int_DA = E_int,A - E_int;D
=>
E_int;B = E_int,C - E_int_BC
E_int,A = E_int,D + E_int_DA
Hence,
E_int_AB = E_int;B - E_int,A
= E_int,C - E_int_BC - (E_int,D + E_int_DA )
= - E_int_BC - E_int_DA

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 = PV
Pressure and volumes can be found in the diagram.

Hence

W_BC = P(V_C - V_B)
= 3101.3 kPa ( 0.4 m² - 0.09 m³)
= 94.2 kPam³
= 94.2 kJ
=>
E_int_BC = Q_BC - W_BC
= 345 kJ - 94.2 kJ
= 250.8 kJ

W_DA = P(V_A - V_D)
= 1101.3 kPa ( 0.2 m² - 1.2 m³)
= - 101.3 kPam³
=>
E_int_DA = Q_DA - W_DA
= - 371 kJ - (-101.3 kJ )
= - 269.7 kJ

=>

E_int_AB = - E_int_BC - E_int_DA
= - (250.8 kJ ) - (-268.7 kJ)
= 18.9 kJ

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