Question

A sample of an ideal gas goes through the process shown. 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).

Answer 1

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