Question

A sample (8.1 mol) of an ideal monatomic gas goes through the process shown in 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 from D to d,itis sobarie with 3711 of energy leaving the system by heat. P (atm) 12 (m') 0.09 0.2 0.4 (a) Determine the difference in internal energy Eint,B-EinA (b) Determine the change in the entropy as the system goes from C to D

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