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# 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.

### 1 Answer

- schmisoLv 76 years agoFavorite 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 = 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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