Question

In a change of an ideal gas from an initial state at 7 kPa and 3 m3 to a final state of 4 kPa and 6 m3, the internal energy change is -20 kJ. Calculate the change in enthalpy in kJ.

Answer 1

Given that the internal energy change is -20KJ the change in enthalpy is : -17 KJ

Given data :

P1 = 7 kPa

P2 = 4 kPa

V1 = 3 m³

V2 = 6 m³

Determine the change in enthalpy

Applying the relationship below

Change in enthalpy = change in internal energy + ( P2V2 - P1V1 )

                                 = -20 + ( 4*6 - 7*3 )

                                 = -20 + ( 24 - 21 )

                                 = - 17 KJ

Hence we can conclude that the change in enthalpy is : -17 KJ

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