For a particular process that is carried out at constant pressure,q = 145 kJ and w = -35 kJ.Therefore:
A)ΔU = 110 kJ and ΔH = 145 kJ
B)ΔU = 145 kJ and ΔH = 110 kJ
C)ΔU = 145 kJ and ΔH = 180 kJ
D)ΔU = 180 kJ and ΔH = 145 kJ

Question

Quizplus · Accepted Answer

The first law of thermodynamics states that ΔU = q + w, where ΔU is the change in internal energy, q is the heat added to the system, and w is the work done by the system. Here, ΔU = 145 kJ + (-35 kJ) = 110 kJ. For a process at constant pressure, the enthalpy change ΔH is equal to the heat added to the system, so ΔH = 145 kJ.

For a Particular Process That Is Carried Out at Constant