Question

For a particular process that is carried out at constant pressure, q = 145 kJ and w = -35 kJ. Calculate δu and δh for this reaction.

Option 1: δu = 145 kJ and δh = 180 kJ
Option 2: δu = 145 kJ and δh = 110 kJ
Option 3: δu = 110 kJ and δh = 145 kJ
Option 4: δu = 180 kJ and δh = 145 kJ

Answer 1

The changes in internal energy (ΔU) and enthalpy (ΔH) for the process are 110 kJ and 145 kJ, respectively, corresponding to Option 3.

Step-by-step explanation:

To solve for the changes in internal energy (ΔU) and enthalpy (ΔH) for the process, we can use the first law of thermodynamics, which is ΔU = q + w, where q is the heat exchanged and w is the work done on or by the system. In this case, q is given as 145 kJ and w is given as -35 kJ (work done by the system). Therefore, ΔU is:

ΔU = q + w = 145 kJ + (-35 kJ) = 110 kJ.

Since the process is carried out at constant pressure, the change in enthalpy ΔH can be found from the formula ΔH = ΔU + PΔV. However, at constant pressure and when the only work done is expansion or contraction work, ΔH is simply equal to q, the heat exchanged at constant pressure. Therefore, ΔH is:

ΔH = q = 145 kJ.

Thus, the correct values are ΔU = 110 kJ and ΔH = 145 kJ, which corresponds to Option 3.