For a particular process that is carried out at constant pressure,q = 140 kJ and w = -30 kJ.Therefore,
A)ΔE = 110 kJ and ΔH = 140 kJ.
B)ΔE = 140 kJ and ΔH = 110 kJ.
C)ΔE = 140 kJ and ΔH = 170 kJ.
D)ΔE = 170 kJ and ΔH = 140 kJ.

Question

Quizplus · Accepted Answer

q = ΔH and w = -ΔE. Since the process is carried out at constant pressure, ΔH = q = 140 kJ. Using w = -ΔE, we can solve for ΔE as: 
ΔE = -w = -(-30 kJ) = 30 kJ
Then, 
ΔH = 140 kJ 
ΔE = 30 kJ 
ΔH = ΔE + PΔV
Therefore,
ΔE = ΔH - PΔV
Since the process is carried out at constant pressure, ΔV = q/P = 140 kJ/1 atm = 140 L*atm/mol.
Substituting the values, we get:
30 kJ = 140 kJ - (1 atm)(140 L*atm/mol)
Solving for ΔH, we get:
ΔH = 170 kJ
Therefore, the best choice is A (ΔE = 30 kJ and ΔH = 140 kJ).

For a Particular Process That Is Carried Out at Constant