When 5.00 mol of benzene is vaporized at a constant pressure of 1.00 arm and at its normal boiling point of 80.1 degrees Celcius, 169.5 kJ are absorbed and P delta V for the vaporization process is equal to 14.5 kJ then....

 

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A) AE = 184.0kJ and AH = 169.5 kJ. AE = 169.5 kJ and AH = 184.0 kJ. B) AE = 155.0 kJ and AH = 169.5 kJ. Đ AE = 1695 k] and AH = 155.0 kJ. !3! !i! A) AE = 184.0kJ and AH = 169.5 kJ. AE = 169.5 kJ and AH = 184.0 kJ. B) AE = 155.0 kJ and AH = 169.5 kJ. Đ AE = 1695 k] and AH = 155.0 kJ. !3! !i!