When 4.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at

Question

When 4.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 135.6 kJ are absorbed and PV for the vaporization process is equal to 11.6 kJ then

When 4.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 135.6 kJ are absorbed and PV for the vaporization process is equal to 11.6 kJ then

E = 147.2 kJ and H = 135.6 kJ. E = 135.6 kJ and H = 124.0 kJ. E = 135.6 kJ and H = 147.2 kJ. E = 124.0 kJ and H = 135.6 kJ.

Explanation / Answer

heat absorbed

H = 135.6 kJ.

PV = 11.6Kj

H    = E + PV

135.6 = E + 11.6

E      = 135.6-11.6   = 124Kj

E = 124.0 kJ and H = 135.6 kJ. >>>>>answer

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