Question

The heat of vaporization of water is 44.01 kJ/mole and the normal boiling point is 100 °C. Calculate the atmospheric pressure in Denver where the boiling point of water is 97.10 °C. Be sure to enter a unit with your answer.

Answer 1

Solution :-

Delta H vap = 44.01 kJ/ mol * 1000 J / 1 kJ = 44010 J per mol

Normal boiling point = 100 C + 273 = 373 K

Boiling point in Denver = 97.10 C +273 = 370.1 K

Atmospheric pressure in Denver = ?

Formula

Ln[P2/P1] = Delta H vap / R [(1/T1)-(1/T2)]

Ln[P2/760 torr ] = 44010 J per mol / 8.314 J per mol K * [(1/373)-(1/370.1)]

Ln[P2/760 torr ] = -0.1112

P2/760 torr = anti ln [-0.1112]

P2/760 torr = 0.89476

P2 = 0.89476 * 760 torr

P2 = 680 torr

So the atmospheric pressure at the Denver = 680 torr or 680 mmHg or

680 torr * 1 atm /760 torr = 0.8947 atm