Question

Calculate the boiling point of water if the atmospheric pressure is 630 mmhg

Answer 1

The boiling point of water can be calculated by the equation:

Where:

P = Pressure in mm Hg

Po = Atmospheric pressure in mm Hg

ΔH= heat of vaporization in kJ/mol

R = Ideal Gas Constant (J/mol-K)

To = normal boiling point in Kelvin

T = boiling point of water (K)

Our known values are:

P = 630 mm Hg

Po = 760 mm Hg

ΔH = 40.66 kJ/mol = 40.66×1000 =40660

R = 8.314 J mol⁻¹ K ⁻¹

To = 373 K

Putting these values in the equation,

$ln \frac{P_{0}}{P}= \frac{\Delta H}{R}(\frac{1}{T}-\frac{1}{T_{0}})$

$ln \frac{760}{630}= \frac{40660}{8.314}(\frac{1}{T}-\frac{1}{373})$

Solving the equation will give:

T=370K

so, the boiling point of water is 370 K.

Answer 2

95 Celsius is the answer.