Question

What is the boiling point of water when the atmospheric pressure exerted on the water is 81 kPa ?

Answer 1

Answer: The boiling point of water at 81 kPa will be 298.17 K.

Explanation:

The boiling point of water is 373 K at atmospheric pressure. So, to calculate the boiling point at 81 kPa, we use the law given by Gay-Lussac.

Gay-Lussac Law states that pressure is directly proportional to the temperature when volume and moles remain constant.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2$

where,

$P_1\text{ and }T_1$ = Initial pressure and temperature

$P_2\text{ and }T_2$ = Final pressure and temperature

We are given:

$P_1=1atm=101.325kPa\\T_1=373K\\P_2=81kPa\\T_2=?K$

Putting values in above equation, we get:

$\frac{101.325kPa}{373K}=\frac{81kPa}{T_2}\\\\T_2=298.17K$

Hence, the boiling point of water at 81 kPa will be 298.17 K.