Molar vaporization enthalpy of liquid water △ vaphm (H2O) =△ FHM (H2O, G)-△ FHM (H2O, L) = 44 kJ/mol.
According to Clausius-krabbe dragon equation:
ln(p 1/P2)=( 1/T2- 1/t 1)*△VAP hm/R
Where T 1 and T2 are the boiling points of liquids at P 1 and P2 pressures, and r is the gas constant.
Let P2 be the standard atmospheric pressure 10 1325Pa, then T2 is the normal boiling point of water, and let T 1 = 383. 15K.
ln(p 1/ 10 1325)=( 1/373. 15- 1/383. 15)* 44000/8.3 144 1
Yide p1=146712.6pa.
So the pressure in the pot is 1467 12.6Pa, which is about 1.448 standard atmospheric pressure.