Answer:
The vapor pressure of water at 26 °C = 26.25 atm
Explanation:
Using Boyle's law
[tex] {P_1}\times {V_1}={P_2}\times {V_2}[/tex]
Given ,
V₁ = 2.00 L
V₂ = 1.93 L
P₁ = ?
P₂ = 750 torr
Using above equation as:
[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]
[tex]{P_1}\times {2.00}={750}\times {1.93}[/tex]
[tex]{P_1}=\frac {{750}\times {1.93}}{2.00}\ torr[/tex]
[tex]{P_1}=723.75\ torr[/tex]
Vapor pressure = Total pressure - Partial pressure of oxygen gas = 750 - 723.75 atm = 26.25 atm
The vapor pressure of water at 26 °C = 26.25 atm
