Question

A 2.00-L sample of was collected over water at a total pressure of 750. torr and 26 °C. When the was dried (water vapor removed), the gas had a volume of 1.93 L at 26 °C and 750. torr. Calculate the vapor pressure of water at 26 °C.

Answer 1

The vapor pressure of water at 26 °C = 26.25 atm

Step-by-step explanation:

Using Boyle's law

`{P_1}* {V_1}={P_2}* {V_2}`

Given ,

V₁ = 2.00 L

V₂ = 1.93 L

P₁ = ?

P₂ = 750 torr

Using above equation as:

`{P_1}* {V_1}={P_2}* {V_2}`

`{P_1}* {2.00}={750}* {1.93}`

`{P_1}=\frac {{750}* {1.93}}{2.00}\ torr`

`{P_1}=723.75\ torr`

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