Question

A 2.01-L sample of O2(g) was collected over water at a total pressure of 785 torr and 25°C. When the O2(g) was dried (water vapor removed), the gas has a volume of 1.92 L at 25°C and 785 torr. How would you calculate the vapor pressure of water at 25°C?

Answer 1

`P_(H_2O)=35 " torr"`

Explanation:

At constant number of mole `n` and Temperature `T`, using the Ideal Gas Law `PV=nRT` we can write:

`P_1V_1=P_2V_2`

We should calculate the pressure of pure `O_2` in a `2.01L` volume:

`P_1=(P_2V_2)/V_1=(785" torr"xx1.92cancel(L))/(2.01cancel(L))=750" torr"`

In the original mixture, the total pressure `P_t` is the sum of partial pressures of `H_2O` and `O_2`:

`P_t=P_(H_2O)+P_(O_2)`

`=> P_(H_2O)=P_t - P_(O_2) = 785 " torr"-750 " torr"=35 " torr"`