# 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?

Nov 3, 2015

${P}_{{H}_{2} O} = 35 \text{ torr}$

#### Explanation:

At constant number of mole $n$ and Temperature $T$, using the Ideal Gas Law $P V = n R T$ we can write:

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

We should calculate the pressure of pure ${O}_{2}$ in a $2.01 L$ 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 ${H}_{2} O$ and ${O}_{2}$:

${P}_{t} = {P}_{{H}_{2} O} + {P}_{{O}_{2}}$

$\implies {P}_{{H}_{2} O} = {P}_{t} - {P}_{{O}_{2}} = 785 \text{ torr"-750 " torr"=35 " torr}$