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?

1 Answer
Nov 3, 2015

Answer:

#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"#