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?

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Answer 1 · 2015-11-03T15:12:12

$P_(H_2O)=35 " torr"$

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 $O_2$ :

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

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