Question

The vapour pressure of water at ${80^{\text{o}}}{\text{C}}$ is 355 torr. A 100 ml vessel contained water saturated oxygen at ${80^{\text{o}}}{\text{C}}$, the total gas pressure being 730 torr. The contents of the vessel were pumped into a 50 ml vessel at the same temperature. What were partial pressures of oxygen and of vapour and the total pressure in the final equilibrium-state? Neglect the volume of water might condense.

Answer 1

Hint: This problem can be solved from the knowledge of Dalton’s Law of partial pressure which states that, the total pressure exerted by a mixture of gases that are unreactive to each other is equal to the partial pressure off the individual gases.

Formula used:
${{\text{P}}_{{\text{total}}}} = {{\text{P}}_{\text{A}}}{\text{ + }}{{\text{P}}_{\text{B}}}{\text{ + }}{{\text{P}}_{\text{C}}}{\text{ + }}{{\text{P}}_{\text{D}}}{\text{ + }}.............$
Where ${{\text{P}}_{\text{A}}}{\text{, }}{{\text{P}}_{\text{B}}}{\text{, }}{{\text{P}}_{\text{C}}}.......$ are the individual pressures of the gases.
\[{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}\]
Where, \[{{\text{P}}_{\text{1}}}\] is the partial pressure of gas 1 at ${80^{\text{o}}}{\text{C}}$ in the vessel of volume $\left( {{{\text{V}}_{\text{1}}}} \right)$ , \[{{\text{P}}_{\text{2}}}\] is the partial pressure of gas 2 in the vessel of volume $\left( {{{\text{V}}_2}} \right)$ .

Complete step by step answer:
Let the total pressure be ${{\text{P}}_{{\text{total}}}}$, and that of oxygen be ${{\text{P}}_0}$ and vapour be ${{\text{P}}_{{\text{vap}}}}$.
According to Dalton’s Law, ${{\text{P}}_{{\text{total}}}}{\text{ = }}{{\text{P}}_{\text{O}}}{\text{ + }}{{\text{P}}_{{\text{vap}}}}$
$ \Rightarrow 730 = {P_O} + 355$
Solving this, we get:
$ \Rightarrow {P_O} = 375{\text{torr}}$
To find out the partial pressure of oxygen we need the Boyle’s Law according to which,
\[{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}\]
\[ \Rightarrow 355 \times 100 = {{\text{P}}_{\text{2}}} \times 50\]
\[ \Rightarrow {{\text{P}}_{\text{2}}} =\dfrac{{355 \times 100}}{{50}}\]
Solving this:
\[ \Rightarrow {{\text{P}}_{\text{2}}} = 750{\text{torr}}\]
Where, \[{{\text{P}}_{\text{1}}}\] is the partial pressure of water at ${80^{\text{o}}}{\text{C}}$ in the vessel of volume 100 ml $\left( {{{\text{V}}_{\text{1}}}} \right)$ , \[{{\text{P}}_{\text{2}}}\] is the partial pressure of oxygen in the vessel of volume 50 ml $\left( {{{\text{V}}_2}} \right)$ .
Therefore total pressure in the vessel of volume 50 ml is,
${{\text{P}}_{{\text{total}}}}{\text{ = }}{{\text{P}}_{\text{O}}}{\text{ + }}{{\text{P}}_{{\text{vap}}}}$
Substituting the values:
$ \Rightarrow {{\text{P}}_{{\text{total}}}} = 750 + 355 = 1105$
$\therefore {{\text{P}}_{{\text{total}}}} = 1105{\text{torr}}$

Note:
The pressure exerted by any gas in a non-reactive mixture of gas is called its “partial pressure.”
The law is valid only in case of ideal gases.
For the law to be valid, the gases should be non-reactive with each other, for example N2 and $ H_2O$, because if the gases react among themselves then their volumes might change which will affect the total pressure.