Question

The vapor pressure of water at $${80}^{\circ }C$$ is 355 torr. A 100 ml vessel contained water-saturated oxygen at $${80}^{\circ }C$$ the total gas pressure being 760 torrs. The contents of the vessel were pumped into a 50.0 ml vessel at the same temperature, what were the partial pressures of oxygen and water vapor, what was the total pressure in the final equilibrated state? Neglect the volume of any water which might condense.

Answer 1

Hint: We are provided pressure of all the molecules. Water vapor at a pure state is also given. To get the total pressure of the mixture at the same temperature, we need to calculate a sum of the vapor pressures of the contents. The water vapor pressure does not change with volume. As temperature remains constant the vapor pressure of the water would be the same.

Formula used: $$P_{total}=P_{O_2}+P_{H_{2}O}$$

Complete step by step answer:
There are different units of pressure used around the world. The pressure is defined as force per unit area which is perpendicular to the surface. SI unit of pressure is Pascal or newton per square meter. The bar is also used but it is based on the metric system. For liquids, it is expressed as mmHg and torr for absolute pressure which is $${\displaystyle \frac{1}{760}}$$ of a standard atmosphere. 1 atm is the standard atmospheric unit of pressure.
Now, according to the question, the vapor pressure of water and the mixture is 355 torr and 760 torrs. Therefore, the value of vapor pressure of oxygen would be,
 $$\begin{gathered} P_{total}=P_{O_2}+P_{H_{2}O} \\ 760=P_{O_2}+355 \\ 760-355=P_{O_2} \\ 405=P_{O_2} \end{gathered}$$
So, at the temperature, the vapor pressure of oxygen is 405 torr.
When volume reduces to 50ml, the pressure of the oxygen would be,
 $$P_1{V}_1=P_2{V}_2$$ , where $$P_1$$ and $$P_2$$ are the pressure and $$V_1$$ , $$V_2$$ are the volumes.
 $$\begin{gathered} P_1{V}_1=P_2{V}_2 \\ {\displaystyle \frac{P_1{V}_1}{V_2}}=P_2 \\ {\displaystyle \frac{405\times 100}{50}}=P_2 \\ 810=P_2 \end{gathered}$$
So, when the volume is 50ml the vapor pressure of the oxygen is 810 torr.
Now the total pressure of water-saturated oxygen at 50 ml volume is,
 $$\begin{gathered} P_{total}=P_{O_2}+P_{H_{2}O} \\ {P}_{total}=810+355 \\ {P}_{total}=1165 \end{gathered}$$
So, the total pressure is 1165 torr.

Note:The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases present in a mixture. This is called Dalton’s law or the law of partial pressures. It can also be applied to the number of moles.