Question

if the partial pressure dropped from 22 to 15 kpa throughout this experiment and the oxygen solubility of water is 0.272 ml l⁻¹ kpa⁻¹ and the volume of water is 40 liters, how much oxygen did the stackarel consume?
a) 0.272 ml l⁻¹ kpa⁻¹ x 40 l x 7 kpa
b) 0.272 ml l⁻¹ kpa⁻¹ x 22 kpa
c) 0.272 ml l⁻¹ kpa⁻¹ x 40 l
d) 0.272 ml l⁻¹ kpa⁻¹ x 40 l x 15 kpa

Explanation:

Step1: Find pressure change

The partial - pressure dropped from 22 kPa to 15 kPa. The change in partial pressure $\Delta P=22 - 15=7$ kPa.

Step2: Recall solubility formula

The amount of oxygen dissolved in water is given by the formula $V = S\times V_{water}\times\Delta P$, where $S$ is the oxygen solubility of water, $V_{water}$ is the volume of water, and $\Delta P$ is the change in partial pressure.
Given $S = 0.272$ ml L$^{-1}$ kPa$^{-1}$, $V_{water}=40$ L, and $\Delta P = 7$ kPa.
The volume of oxygen consumed is $V=0.272$ ml L$^{-1}$ kPa$^{-1}\times40$ L$\times7$ kPa.

Answer:

A. $0.272$ ml L$^{-1}$ kPa$^{-1}\times40$ L$\times7$ kPa