Question

The solubility of a gas is 0.64 g/L at a pressure of 112 kPa. What is its solubility if the pressure is increased to 260 kPa while the temperature is held constant? Answer choices: 1.5 g/L 3.0 g/L.28 g/L .56 g/L

Answer 1

Step-by-step explanation

Note that the basic relationship between the solubility and temperature can be given by: The higher the temperature is, the easier a solid will be able to dissolve. Likewise the lower is the temperature the harder is for a solid element to dissolve.

However, for a gas, the higher the temperature is, the more there is a decrease in the gas solubility. The lower the is temperature the higher is a gas solubility in water.

Therefore,

`\begin{gathered} \frac{112\text{ }kPa}{260\text{ }kPa}=\frac{x}{0.64\text{ }g\text{/}L} \\ \\ x=\frac{112\text{ }kPa*0.64\text{ }g\text{/}L}{260\text{ }kPa}=0.28\text{ }g\text{/}L \end{gathered}`