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The solubility of a gas is 2 g/L at 50 kPa of pressure. What is the solubility of a gas
at 10 kPa of pressure? Assume the temperature is constant. *

Sagot :

sebassandin

This problem is providing us with the initial solubility and pressure of a gas and asks for the solubility once the pressure is changed to 10 kPa. At the end, the answer is found to be 0.4 g/L.

Ideal gases

In chemistry, the concept of ideal gas provides us first approach to the understanding of gases, as a model with no interactions and perfectly spherical-shaped molecules, however, this is not an actual a fact in nature.

Thus, we can relate pressure, volume, temperature and moles (or grams) by using the widely-known ideal gas equation:

PV = nRT

Which in the case of the solubility of the gas, can be written as:

P=Sol*RT*MM

Which at constant temperature can be just:

[tex]\frac{P}{Sol} =k[/tex]

Since R, T and MM (molar mass of the gas) are constants, and for two states:

[tex]\frac{P_1}{Sol_1} =\frac{P_2}{Sol_2}[/tex]

In such a way, one can calculate the solubility when the pressure changes as follows:

[tex]Sol_2=\frac{Sol_1*P_2}{P_1}[/tex]

Finally, we plug in the numbers to obtain:

[tex]Sol_2=\frac{2g/L*10kPa}{50kPa} =0.4g/L[/tex]

Learn more about ideal gases: brainly.com/question/8711877

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