Question

a student has two unopened 33 cl cans containing carbonated water. can a has been stored in the garage (32°c) and can b has been stored in the fridge (8°c). the student opens one can at the time, both cans make a fizz.
a student sees tiny bubbles clinging to the inside of an unopened plastic bottle full of carbonated soft drink. the student squeezes the bottle.

Explanation:

First Experiment

Step1: Recall gas solubility and temperature relationship

The solubility of a gas in a liquid decreases with an increase in temperature. Carbonated water has dissolved \(CO_2\) gas. Can A is at a higher temperature (\(32^{\circ}C\)) than can B (\(8^{\circ}C\)). At higher temperatures, more \(CO_2\) will escape when the can is opened.

Second Experiment

Step1: Recall gas solubility and pressure relationship

According to Henry's law, \(C = kP\) (where \(C\) is the concentration of the dissolved gas, \(k\) is the Henry's law constant, and \(P\) is the partial pressure of the gas above the solution). When the bottle is squeezed, the pressure inside the bottle increases. As pressure increases, the solubility of \(CO_2\) (a gas in the soft - drink) increases. So, the bubbles (which are \(CO_2\) gas) will shrink as more \(CO_2\) dissolves back into the soft - drink.

Answer:

For the first experiment: Can A will make a louder and stronger fizz than can B.
For the second experiment: The bubbles will shrink, and some may vanish.