A single drop of water is injected into a balloon at 25C that contains ammonia, at 1 atm. What will happen to the V of the balloon?Assume Henry's constant is ~0.02 atm/M
To determine what will happen to the volume (V) of the balloon when a single drop of water is injected into it, we need to consider Henry's Law.
Henry's Law states that the concentration of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. Mathematically, it can be represented as:
C = k * P
where C is the concentration of the gas in the liquid, k is Henry's constant, and P is the partial pressure of the gas.
In this case, we are dealing with the gas ammonia in the balloon. When the water drop is injected into the balloon, ammonia gas molecules will dissolve into the water, resulting in an increase in the concentration of ammonia in the liquid.
The partial pressure of ammonia in the balloon remains constant at 1 atm. Therefore, according to Henry's Law, the concentration of ammonia in the liquid (water) will increase.
Now, let's connect this information to the volume of the balloon. When the concentration of a gas dissolved in a liquid increases, the volume of the gas in the liquid also increases. This is because the gas molecules take up space within the liquid.
Therefore, when the concentration of ammonia in the water increases due to the injection of a water drop, the volume of the ammonia gas within the balloon will also increase. Consequently, the volume of the balloon will expand.
In summary, injecting a single drop of water into the balloon containing ammonia will cause an increase in the concentration of ammonia gas in the water. This will result in an increase in the volume of the ammonia gas within the balloon, causing it to expand.