A physical chemistry problem; let’s discuss it together
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In a container with a constant temperature of 100 degrees and a volume of 2 cubic decimeters, there is 0.04 mol of water vapor. If an additional 0.02 mol of H2O (in liquid form) is added to this container, then at equilibrium the state of H2O in the container will necessarily be ( ) 1. Gaseous 2. Liquid 3. A balance between gas and liquid phases. The answer is 1, but I don’t understand why. Could some expert explain this?
First, what is the phase of water at 100 degrees? You could guess the answer to this question even without thinking! With constant temperature and volume, partial pressure is directly proportional to the number of moles; if the number of moles increases, the partial pressure must also increase, and the only way for this to happen is for water to turn into steam! This requires some thinking!:lol
Is it possible for a gas-liquid equilibrium or a situation where there is no separation between gas and liquid to occur?
Do you think water at 100 degrees will be in a liquid state? 100 degrees is its boiling point!!! How much is 0.06 mol of water? In a 2M3 container! :lol
Okay! I overthought it! :lol
It’s 2 cubic decimeters, which is only 2 liters, not 2 square units. Read the question carefully.
This post was last edited by qugd on 2015-6-30 at 12:50. According to the gas equation PV=nRT, when volume and temperature remain constant, the amount of substance also remains constant; thus, only pressure needs to be calculated to make a determination. If the pressure is 1 atmosphere or higher, a two-phase state of gas and liquid will definitely exist at 100 degrees; if the pressure is less than 1 atmosphere, then there will definitely only be a gas phase. P=0.06*0.082*373/2 = 0.92atm-A; it must be a gas. Why not go and calculate it? It’s such a simple question. It is also possible to assume an atmospheric pressure for system 1 and calculate the amount of gas at equilibrium within the system to make a judgment.
It seems that judging whether a state is gaseous just by looking at 100 degrees isn’t necessarily reliable.:L The calculation method mentioned above is the proper way to solve this problem; I’ve learned something new again.
Then go ahead and award the Best Support prize quickly – something practical.