Understanding Pressure Equalization in the Classic Soda Can Experiment

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Discussion Overview

The discussion revolves around the classic soda can experiment involving pressure equalization when a heated can is inverted into an ice bath. Participants explore the underlying principles of pressure and temperature changes, as well as the relationships described by the ideal gas law.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant describes the process of heating water in a can, inverting it into an ice bath, and the resulting crushing of the can due to pressure differences.
  • Another participant questions what other variables could decrease alongside temperature to maintain the ideal gas law relationship.
  • A different participant suggests that volume could also decrease, raising the question of how this affects the pressure and temperature relationship.
  • One participant asserts that the volume decreases until the pressure inside the can equals atmospheric pressure.
  • A later reply indicates a possible understanding of the concepts discussed, expressing gratitude for the assistance received.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the relationship between pressure, volume, and temperature, with some uncertainty remaining about how these variables interact in the context of the experiment.

Contextual Notes

Participants do not fully resolve the implications of the ideal gas law in this scenario, particularly regarding the transitions between the states of pressure and volume during the experiment.

Who May Find This Useful

Individuals interested in thermodynamics, experimental physics, or those looking to understand the principles of gas behavior under changing conditions may find this discussion informative.

woosh9013
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I have a question regarding the classic experiment with a [soda] can involving pressure. Here is a simplified version of the experiment, without going into too much detail:

1. Heat some water up.
2. Put the water in the can.
3. Invert the can inside an ice bath.

The result is that the can is crushed.

I mostly understand why this happens, but there is one part that I don't get. Here is my explanation:

The air inside the can is rapidly cooled when inserted into the ice bath. As a result, the decrease in temperature causes a decrease in pressure. The pressure on the outside of the can is greater than the pressure on the inside of the can. Since this is true, the air outside exerts a force crushing the can. What I don't get is how the pressure becomes equalized.

Equations used to describe the air inside of the can

Equation 1
Before the water is heated and added to the can
PV=nRT (normal T and P)

Equation 2
Once the can filled with heated water and inverted into the ice bath
(0.5)PV=nR(0.5)T (drop in T causes the drop in P; let's just say T drops by a factor of 0.5, so then the P side must drop by the same factor)

Equation 3
This is the part I don't get. Once the pressure on the inside is equal to the pressure on the outside of the can, the crushing force exerted on the can no longer exists. So if the pressure initially had a factor of 1 before the experiment, then once the experiment is over and the pressures become equal, shouldn't it also have a factor of 1? How does the equation jump from Equation 2 to Equation 3. Doesn't something on the right side have to increase to compensate for the normalization of the pressure?
PV=nRT

If I've messed up somewhere in my thought process, let me know.
 
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In addition to decreasing pressure, what other variable could you decrease to compensate for a decrease in temperature?
 
The volume, but even if the factor of 0.5 is split between volume and pressure then something on the right side still has to increase. Right?

Unless only the volume decreased, because then the value of volume would stay low after the experiment... hmm...
 
Last edited:
Yes, the volume decreases until the pressure is equal to the atmospheric pressure.
 
I think I get it now. Thank you very much for all your help.
 

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