Why is Liter*atm equivalent to Joules?

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SUMMARY

The discussion centers on the equivalence of liter-atmospheres (L*atm) and joules (J) in the context of gas expansion and work done. The conversion factor established is 1 L*atm = 101.32 J. Participants clarify that work can be expressed as W = PΔV, where pressure (P) is defined as force per unit area, and ΔV represents the change in volume. The conversation emphasizes understanding the physical concepts behind these units rather than merely relying on equations.

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  • Understanding of basic physics concepts such as work, pressure, and volume.
  • Familiarity with the ideal gas law and the gas constant (R).
  • Knowledge of unit conversions, specifically between L*atm and joules.
  • Basic grasp of thermodynamics, particularly the work done by gases.
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  • Study the derivation of the work formula W = PΔV in detail.
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patrickbotros
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Hey forum! My name's Patrick. :biggrin:
I've been staring at this FOREVER, but I cannot seem to come up with an answer on my own. I've taken 1.5 semesters of physics, but I cannot figure this out. The reason I posted this in Chemistry is because I realized that the R constant can be written as 8.314 J/(mol*K) or as .0821 L*atm/(mol*K). So why is the Joule just a constant multiplier away from L*atm. I think my dilemma is that I cannot think of a SINGLE thing that would have the units L*atm. In your answers, please don't just draw up a formula that you think I should know and then verify that the units work out. I want a CONCEPT that verifies this, not an EQUATION.
Thanks so much in advance! I feel like this is the smartest website on the internet! #IwillNotMissYahooAnswers :smile:
 
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Hey Patrick. Welcome!

At the risk of being facetious, have you confirmed that the units are consistent? That, indeed, L*atm, when expressed in standard units, will be equivalent to Joules?

Then imagine a gas at some pressure (expressed in atm) expanding by some volume (in L). It seems reasonable to express the work done in units of L*atm. (Though I would stick with standard units.)
 
The conversion from L*atm to J is 101.32 J=1(l*atm). And I don't quite understand why that is reasonable, because I don't understand why I'm allowed to say that a gas is expanding. I think if you could sell me on this point then I would understand it: Why would the units of a gas expanding in some fixed volume be L*atm?
 
Do you agree that the work done by an expanding gas can be expressed as ##P\Delta V##?
 
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patrickbotros said:
gas expanding in some fixed volume

Gas doesn't expand in a fixed volume, gas expands changing volume. And it expands pushing things around apart.
 
Doc Al said:
Do you agree that the work done by an expanding gas can be expressed as ##P\Delta V##?
Hmmm. I'm not sure if I know why that is. I see that the units would work out:
Work=Force*ΔPosition, so Work=(Force/(unit area))*(ΔVolume)=Pressure*ΔVolume, right? It still seems a bit iffy, though. :H:frown::sorry: I don't know what concept says that W=PΔV, so if you could tell me a little more of what you know about that formula, I'd REALLY appreciate it. I know that work is area under the force vs. time curve, but I can't figure out how to connect that fact to pressure and volume.
Thanks so much for all your help!
 
What is the most basic definition of work (hint: it involves force and displacement)?

What is the definition of the pressure?

Imagine a piston moving in a cylinder. Assume the piston has a head surface A and is in contact with a gas of pressure P.

What is the force acting on the cylinder?

Now imagine this piston moved by L, compressing the gas. What was the work done?

What is A×L?
 
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Borek said:
What is the most basic definition of work (hint: it involves force and displacement)?

What is the definition of the pressure?

Imagine a piston moving in a cylinder. Assume the piston has a head surface A and is in contact with a gas of pressure P.

What is the force acting on the cylinder?

Now imagine this piston moved by L, compressing the gas. What was the work done?

What is A×L?
Hey THANKS! It took me a few minutes, but I totally get it! Pressure=Force/A, So Work=Pressure*A*L, Work=Pressure*∆V!
P.S. I like your hair.
 
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