Van der Waals equation 1.2

Whether X_{\text{measured}} is bigger or smaller than X_{\text{ideal}} ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

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In the ideal gas equation derivation, the volume used in the equation refers to the volume that the gas molecules can move within. So

V_{\text{ideal}} = V_{\text{available for a real gas' molecules to move within}}

Then, when deriving the pressure, it is assumed that there are no intermolecular forces among gas molecules. So

P_{\text{ideal}} = P_{\text{assuming no intermolecular forces}}

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These are the reasons that

V_{\text{ideal}} < V_{\text{measured}}

P_{\text{ideal}} > P_{\text{measured}}

P_{\text{ideal}} V_{\text{ideal}} = nRT

\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) \left(V_\text{measured}-nb\right) = nRT

— Me@2018-05-16 07:12:51 PM

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… the thing to keep in mind is that the “pressure we use in the ideal gas law” is not the pressure of the gas itself. The pressure of the gas itself is too low: to relate that pressure to “pressure for the ideal gas law” we have to add a number. While the volume occupied by the real gas is too large – the “ideal volume” is less than that. – Floris Sep 30 ’16 at 17:34

— Physics Stackexchange

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2018.05.16 Wednesday (c) All rights reserved by ACHK

Tree rings, 2

69b73-growth_rings

This file is licensed under the Creative Commons Attribution 2.0 Generic license. Author: Lawrence Murray from Perth, Australia

Time-traveling to the past is like “making an outside ring more inside”, which is logically impossible.

— Me@2011.09.18

8fd2c-pastpresentfuture

Me@2010

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2018.05.16 Wednesday (c) All rights reserved by ACHK