Why do we add, and not subtract, the correction term for pressure in [Van der Waals] equation?

Since the pressure of real gases is lesser than the pressure exerted by (imaginary) ideal gases, shouldn’t we subtract some correction term to account for the decrease in pressure?

I mean, that’s what we have done for the volume correction: Subtracted a correction term from the volume of the container V since the total volume available for movement is reduced.

asked Sep 30 ’16 at 15:20

Ram Bharadwaj

— Physics Stackexchange

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Ideal gas law:

However, since in a real gas, there are attractions between molecules, so the measured value of pressure P is smaller than that in an ideal gas:

Also, since the gas molecules themselves occupy some space, the measured value of the volume V is bigger that the real gas really has:

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If we substitute onto the LHS, since , the LHS will be smaller than the RHS:

So in order to maintain the equality, a correction term to the pressure must be added:

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If we substitute onto the LHS, since that volume is bigger that actual volume available for the gas molecules to move, the LHS will be bigger than the RHS:

So in order to maintain the equality, a correction term to the pressure must be subtracted:

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In other words,

— Me@2018-05-13 03:37:18 PM

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Why? I still do not understand.

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How come

but

?

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How come

but

?

— Me@2018-05-13 03:22:54 PM

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The above is wrong.

The “real volume” has 2 possible different meanings.

One is “the volume occupied by a real gas”. In other words, it is the volume of the gas container.

Another is “the volume available for a real gas’ molecules to move”.

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To avoid confusion, we should define

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Or even better, avoid the terms and altogether. Instead, just consider the relationship between and that between .

Whether is bigger or smaller than ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

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— Me@2018-05-13 04:15:34 PM

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