# Van der Waals equation 1.1

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

— Physics Stackexchange

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Ideal gas law: $P_{\text{ideal}} V_{\text{ideal}} = nRT$

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: $P_{\text{measured}} = P_{\text{real}}$ $P_{\text{measured}} < P_{\text{ideal gas}}$

Also, since the gas molecules themselves occupy some space, the measured value of the volume V is bigger that the real gas really has: $V_{\text{measured}} > V_{\text{real}}$ $P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $P_{\text{measured}}$ onto the LHS, since $P_{\text{measured}} < P_{\text{ideal}}$, the LHS will be smaller than the RHS: $P_{\text{measured}} V_{\text{ideal}} < nRT$

So in order to maintain the equality, a correction term to the pressure must be added: $\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) V_{\text{ideal}} = nRT$ $P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $V_{\text{measured}}$ 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: $P_{\text{ideal}} V_{\text{measured}} > nRT$

So in order to maintain the equality, a correction term to the pressure must be subtracted: $P_{\text{ideal}} \left(V_\text{measured}-nb\right) = nRT$

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In other words, $V_{\text{measured}} > V_{\text{real}}$ $V_{\text{ideal}} = V_{\text{real}}$ $V_{\text{measured}} > V_{\text{ideal}}$

— Me@2018-05-13 03:37:18 PM

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Why? I still do not understand.

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How come $P_{\text{measured}} = P_{\text{real}}$

but $V_{\text{measured}} \ne V_{\text{real}}$?

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How come $V_{\text{real}} = V_{\text{ideal}}$

but $P_{\text{real}} \ne P_{\text{ideal}}$?

— Me@2018-05-13 03:22:54 PM

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The above is wrong.

The “real volume” $V_{\text{real}}$ 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 $V_{\text{real}} \equiv V_{\text{measured}}$ $P_{\text{real}} \equiv P_{\text{measured}}$

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Or even better, avoid the terms $P_{\text{real}}$ and $V_{\text{real}}$ altogether. Instead, just consider the relationship between $(P_{\text{ideal}}, P_{\text{measured}})$ and that between $(V_{\text{ideal}}, V_{\text{measured}})$.

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.

— Me@2018-05-13 04:15:34 PM

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