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.

  • $\begingroup$ Could you provide the appropriate equations please. $\endgroup$ – lemon Sep 30 '16 at 15:21
  • $\begingroup$ P + a(n/V)^2 * V - nb = nRT $\endgroup$ – Ram Bharadwaj Sep 30 '16 at 15:24

There are two terms in the Van der Waals correction:

$$\left(P + a\left(\frac{n}{V}\right)^2\right)\left(\frac{V}{n}-b\right)=RT$$

The first of these ($a$) is an attraction term: the molecules attract each other, which makes the pressure lower. It's the sum of the actual pressure and the attraction term that gives you the "ideal gas pressure".

The second term ($b$) is a volume term: the "apparent volume" that the molecules can move in is smaller than the volume of the container, so we need to subtract a number from the size of the container to get the "ideal volume".

And this is why you have the two terms, with opposing signs.

  • 1
    $\begingroup$ Since molecules attract each other in the gas, the real pressure will be lower than the ideal pressure. Given that, I still don't understand why we ADD the pressure correction term. Please help. $\endgroup$ – Ram Bharadwaj Sep 30 '16 at 15:45
  • $\begingroup$ Since molecules attract each other in the gas, the real pressure will be lower than the ideal pressure. So we add a term to compensate for this reduction in pressure ( I guess that's what you're saying). But going by this logic, shouldn't we add a term in the volume correction as well to account for the reduction in volume available for movement? Please help. $\endgroup$ – Ram Bharadwaj Sep 30 '16 at 15:51
  • 1
    $\begingroup$ The term in the first set of brackets corresponds to "pressure if this was an ideal gas". The actual pressure is lower - to get to the "ideal gas equation" we need to correct pressure upwards. The second term is "volume if this was an ideal gas". The actual volume of the container is larger (because the "available" volume is reduced by the volume of the molecules). I don't know how else to say this... $\endgroup$ – Floris Sep 30 '16 at 15:57
  • 3
    $\begingroup$ You can understand this gas law as this :$$P = \frac{N k T}{V - V_0} - P_0,$$ where $V_0$ is the effective volume of all the molecules packed together, that you need to subtract from the physical volume $V$ available. Also, $-\; P_0$ is the effective tension produced by the attraction of the molecules, that reduces the total pressure of the whole gas, so that $P < N k T/(V - V_0)$. $\endgroup$ – Cham Sep 30 '16 at 19:16
  • 1
    $\begingroup$ The ideal gas equation works for an ideal gas. A real gas is "ideal but for these effects". By taking account of these effects we can relate the real and ideal gas behaviors. $\endgroup$ – Floris Oct 1 '16 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.