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How do you explain the fact that the observed pressure goes DOWN with attractive forces but volume doesn't go up? Does PV = constant not apply here? Volume also decreases?
van der waals equation
- Thread starter echoyjeff222
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How do you explain the fact that the observed pressure goes DOWN with attractive forces but volume doesn't go up? Does PV = constant not apply here? Volume also decreases?
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@echoyjeff222
I'm a bit confused by that last sentence as well. Doesn't choice C imply that volume is INCREASED (not reduced). It went from 22.4L (ideal) to 24.5 (non-ideal).
Volume should increase because temp increased.
PV = constant but that is assuming temperature is the same.
PV = nRT
If temperature changes, the "nRT" term changes.
Btw: Where is this question from?
I'm a bit confused by that last sentence as well. Doesn't choice C imply that volume is INCREASED (not reduced). It went from 22.4L (ideal) to 24.5 (non-ideal).
Volume should increase because temp increased.
PV = constant but that is assuming temperature is the same.
PV = nRT
If temperature changes, the "nRT" term changes.
Btw: Where is this question from?
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the ideal gas law is PV=NRT
for a real gas corrected to show the ideal behavior, van der waal equation (Pobs + aN^2/V^2)(V-Nb)=NRT
Since the attractive force of the molecules pull on each other, you have to add a term aN^2/V^2 to adjust for ideal behavior. Since we have defined the pressure to be 1atm. so the adjusted term Pobs+aN^2/V^2 will be greater than 1 atm. Since the ajusted term is greater than 1 atm, V-Nb will be smaller than ideal behavior evaluated at 1atm to keep NRT constant. Since b is specific for a specific molecule, it does not change. The change will be only applied to V, so V is decreased
for a real gas corrected to show the ideal behavior, van der waal equation (Pobs + aN^2/V^2)(V-Nb)=NRT
Since the attractive force of the molecules pull on each other, you have to add a term aN^2/V^2 to adjust for ideal behavior. Since we have defined the pressure to be 1atm. so the adjusted term Pobs+aN^2/V^2 will be greater than 1 atm. Since the ajusted term is greater than 1 atm, V-Nb will be smaller than ideal behavior evaluated at 1atm to keep NRT constant. Since b is specific for a specific molecule, it does not change. The change will be only applied to V, so V is decreased
