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Question on gas law

  1. May 26, 2013 #1
    After watching this video at the end when he gave the example of PV=nRT and he explained that T and n decreases he went on to explain P and V. Prior to this I would have thought that the ideal gas equation required all but one unknown variable to check if the unknown variable has either increased, decreased or remained the same.

    However, from his example he shown that this is not the case. So can we relate the effects on temperature say now we have a can of boiling water and we cool it down. So now we know that T and n decreases. So the left hand side P times V should decrease. So would I use the variable T of average velocity. So I would say that the pressure increases as the number of collisions per unit time decreases and so does the force per collision. so as a result of that there is a net force pushing the can out in. So the V increases too.

    But now thinking about this, after increasing my temperature and number of moles of gas, the PV definitely should increase. So I'm thinking if volume increases could the pressure remain constant? Cos now if I combined the laws and give as n and T increase and V would increase where P remains constant. Or I could also have a case where P is not constant so the 'stress' is more evenly distributed between the decrease in pressure and volume as he described in the example.

    So actually, would I still need only one variable? So in the video now that PV=nRT and since nRT decreases so PV would have to decrease. So now we know for sure that the V decreases as it is shown that it contracts. But how can we also be sure that the pressure drops? Perhaps the pressure remained the same or it could be that the volume and pressure decreases as he'd described.

    I think using the temperature and number of moles decreasing to explain that P drops would be the intermediate explanation. So it would be something like: as the T and n decreases the pressure would drop. Then after saying that, we should explain that the atmosphere exerts a net force causing it to shrink. So when it shrinks the pressure would increase to either A) the amount it once was or B) increase to a amount lesser than the initial amount. So in this manner of explanation the ideal gas law's variable on the left hand changes constantly but they still multiply to get the same reduced nRT. But the final state of the PV could have 2 possible end product.

    But could there be a 3rd product where the V decreases so much that P actually increases? I don't think so right? Because that would mean the can continues to contract even after there is no net force. So this would break newtons 2nd law because if it shrinks any further the internal pressure>external pushing it out again. So actually only 2 scenarios are possible even thought mathematically 3 scenarios are possible?

    And actually what would be the reason for why the P would not remain constant? I agree that mathematically it makes sense. But if I look at the forces, the pressure is smaller so they would want to have equal external and internal pressure. So shouldn't it contract until Pexternal=Pinternal? What would allow the Pexternal>Pinternal? Could it be that the can being slightly rigid would stop it from getting crushed so the V doesn't continue decreasing? If so what enables the rigidity of the can to prevent it from being further compressed? Because I thought in terms of forces, the can shouldn't be able to push the external pressure away by itself (the only forces acting on it are the external atmosphere and internal gases) so I don't really understand how and why this 'rigidity' comes in actually.

    So essentially, how can we say for sure than both the pressure and volume decrease and not just volume decreases?

    Thanks :) and let me know if I'm vague at some parts.
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. May 26, 2013 #2
    If the can was strong enough to withstand the atmospheric pressure then only P would decrease, as the temperature decreased. V would be constant.
  4. May 26, 2013 #3
    So its also possible for V to remain constant and P drops right? But it would be impossible for either of them to increase right? Cos if we say P increases, it would mean that the external P is smaller so it should expand right?

    But actually what is this rigidity factor here? Because from just my basic physics background I'm thinking the only two forces are from the atmosphere and the internal one. So now if my volume remains the same only the pressure drops. So won't there be net force compressing it?
  5. Jun 1, 2013 #4
    Anyone got any ideas for this?
  6. Jun 1, 2013 #5


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    Staff: Mentor

    You can picture the gas enclosed within a cylinder by a piston. The piston can be drawn outwards, increasing V and decreasing P (and T).

    It is best that you ask a question so readers can address it.
  7. Sep 21, 2013 #6
    Sorry I meant to say that now we know the right hand side nRT decreases. So the PV would also decrease. So mathematically, either/both P or V can decrease to match the drop in magnitude on the nRT side.

    But also mathematically, P can increase while V decreases a lot and so can V increase and P decrease by a lot. Are these 2 cases possible?

    I thought, P can increase while V drops by a lot. But applying physical laws here, this would not happen as this would mean it could push the can out more than previously. So the volume should expand.

    Also, mathematically V can increase while P decreases a lot. Again this would not happen as if P decreases the volume cannot increase as now the external pressure is able to push the can inwards decreasing the volume.

    However, P is inversely proportional to V so I still have some doubts on whether my explanation is valid or not actually.
  8. Sep 22, 2013 #7
    Yes they are.

    This ignores the fact that the container may be strong enough to hold off the external pressure. That's not the case with the soda can, obviously, but that's not a particularly strong thing to begin with.
  9. Sep 22, 2013 #8
    Hmm but actually if we assume the soda can to be non rigid, I don't think these two cases would occur. Because thinking about this more deeply, if the end product case was that the pressure of the can was less than external atmospheric pressure while the volume increased it doesn't make sense. Because it he can should just get crushed because of that.

    Do you think so too?
  10. Sep 22, 2013 #9
    In the experiment with the soda can the volume does not change unless the can is crushed, and it is crushed by a mechanism that is external to the ideal gas law.

    If we are to stay in the realm of the ideal gas law, and consider changing volumes, we need some more well-behaved equipment. Such as a cylinder with a piston as suggested earlier.
  11. Sep 23, 2013 #10
    Oh okay say we have a cylinder containing water with a movable piston. So we heat it up while fixing the volume until all the water has evaporated. Then we cool the cylinder and allow the piston to Kobe by itself. So in this case, it would unnatural for the pressure to decrease by a lot while the volume increases or vice versa whereby the pressure increases and the volume decreases a little?
  12. Sep 23, 2013 #11
    Since we have the piston, and can control the temperature, we can first lower the piston and then cool the system. The end result will be that the final volume is smaller, and the temperature is lower; pressure, depending on how low the piston is set, will be greater than, equal to, or lower than the original pressure.

    Letting the piston go by itself as we cool the cylinder will result in some complicated behavior. You need more than the ideal gas law to analyze that. Don't go there just yet.
  13. Sep 23, 2013 #12
    Hmm I thought the pressure can't be more than when the gas was heated? Because now that nRT decreases when the gas is cooled because n drops and so doesn't T, PV should decrease too. So if the movable piston is perfect only the volume would decrease and the pressure would remain the same.

    But if the piston isn't perfect and as a result the gas experiences changes in pressure too I would think that only the pressure can decrease and not increase. I think it shouldn't be able to increase because if so, the morale piston would be pushed away causing expansion. But than again we said this piston isn't perfect so I'm not entirely sure that the pressure can increase. Why so you say that it is possible that the pressure increases?

    Thanks :)
  14. Sep 23, 2013 #13
    If we use the piston to compress the gas, i.e., decrease the volume available to it, and keep the temperature constant, the pressure has to go up. This follows from the ideal gas law. If we then let the temperature drop as well, the end pressure will depend on both the end temperature and the end volume.

    I am not sure what you mean by "perfect piston". Keep in mind that in this setup piston is not free to move; we are in full control over its motion. It only moves when we want it to, and it only moves the way we desire.

    Letting the piston to move on its own, as I said earlier, would complicate things greatly, and you do not seem to be in position to cope with the complications yet.
  15. Sep 23, 2013 #14
    Ohh I understand now. If we just keep pushing down then surely the pressure could increase from before. However can we try out a simple example where the piston is movable?

    I think that if its movable when some work is performed then heat could enter or leave the system. But can we ignore that for now? Because I'm wondering what would go on if we heated up the cylinder at constant volume, then we cool it down and let the piston go as it would naturally. I think that if this was a perfect scenario where Pressure is inversely proportional to Volume, then the volume would only decrease with the pressure remaining constant.

    But since we are using this movable piston as a representation for the can, I thought the Pressure and Volume should from change together. So for this case I'm not sure if its possible for the Volume to decrease and Pressure to increase.

    Thanks :)
  16. Sep 23, 2013 #15
    Assuming no force except atmospheric pressure is applied to the piston, the initial and the final internal pressures must be equal. Do you see why?
  17. Sep 23, 2013 #16
    I think its because since PV=nRT so since its a movable piston V is proportional to nxT where P is constant. So as the T and n decreases only the V would decrease?

    But what if the piston does not perfectly decrease in volume while holding P constant, would it be possible for P to increase compared to when the cylinder was hot?

    Thanks so much voko :)
  18. Sep 23, 2013 #17
    The pressure stays constant because it is equalized with the atmospheric pressure via the movable piston.

    The piston does not decrease "in volume". The cylinder does.

    I am not sure what kind of imperfection you are talking about. If you mean some deviation from the ideal gas law, a more accurate equation of state could be used. They may be very complex. In any case, pressure is constant here, because it is determined by the external environment (atmosphere).

    If you mean that some other forces may be acting on the piston, in addition to atmospheric pressure, then, indeed pressure may become greater (or lower). But this is exactly what I said earlier.
  19. Sep 23, 2013 #18
    Oh yes that's right. The pressure remains constant due to Pext=Pint so since n and T decreases only V can decrease. Oh yeah I meant to say that the gas in the cylinder not the piston my bad.

    i'm actually wondering about a piston that doesn't completely allow equalization of pressure between the exterior and interior. Something like having a partially rigid structure such that both the pressure and volume of the gas in the cylinder changes. And in this kind of a piston I'm not too sure if the pressure of the gas can actually increase from before?
  20. Sep 23, 2013 #19
    I said "yes" multiple times to that question.
  21. Sep 23, 2013 #20
    Hmm but why is that possible. It's not the same case as where we push the piston down so that the pressure changes but more like having that enclosed can and dipping it into water.

    In the enclosed can case, the volume changes but I'm not sure if its naturally possible for the can to get crushed so much that the pressure actually increases from before icing the can actually.

    Thanks so much for the quick responses :)
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