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AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] Watch

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    (Original post by Mehrdad jafari)
    Guys, I also have a feeling that a six marker could be on commenting on the assumptions made before deriving the equation of the kinetic theory of gases. The textbook does comment on two of the assumption but no further. Here are some comments relating to the assumptions made:

    1) The volume of each molecule is negligible compared to the volume of the gas.
    • Comment: that is to say the molecules of the gas are very far apart because in liquid state the molecules are sufficiently close to each other for the attractive forces between molecules to be important.

    2) The time of each collision between molecules or each molecule with the container walls is negligible compared with the time between successive collisions.
    • Comment: ( i still don't really know why this assumption is made. I would be very happy if any one could let us know.)

    3) The molecules do not attract each other.
    • Comment: if the particles attracted each other they would not exert force on the walls of the container as they would all tend to clump together in the middle of the container.

    4) the particles move in continual random motion.
    • Comment: the forces exerted by the particles is evenly distributed over the entire container surface that's why particles are assumed to have random motion in 3 dimensions.

    5) The collisions of particles with each other and with the container surface are perfectly elastic, so that no kinetic energy is lost.
    • Comment: the total kinetic energy of particles is the internal energy of the gas, therefore kinetic energy is assumed to be conserved so that the total energy of the gas in the total kinetic energy of the molecules.


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    For why the time elapsed for each collision is negligible compared to the time between collisions. This is due to what you have used in your equation. In the derivation of the kinetic equations you say that delta t is the time taken for the molecule to travel from one side of the container to the other and back. Hence dt is not the time duration of each collision, but an average time which requires that the actual time of collision is small compared to the time between. I don't think they will ask that in the exam. It's to do with your assumption in the motion of the particle itself
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    (Original post by Protoxylic)
    For why the time elapsed for each collision is negligible compared to the time between collisions. This is due to what you have used in your equation. In the derivation of the kinetic equations you say that delta t is the time taken for the molecule to travel from one side of the container to the other and back. Hence dt is not the time duration of each collision, but an average time which requires that the actual time of collision is small compared to the time between. I don't think they will ask that in the exam. It's to do with your assumption in the motion of the particle itself
    "Hence delta t is not the time duration of each collision, but an average time which requires that the actual time of collision is small compared to the time between."
    Would you explain that a bit more. I remember Amanzz said something similar to this but if dt is not the time of each collision then the assumption is somewhat wrong.


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    (Original post by Mehrdad jafari)
    "Hence delta t is not the time duration of each collision, but an average time which requires that the actual time of collision is small compared to the time between."
    Would you explain that a bit more. I remember Amanzz said something similar to this but if dt is not the time of each collision then the assumption is somewhat wrong.


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    dt is the time between collisions. So this is the overall time it takes to change the momentum of the particle from it's initial position (wall A) by 2mux in the x direction. So that when time dt is elapsed, the particle is at the same position as it was, but now with the opposite momentum. This requires that the time of the collision is very small in comparison to the length of time between the start and end points of which you're describing the change in momentum/force. Otherwise if the collision takes place over a larger time interval, then it HAS to be true that the speed is not constant (which is the case in reality) and this dt cannot be calculated using the magnitude of vx as it isn't constant, it changes with time as there is a force acting on it at the boundary.
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    (Original post by IWantSomeMushu)
    Quick question!

    If you have a question which involves rearranging equations, i.e. v=Hd to d=v/H or something more complicated involving combining multiple equations and even substituting them into one another, can you simply show the final rearranged equation in symbols and then use the values to get your answer? Or will you lose method marks?
    I'd say depends on the question - a low mark question on a simple rearrangement of a 3 term equation, going straight to your final rearrangement (eg the v=Hd could be written as d =v/H) shouldn't lose you any marks as far as I know. A complicated substituting things in etc, I'd definitely show the steps.

    (Original post by Mai.H)
    There's only been one thermal 6 marker in the past do you think that makes a thermal 6 marker more likely? Not that it matters too much!
    Not necessarily, thermal is a pretty small topic and except for experiments (which they don't seem so keen on asking about at A2 as they were at AS) there's not so many possible six markers really whereas nuclear has lots. However 1 in 5 so far and the last in 2011 does make it seem like it's got a good chance of popping up this year!
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    (Original post by Protoxylic)
    For why the time elapsed for each collision is negligible compared to the time between collisions. This is due to what you have used in your equation. In the derivation of the kinetic equations you say that delta t is the time taken for the molecule to travel from one side of the container to the other and back. Hence dt is not the time duration of each collision, but an average time which requires that the actual time of collision is small compared to the time between. I don't think they will ask that in the exam. It's to do with your assumption in the motion of the particle itself
    Dude you seem so clever! , always see you in these threads! I take it you want to study chemistry or something at university?


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    (Original post by SuperMushroom)
    Dude you seem so clever! , always see you in these threads! I take it you want to study chemistry or something at university?


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    Offer holder for NatSci at Cambridge . My passion is in Chemistry, however. But (grades permitting) I get to continue with Physics, Chem, Maths and take up Material Sciences at Cambridge
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    (Original post by Protoxylic)
    dt is the time between collisions. So this is the overall time it takes to change the momentum of the particle from it's initial position (wall A) by 2mux in the x direction. So that when time dt is elapsed, the particle is at the same position as it was, but now with the opposite momentum. This requires that the time of the collision is very small in comparison to the length of time between the start and end points of which you're describing the change in momentum/force. Otherwise if the collision takes place over a larger time interval, then it HAS to be true that the speed is not constant (which is the case in reality) and this dt cannot be calculated using vx as it isn't constant, it changes with time as there is a force acting on it at the boundary.
    Ok, if dt is not the time during which each particle and the container wall are in contact then we cannot substitute this into the momentum formula because the force found is not the force exerted by the each particle on the wall but is the deriving force supplied by the energy of the particle to change its momentum, if I'm not wrong


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    (Original post by Mehrdad jafari)
    Ok, if dt is not the time during which each particle and the container wall are in contact then we cannot substitute this into the momentum formula because the force found is not the force exerted by the each particle on the wall but is the deriving force supplied by the energy of the particle to change its momentum, if I'm not wrong


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    Basically dt is not the true time duration of each collision, it is the time between collisions which requires the assumption that the time duration of each collision is small.
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    Why is the change in temperature 327-84? Kind of confused
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    (Original post by Protoxylic)
    Basically dt is not the true time duration of each collision, it is the time between collisions which requires the assumption that the time duration of each collision is small.
    That makes perfect sense but do you not think assuming that would give us the wrong overall pressure?


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    (Original post by Mehrdad jafari)
    That makes perfect sense but do you not think assuming that would give us the wrong overall pressure?


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    The pressure is of the order of KPa normally, so this assumption is justified by the fact that the difference between the actual pressure and the calculated pressure isn't that much I presume
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    (Original post by Protoxylic)
    Offer holder for NatSci at Cambridge . My passion is in Chemistry, however. But (grades permitting) I get to continue with Physics, Chem, Maths and take up Material Sciences at Cambridge
    Ahh awesome!! Well all the best hopefully you get in


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    Do any of you guys know why 1c uses the change in temperature as 327-84? http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
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    Just started revision for this. How likely is it to get A* after a day's revision?
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    (Original post by Protoxylic)
    The pressure is of the order of KPa normally, so this assumption is justified by the fact that the difference between the actual pressure and the calculated pressure isn't that much I presume
    Well, that's true, we tend to give the priority to the experimental observations rather than the theoretical understanding of it.


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    (Original post by SuperMushroom)
    Ahh awesome!! Well all the best hopefully you get in


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    Cheers! And I hope you get the grades you need for wherever you want to go
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    Absolutely bricking it! I know the spec inside out I can just tell its gonna be a bad exam


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    Someone ask me some turning point questions


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    (Original post by JJBinn)
    Why is the change in temperature 327-84? Kind of confused
    Because that's the temperature that the mould decreases to, from its initial temperature of 327 degrees.


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    (Original post by ubisoft)
    Just started revision for this. How likely is it to get A* after a day's revision?
    Just how likely you think it is


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