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My sir told me that when bonds are formed or broken , average separation between particles comes closer or decreases when they are broken.I see there is a lot of change in K.E but how is there change in P.E is not what I am able to understand.My sir also told that in case of liquids , P.E would be less as compared to gases. Pls help me understand how does that happen.

Online it says that since the gaseous atoms are attracted to each other , they have P.E.Now P.E =mgh right?So if we say bonds come closer , We know that M is mass of body , g is gravity , H is distance from the earth.As you go up , P.E increases and down,P?e decreases.Just saying that they are attracted to each other (in all cases liquids , gases and solids), they posses P.E.But how does the distance of force of attraction that the molecules have matter?.That i what I am not able to understand.I hope you have got it.

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  • $\begingroup$ The one on the fridge doors have higher P.E.Energy is released while attaching or detaching the magnet yes. $\endgroup$ – srijan Sri Oct 29 '20 at 7:37
  • $\begingroup$ I don’t understand how can this be compared to my question. $\endgroup$ – srijan Sri Oct 29 '20 at 7:37
  • $\begingroup$ Please tell what have I not elaborated ? $\endgroup$ – srijan Sri Oct 29 '20 at 7:39
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    $\begingroup$ See e.g. courses.lumenlearning.com/boundless-chemistry/chapter/… Skip the text, if too difficult, and see the chart of bond energy versus distance. See also en.wikipedia.org/wiki/Potential_energy $\endgroup$ – Poutnik Oct 29 '20 at 9:54
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    $\begingroup$ Note that $E_p= mgh$ is simplification of $$Ep=\kappa Mm \left(\frac{1}{R}-\frac{1}{R+h}\right)$$ where M and R are the Earth mass and radius, $\kappa$ is gravitational constant, m and h are the object mass and relative height wrt Earth surface $\endgroup$ – Poutnik Oct 29 '20 at 10:01
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Your main mistake is to believe that the potential energy is only due to gravity $(E_\mathrm{pot} = mgh).$ But there exist other forms of potential energy. Potential energy may have another origin. It may have an electric origin.

When you take a H atom (containing one proton and one electron) in order to separate the proton and the electron, you must give some energy to the H atom (as light for example). When the proton and the electron are separated, far from one another, they usually do not have any kinetic energy $(E_\mathrm{kin} = mv^2/2),$ they do not have more gravific energy $(E_\mathrm{pot} = mgh).$

But they do have electric potential energy. Potential energy means that the sample has "something" that is invisible, but could be delivered to the surrounding if it has an opportunity to do it. A bead on a table has a gravific potential energy, because if you give it the possibility of going to the ground, it will deliver energy which is first in the form of kinetic energy $(E_\mathrm{kin} = mv^2/2),$ then this energy is transformed into heat when touching the ground (if it does not rebound).

That is about the same for the proton and the electron. When infinitely separated and not in movement, the only energy they have got is the electric potential energy. If you give them the opportunity of approaching one another, they will be attracted, loose their electric potential energy which is transformed into kinetic energy or into light during this approach.

When the H atom is created and stabilized, the atom may still have some kinetic energy, but does not have electric potential energy any more. The gravific energy $(E_\mathrm{pot} = mgh)$ of the proton and electron may have changed a bit during the process, but this change is not significant.

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  • $\begingroup$ Why is it that the gravific P.E does not change much ? In your last sentence. @Maurice $\endgroup$ – srijan Sri Nov 16 '20 at 16:30
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    $\begingroup$ The variation of the gravific potential energy is $mg\Delta h$, with $m$ is the mass of the electron ($\ce{9·10^{-31}} kg$, and $ \Delta h$ is not far from the dimension of the atom, about $10^{-10} m$. So the energy change is about $\ce{10^{-41} }J$. This is so small that it can not be measured. $\endgroup$ – Maurice Nov 16 '20 at 16:49

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