0
$\begingroup$

My sir told me that Total energy of system = K.E + P.E in starting.

Then Change in energy = F(External force on body ) * displacement of walls.

Then from here , change in energy = W+q. (Don’t understand how we reached till here )

This change in energy is also called change in internal energy. Then it should be equal to Kinetic + potential energy of the molecules inside the system that we are considering .

Q1 is that if it is equal to kinetic + potential energy, then in case of ideal gases.P.E = 0.So K.E will be equal to change in energy $\delta U$.

Q2 is that if it change in energy , then it should be change in kinetic energy as well right.

Ok then my sir also tells me that when a reaction happens in non ideal gas scenario , example of 1mole N2 + 3 mole H2 = 2mole NH3.Now my sir says , there will be a lot of potential energy here.So we can’t use the formula kinetic energy = q+w here.

When Na(s) in converted to Na(g) we consider , delta H of sublimation of Na and then potential energy change I.e internal energy change = Na+(g) + e. Is this internal energy change $\delta U$.

All these doubts arrived from this question.enter image description here

Because here they ask for internal energy change , so does it mean that it is $\delta H_1$ + $\delta H_2$ + Internal energy of 1 + internal energy of 2.

I am sorry if this in not written in good manner but am so much in tension right now that I just gave out everything I knew

Improve this question
$\endgroup$
6
  • $\begingroup$ you should focus on one general question instead and be precise $\endgroup$ – user99515 Dec 21 '20 at 12:57
  • 1
    $\begingroup$ Change of energy is not simply F*displacement of the wall, as you write. This formula gives only the work entering or getting out of the system. This work increases the internal energy. But it is not the only way of increasing the internal energy. Heat (or heat transfer) is another way of increasing the internal energy. And of course changing the chemical nature of the substances present in the container is third way of changing the internal energy. What is Q1 and Q2 in your question ? $\endgroup$ – Maurice Dec 21 '20 at 13:04
  • 1
    $\begingroup$ See Wikipedia - Internal energy changes and Enthalpy. Why to write again, what has been written about many times ? $\endgroup$ – Poutnik Dec 21 '20 at 14:20
  • $\begingroup$ See also Wikipedia - First law of thermodynamics and hyperphysics - First law of thermodynamics. The latter site provies a set of easy, cheat-sheet-like format of short simple pages. $\endgroup$ – Poutnik Dec 21 '20 at 14:55
  • $\begingroup$ @Poutnik Can we say enthalpy change as heat transfer when pressure is constant but not volume.? $\endgroup$ – srijan Sri Dec 21 '20 at 17:09
2
$\begingroup$

I'm glad that you have got good concepts but the problem is you're unable to connect the discontinuous dots. See, in order to answer your first two questions I would like to bring attention towards the difference of microscopic energy (energy possesed by the molecules present inside the system) and macroscopic energy (energy you get to see in the system). As a example, a cup of coffee is placed over ground all will say it has zero K.E and 0 P.E. But the molecules present inside the system posses both K.E and P.E. You are right with the fact that change in internal energy for ideal gas means change in K.E but the change in K.E means, difference between the total K.E(summation of K.E of every molecules present in the system) of the sytem before and after the action. In case of non ideal gas, here both P.E and K.E both will contribute. In case of sublimation of Na, here solid is changing to gas you can't neglect volume change also you have to keep it in mind that change in enthalpy is equal to change in internal energy if the PV-work is zero and the process is isobaric.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Am I right with the Na part I wrote as a whole. $\endgroup$ – srijan Sri Dec 21 '20 at 16:35
  • $\begingroup$ No, you can't consider enthalpy of sublimation with internal energy without proper condition is provided. $\endgroup$ – Summit Dec 21 '20 at 17:21

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