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I was sent a problem which goes roughly like this:

There are 6 moles of diatomic nitrogen in a container at temperature T. Some heat is given to the gas such that 2 moles disociate into atomic nitrogen keeping the temperature constant. Calculate the heat given to the gas.

And the solution says that the change in internal energy would be equal to the heat provided and the change in internal energy can be calculated by comparing the average kinetic energies of the gases which are different since one is diatomic and one is monoatomic.

Is this explanation/solution correct? If yes/no then can you please explain why?

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    $\begingroup$ Internal energy will include a contribution from any interactions between the particles in the sample. If the gases are assumed ideal then you ignore these interactions and the internal energy is assumed equal to the kinetic energy. $\endgroup$
    – Buck Thorn ♦
    Commented Feb 3, 2022 at 17:09

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Yes, it is correct for the sake of problem. You can say that the total heat ( which causes change in U, internal energy) is solely responsible for the change in average kinetic energy, by assuming that there is no change in other type of potential energies. Hence, ∆KE becomes equal to ∆U as KE is also a constituent of U.

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  • Internal energy : It is the total energy contained within the system .
    i.e., It is the sum of all types of energies possessed by a system .

  • Kinetic energy : It is the energy possessed by a body by virtue of its motion.

Is this explanation is correct ?

Yes of course , but on one condition . We should treat the gas as an ideal gas. If it is an ideal gas then interaction potential energy in the system will be zero. Therefore , the change in internal energy is equal to the change in Kinetic energy of the system.
The work done on the system will be zero as the volume of container remains unchanged. So from the first law of thermodynamics the change in internal energy is equal to the heat supplied to the system.

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  • $\begingroup$ I wonder if there is ambiguity sometimes in what counts as "kinetic" energy. Sometimes I think people use it to include rotational and vibrational energies, and sometimes I think they use it just to mean translational energy. $\endgroup$
    – Curt F.
    Commented Feb 6, 2022 at 18:15
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    $\begingroup$ Yes , but I here included vibration and rotation too $\endgroup$
    – Infinite
    Commented Feb 7, 2022 at 13:42

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