According to my textbook ,

1st law of thermodynamics is

Change in internal energy of ideal gas = K.E + P.E = Work done = $F_{ext}$ * displacement$_{walls}$

= n * degree of freedom * R * $\delta$T / 2 + 0.

Change in internal energy for any gas = W$_{macroscopic}$ + W$_microscopic$ or Thermal. W= work done here

= W + q

= $\frac{n * degree of freedom * R *$ \delta$ T_{diff}}{ 2}$ = P$_{ext}$($V_i$ - $V_f$) \delta$ is macroscopic work + q is microscopic work(No formula )

My Questions :

  1. Here , if we are finding change in internal energy , why do we write macroscopic + micro work. Both should be microscopic right.

  2. So , $\delta$ E for ideal gas was = $\frac{n * degree of freedom * R *$ \delta$ T_{diff}}{ 2}$ and so is same for the 1st law. Shouldn’t we have had added + P.E ?

  3. Can we say this formula of 1st law on external + internal system because of the formula of work done ?.

  • 1
    $\begingroup$ Where did you encounter these phrases "macroscopic work" and "microscopic work"? The usual classifications of energy-during-transfer are "work" and "heat". $\endgroup$ May 16, 2021 at 11:21
  • $\begingroup$ @DanielHatton From an online lecture sir. $\endgroup$
    – Rider
    May 16, 2021 at 11:22
  • $\begingroup$ @DanielHatton For microscopic work , it states it as thermal work. And macroscopic is work done or W $\endgroup$
    – Rider
    May 16, 2021 at 11:24

1 Answer 1


When energy is being transferred, it can be classified as "work" or "heat" (it looks like this textbook/online lecture is using the name "macroscopic work" for what's usually called "work", and the name "microscopic work" for what's usually called "heat"). When energy is being stored, it can be classified as "(bulk) kinetic energy", "(bulk) potential energy", or "internal energy". Energy can be transferred in a "macroscopic" way (work), but still end up stored in a "microscopic" way (internal energy) when it reaches its destination and vice versa.


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