# Is heat transfer = potential energy in formula of 1st law of thermodynamics?

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 ?.

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