# Thermodynamic work done on a free-falling object

By definition the thermodynamic work is the opposite of the mechanical work (engineers'convention).

Consider then an object of mass $$m$$ falling at constant velocity above the Earth surface (imagine that someone is pushing in order to balance the gravitational pull) for an height $$h$$. We know that the potential energy decreases by $$-mgh$$, hence the internal energy should also decrease by the same amount however, by applying the First Law of Thermodynamics I get that $$\Delta U = 0$$ because there is no net work done on the object.

I would like to know what I am doing wrong.

• "By definition the thermodynamic work is the opposite of the mechanical work (engineers'convention)". Please cite a reference. I'm not aware of any difference. Jan 26, 2023 at 12:57
• See for instance Zemansky's "Heat and Thermodynamics", especially the beginning of Chapter 2. Jan 26, 2023 at 13:15
• I have Zemansky. Can you be more specific as to where in chapter 2. Jan 26, 2023 at 13:24
• @BobD I am sorry, I meant Chapter 3, to be more specific page 50 at the end of the paragraph: "This convention disagrees with engineering practice,...". Jan 26, 2023 at 13:26
• OK, found it. See my answer below. Jan 26, 2023 at 13:49

By definition the thermodynamic work is the opposite of the mechanical work (engineers'convention).

Work is defined the same way in thermodynamics as mechanics. The difference is how the definition of work is applied in thermodynamics vs mechanics as described in your reference to Zemansky.

Mechanics is generally only concerned with "external work", i.e. the effect of work on changes in kinetic and potential energy of the system as a whole with respect to an external frame of reference.

For closed systems, thermodynamics is generally concerned with the effects of work within the system which causes changes in microscopic molecular potential and kinetic energy, i.e., changes in internal molecular energy.

However, the total change in energy of the system, $$\Delta E_{tot}$$ is the sum of the change in internal energy and external energy of the system based on the general form of the first law:

$$\Delta E_{tot}=\Delta U+\Delta KE+\Delta PE=Q-W$$

Where $$\Delta KE$$ and $$\Delta PE$$ are the changes in kinetic and potential energy of the system as a whole with respect to an external (to the system) frame of reference. See the figure below for a closed system.

We know that the potential energy decreases by $$-mgh$$, hence the internal energy should also decrease by the same amount however

The decrease is in the potential energy of the system as a whole, as discussed above. The loss of gravitational potential energy of a falling object (neglecting air resistance) has no effect on the internal kinetic or intermolecular potential energy of the falling object, i.e., the kinetic and potential energy associated with molecular motions and intermolecular forces of the object (system).

by applying the First Law of Thermodynamics I get that $$\Delta U = 0$$ because there is no net work done on the object.

There is net work done, but it's done by gravity on the system as a whole giving the system kinetic energy as a whole. It's the $$\Delta KE$$ in the general first law equation above. But there is no change in internal molecular kinetic and potential energy since that only applies to the kinetic and potential energy associated with molecular motions and intermolecular forces.

Hope this helps.

The complete form of the first law is: $$\Delta (U + E_K + E_P + \cdots) = Q + W,$$ where $$U$$ is internal energy, $$E_K$$ is kinetic energy and $$E_P$$ is potential energy. The left-hand side includes all forms of energy in the system; the right-hand side is the net inputs of heat and work. For a stationary system, like a cylinder/piston that we often use in thermodynamics, $$\Delta E_K=\Delta E_P=0$$, but in the case of the example that you give we have $$\Delta E_K +\Delta E_K = Q = W = 0$$ The net result is that potential energy is converted to kinetic. $$\Delta U$$ is zero in this case. Internal energy will increase when the mass impacts on the ground. Unless the collision with the ground is perfectly elastic some energy will be converted into internal but the body will rebound with less kinetic energy than when it hit the ground.

(I am ignoring in this analysis the drag due to air, which would convert some of the kinetic energy into internal of the air and of the falling body).