# If Internal energy can be changed by doing work also, then how come internal energy is zero in a isothermal process involving non zero work?

In the equation of First law of thermodynamics

$$∆U = q + w$$,

where $$U$$ is the internal energy, $$q$$ is the heat supplied and $$w$$ is the work done.

It is said that for an isothermal process, since the temperature remains constant, the change in internal energy is zero even if the work done is non zero.

On the other hand, thermodynamics also says that internal energy can be changed by doing work.

So supposing for a particular process the work is not zero in the above equation, then shouldn't the internal energy also be non-zero?

I don't get it why such type of consideration is taken while solving problems of the type given below:

1. An ideal gas undergoes isothermal compression from 5 m³ to 1 m³ against a constant external pressure of 4 N m-2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol-¹K-¹, the temperature of Al increases by (a) 2 K (b) 1K (c) 2/3 K (d) 3/2 K

In the solution, the heat released has been equated to the work done which is

$$-P(\mathrm{external})(V_2-V_1)=-4(1-5)=\pu{16 J}$$ <--------I don't get this part.

What actually am I missing?

TL;DR: You can't keep the temperature of a perfect gas constant if only work is performed. Heat must be exchanged also.

Answers involving the thermodynamics of ideal gases are logically self-consistent. Start from some logical premises:

If the gas is ideal, then its internal energy depends only on T. Corollary: if the internal energy has changed, T has changed, and vice-versa.

Changes in internal energy are possible through either exchange of heat, or work, or both.

Starting from the above, you can state that:

(1) if a net amount of heat is exchanged with only one other body, but no work is done at all, then T must have changed, because the internal energy must have changed.

(2) if the internal energy has not changed, then T has not changed, and either W=-Q or W=0 and Q=0.

• It is said that for an isothermal process, since the temperature remains constant, the change in internal energy is zero even if the work done is non zero. I think this question is another example of the confusion that basing the teaching of ThermoD on the ideal gas creates. Here is Truesdell: "The tragicomedy of thermodynamics has been the persisting tendency to entangle in special constitutive relations, so that the purely energetic content of the theory becomes invisible." Why does this tendency persists in teaching, who is benefiting from it? Surely, not the students. Commented Apr 5 at 22:33
• @hyportnex Truesdell? Who's that? How is this relevant? Can you explain? Commented Apr 7 at 5:23
• see: Clifford Ambrose Truesdell. He wrote a truly awesome history of TD: The Tragicomical History of Thermodynamics, 1822-1854, he was originator/founder of Rational Thermodynamics, see his several books on the subject. especially, this Rational Thermodynamics. He had a very sharp tongue and made lots of enemies but his writings are never boring... Commented Apr 7 at 12:18
• and with apologies to both, the quote as written was taken from Sorensen, while referring to Truesdell and to his work, and the full context from which the quote taken is on page 557; I am a Bronstedian... Commented Apr 7 at 12:36
• @hyportnex Thanks for the links and refs. I'll have to take a look. Commented Apr 7 at 14:05