0
$\begingroup$

ΔU = Q - W

Q: Heat received by the gas from the surroundings (positive if received, negative if given to the surroundings) W: Work done by the gas on the surroundings (positive if done on the surroundings, negative if received from the surroundings)

In an adiabatic process, the heat received from the surroundings is zero, so the change in internal energy is determined by the amount of work exchanged with the surroundings.

"Considering the case of an ideal gas during an adiabatic process, the temperature of the ideal gas is determined solely by its internal energy. Therefore, if work is received from the surroundings, Q=0, the internal energy increases, and the temperature rises."

It seems like I can understand it this way, but various questions are bothering me...

What is the internal energy of an ideal gas? It is said to be the kinetic energy of the ideal gas. Then, what is the thermal energy of an ideal gas? This is said to be the average kinetic energy of the ideal gas. If we understand it this way:

---> In an adiabatic process, if an ideal gas receives work from the surroundings, the internal energy increases, and since internal energy is kinetic energy, and average kinetic energy is thermal energy, the thermal energy of the ideal gas increases. Therefore, the temperature increases. It seems that since an adiabatic process does not involve heat energy supply from the surroundings, it doesn't matter what happens to the internal thermal energy. So, it can be understood this way, but I'm not sure if it's correct.

Can we not think of the increase in internal energy as an increase in internal thermal energy?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ The term Thermal Energy is sometimes used in the in reference to internal kinetic energy. But it is also sometimes erroneously used in reference to heat and temperature, which is why Mark Zemansky, author of textbook Heat and Thermodynamics, recommends avoiding the term. $\endgroup$
    – Bob D
    Commented Jun 17 at 9:15
  • 1
    $\begingroup$ The internal energy $U$ of a gas (ignoring gravitational or electric/magnetic interactions) has three parts and can be written as $U=TS-pV+\mu N$ where the symbols are temperature $T$, entropy $S$, pressure $p$, volume $V$, chemical potential $\mu$ and chemical moles $N$. You are free to call the constituent terms $TS$, $pV$, $\mu N$ as thermal energy, volumetric energy, chemical energy, resp., and no harm will be done, but for historical reasons some people are against such nomenclature while others are for it. $\endgroup$
    – hyportnex
    Commented Jun 17 at 9:52

1 Answer 1

Reset to default
2
$\begingroup$

Can we not think of the increase in internal energy as an increase in internal thermal energy?

To quote the introduction to the Wikipedia article on "Thermal Energy":

The term "thermal energy" is used loosely in various contexts in physics and engineering, generally related to the kinetic energy of vibrating and colliding atoms in a substance. It can refer to several different physical concepts. These include the internal energy or enthalpy of a body of matter and radiation; heat, defined as a type of energy transfer....

As you can see, the problem with the term "Thermal Energy" is that it is "used loosely".

So while you may "think" of an increase in the kinetic energy component of internal energy as an increase in internal thermal energy, if you use the term thermal energy in place of internal energy, others may think you are referring to something else. On the other hand, the term "internal energy" is well defined.

So for the sake of avoiding confusion, I would recommend you avoid substituting the term "thermal energy" for "internal energy".

Hope this helps.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.