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I am having trouble understanding why in an isothermal process, the change in internal energy is zero. I know that $\Delta U$ or $\Delta E=q+w$, and so in isothermal process $q=0$. But how does one show that $w =0$? Or is it necessary that if we are talking of isothermal process, we are not doing work on the system; why or why not? Or is it necessary that if $w$ is not equal to 0, then the process can't be isothermal; why? Please explain this. I found a similar question here but I was not able to understand anything from it.

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    $\begingroup$ Q is not zero in an isothermal process. $\endgroup$ – march Oct 14 '15 at 3:47
  • $\begingroup$ Can you provide a link to the similar question you mention? $\endgroup$ – Kyle Kanos Oct 14 '15 at 13:34
  • $\begingroup$ physics.stackexchange.com/questions/113586/… $\endgroup$ – Freelancer Oct 14 '15 at 15:15
  • $\begingroup$ @march I thought that isothermal means temperature is constant which means the system does not exchange or take heat energy from the surroundings otherwise its temprature would have increased that is q=0//so how can you say q is not equal to zero?? $\endgroup$ – Freelancer Oct 14 '15 at 16:09
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    $\begingroup$ I think the answers below have cleared things up for you, but just in case: $Q$ is the amount of energy added via heat. Energy being added via heat does not necessarily lead to a change in temperature; that is a common misconception that is important to clear up. The reason is outlined below: for an ideal gas, $U$ is proportional to $T$, so if $\Delta T=0$, then $\Delta U =0$. This means that whatever energy comes in via heat has to go out via work, or vice versa. $\endgroup$ – march Oct 14 '15 at 20:45
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Isothermal means constant temperature, which in turn often means constant internal energy $\Delta U=0$. The reason is that temperature often "governs" the energy content or at least is a measure of it. Changes in internal energy very often (that is, very often in typical problems) only comes from changes in thermal energy. If there are no phase transformations (no latent heat exchanges), no chemical or magnetic changes, no difference in motion (no kinetic energy changes), etc., which all is very often the case when we consider a system, then the only thing left to cause energy changes is thermal energy. And that is directly bound to the temperature,which is our measure of thermal energy.

So, just because the energy content doesn't change, it doesn't mean that no energy is added. Energy can certainly be added as long as the same amount of energy is removed at the same time. Heat $q$ can be nonzero, no problem. If that is the case, then let's see what the energy conservation equation says:

$$\Delta U=q-w\Leftrightarrow 0=q-w\Leftrightarrow q=w$$

You see, there is nothing preventing a net heat inflow, as long as there at the same time is a net work done by the system. Our the other way around: there can easily be a net heat outflow if just there also is work done on on the system.

The isothermal condition doesn't prevent any of those energy transfers to be present, it only gives the requirement that $q$ and $w$ must be equal.

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  • $\begingroup$ I think I said in my comment that "isothermal means temperature is constant which means the system does not exchange or take # heat energy# from the surroundings"I agreed on this heat energy part only which is wrong also as you pointed out ... But still you didn't state how or why ∆U=0,in isothermal process other things I have understood like how ∆U can be zero by q = -w etc...etc..but just elaborate the part why in isothermal ∆U is equal to zero $\endgroup$ – Freelancer Oct 14 '15 at 17:18
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    $\begingroup$ @Freelancer, ah I see, I didn't get the actual question then. I have now updated the answer to address this. Let me know if this is fulfilling. $\endgroup$ – Steeven Oct 14 '15 at 17:32
  • $\begingroup$ No,again it is the same thing all the other things you talked about are fine but I am not able to understand properly how constant temperature means constant internal energy !! Can you elaborate it a bit more?? $\endgroup$ – Freelancer Oct 15 '15 at 2:49
  • $\begingroup$ Okay, no problem, let's think it over in a few steps. What kinds of energy would be a part of the internal energy? $\endgroup$ – Steeven Oct 15 '15 at 23:25
  • $\begingroup$ I don't know about all the energies that are part of internal energy but I know that translational energy ,rotational energy,vibrational energy,bonding energy are included in this !! $\endgroup$ – Freelancer Oct 16 '15 at 3:24
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The quick answer is $\Delta U \neq 0$.

Let's look at some details. In the special case where you are dealing with ideal gas. $$U = \frac{3}{2} nRT$$ Thus $$\Delta U = \frac{3}{2}nR\Delta T $$ Since the process is isothermal, $\Delta T$ is zero. Therefore $\Delta U = 0$. So it is not true that $q = 0$(that would be called adiabatic). Rather, $q = -w$.

The above analysis fails if the gas is NOT ideal. Since $U = \frac{3}{2}nRT$ is generally not true. But usually the ideal gas approximation works fine.

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  • $\begingroup$ How did you get U=3/2×nRT I think you used an example can you generalise it ?? I only know basic things like ∆U=q+w etc..etc.. Can you please talk in theese terms?? $\endgroup$ – Freelancer Oct 14 '15 at 15:19
  • $\begingroup$ Here is a nice Khan Academy video for deriving $U = \frac{3}{2} nRT: khanacademy.org/science/physics/thermodynamics/… If you take this equation as a given law for ideal gas, your goal is to keep temperature CONSTANT. If you pump in some heat and fix the volume of the gas, then the temperature is going to increase. Thus intuitively the volume has to expand and $w = -q$ in order to guarantee that $T$ does not change. $\endgroup$ – Zhengyan Shi Oct 14 '15 at 16:37
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To Freelancer, I think what you need is a more intuitive answer. I try to give an answer without requiring you to have any knowledge in integral, thermodynamics, physical chemistry and statistical mechanic to understand it. The answer is actually quite simple; "Temperature is in fact an indicator of a system internal energy". In other words, when you measure the temperature of a system, let says a cup of water, the reading which is shown on thermometer is just internal energy times some constant, and this "internal energy times constant" stuff is what you called temperature (the actual equation according to statistical mechanic is far more complex, i had over-simplied it, but the idea is more or less the same). No equation...as i promised.

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  • $\begingroup$ "Temperature is in fact an indicator of a system internal energy." This is demonstrably false. A system undergoing a phase transition (e.g., liquid water being vaporized) has increasing internal energy but constant temperature. $\endgroup$ – tjd Sep 30 '19 at 14:25
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General expression for change in internal energy for 1 component 1 phase system:

$$dU = C_V dT + \left[T \left(\frac{\partial p}{\partial T}\right)_V - p \right] dV.$$

Considering ideal gas, the ideal gas law can be used, in which case the 2nd term on the right becomes zero (you can try it). Then, dU is just a function of temperature ONLY for ideal gas. Thus, in an isothermal process, dU is zero. Remember, because the ideal gas equation is used, the 2nd term becomes zero. For other systems (non-ideal gases or liquids), the equation of states (EOS) that describe their behavior are different!

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  • $\begingroup$ This answer provides the general relationship between energy and temperature in a simple system -- applicable to solids, liquids, and gases, even during phase transitions. The equation provided definitively shows that isothermal processes have constant energy only in special cases (e.g., ideal gases). As such, this answer should be the one marked as correct. $\endgroup$ – tjd Sep 30 '19 at 14:42
  • $\begingroup$ I've edited your answer to use MathJax, which is the generally preferred way to write mathematical expressions here. If you are unfamiliar with it, you can find a tutorial here. $\endgroup$ – J. Murray Sep 30 '19 at 14:46

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