Change in internal energy is 0 in isothermal process 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.
 A: 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.
A: 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!
A: 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.    
A: 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. 
