I read that internal energy is a function of temperature only and not pressure. But say, we compress a volume of gas. Won't the particles start moving more quickly? Meaning an increase in their internal energy.
Is this an increase in the internal energy due to pressure?
Or an increase in temperature due to pressure which in turn increases the energy?
How do I see this?
Best Answer
For an ideal gas the internal energy only depends on the temperature of the gas. How the temperature relates to pressure is easily seen in the ideal gas law $$PV=NkT$$ So I suppose one could make the argument that the internal energy for the ideal gas depends on the quantity $$\frac{PV}{Nk}$$ and it's up to you how you want to explain the dependency. The problem with this though is that, depending on the process, these variables are constrained to evolve in certain ways.
Indeed, for your compression example, how you are compressing the gas matters. I will cover some typical examples:
If the gas is compressed in such a way so that its pressure is inversely proportional to it's volume, then by the ideal gas law the temperature remains constant. It turns out that in this case the heat that leaves the gas is exactly balanced by the work you do on it. The internal energy does not change.
If the gas is compressed in such a way so that its pressure remains constant, then by the ideal gas law the temperature drops in proportion to the volume. In this case more energy leaves the system as heat than what you put in as work. The internal energy decreases.
If the gas is compressed in such a way so that no heat enters or leaves the system, then all the work you do goes into increasing the internal energy, and hence the temperature of the gas. This process is probably what you had in mind. Here the pressure and volume both change, but the pressure increases in a larger proportion than the volume decreases$^*$. Therefore, by the ideal gas law the temperature increases.
$^*$ In fact, for this process the value $PV^\gamma$ is constant where $\gamma>1$. So, if $V$ decreases by a factor $x>1$, then $P$ must increase by a factor of $x^{\gamma}$. This means that $PV$, and hence $T$, must increase by a factor of $x^{\gamma-1}>1$.