# Isothermal expansion/compression

1. Aug 23, 2010

### hydrogen1

Maybe somebody can explain this to me and I will have a better understand of the Isothermal expansion problem. Here is what I know:

$$\Delta$$U(T)=0
but internal energy is also a state function. So if I expand from P1,V1 to P2,V2 isothermally then how can internal energy not change. Internal energy is a state function and depends on the state which has changed.

2. Aug 23, 2010

### r_tea

So, what I think you're saying is "U should entirely specify the state of my system because it is a state function, hence if I have some state where U=something and another state where U=same thing, then those two states should be the same, but they clearly aren't because they have different P,V".

Well, "objectively" speaking, P,V,U (G,H...there's a lot of them) are all equally useful state functions, and in the case you've described you need to know *two* state functions in order to specify a system (i.e. you have a state-space of dimension two). The typical state functions people choose for this are P,V, so knowing P,V for an ideal gas will allow you to entirely specify all other state functions. Thus, only knowing U isn't enough to entirely specify the state--you need something else. Furthermore, that other state function must be linearly independent of U (i.e., we can't pick T, because in an ideal gas it's just proportional to U). Thus U alone doesn't specify the system, but, for example, U and P do. In the isothermal expansion example you brought up, though U didn't change, P did change so the state is still uniquely specified.

Your state function needs to be one-to-one only if your state space if one-dimensional, ie. that state function entirely specifies the state of the system.

3. Aug 23, 2010

### hydrogen1

Yeah, I understand that one phase of one component means two state functions will fix all others. So what I'm saying is P1,V1 defines state one and P2,V2 defines state two. I feel like U1 and U2 should be different then. However, they are not U2-U1=0. Is this just because the states are different in such a way that U2=U1?

4. Aug 23, 2010

### r_tea

Right, exactly--the states are different in the exact special way that U1=U2 (which is exactly the definition of an isothermal process for an ideal gas).

What I meant to say was that there's no reason to intuitively believe that U1 should be different than U2, because U alone doesn't specify the state of the system. If, for example, after an isothermal expansion (in which the state has clearly changed) (U1,P1)=(U2,P2), or (U1,V1)=(U2,V2), or, using the typical state variables, (P1,V1)=(P2,V2), then we'd have a pretty big problem. But this isn't the case, so we're okay.

The reason why U1=U2 is because that's the definition of an isothermal process for an ideal gas. Recall that for an ideal gas, U=0.5fNkT, i.e. $$U \propto T$$, so if we want $$\Delta T=0$$ that also means $$\Delta U=0$$.

If you had a non-ideal gas (with interactions), then an isothermal process doesn't necessarily mean $$\Delta U=0$$.

Last edited: Aug 23, 2010
5. Aug 24, 2010

### Andrew Mason

In an ideal gas, internal energy is a function of temperature only. So there is no change in internal energy of an ideal gas during an isothermal (constant temperature) process.

If this is an ideal gas and the temperature has not changed, then P1V1 = P2V2. So, although the pressure and volume may have changed, the internal energy (which is determined by the product PV = nRT) has not.

AM