# Help with internal energy please!

1. Oct 24, 2006

### student85

ok so DELTA U = Q - W

Where U is internal energy, Q is heat transferred from or to the system and W is the work done BY the gas right?

I always mix up U and Q, the concepts. My teacher said U depends entirely on temperature, is this right??
Also, if Q =ncDELTA T, isnt Q also dependent on a change of temperature?

We saw that for an isotermic process, Q = W......WHY!?!?! If it is isotermic, isnt delta T = 0 , so that Q = is 0? Instead my teacher wrote that for an isotermic process, delta U is zero?

Can someone explain this?
Thanks a lot for reading and helping.

2. Oct 25, 2006

### quasar987

U is the internal energy of a gaz. We know what this is; it's just the sum of kinetic + potential energy of its constituents. Right?

W is the work done by the system. If work is done by the system, we know that its internal energy diminishes. You also agree with that? This is only classical mechanics so far. (note that if work is done on the system, W is negative)

Now, we recognize that the internal energy of a system can change even when no work is done by (or on) the system. We call this process "heat exchange" and we define the heat absorbed by the system Q precisely by that amount by which the energy has changed not as a result of work being done by the system:

$$Q \equiv \Delta U+W$$

(If the system cannot interact with its environement, we say that the system is thermally isolated, and in this case, the energy of the system is conserved and by newton's laws, whatever work the system does on its environement it looses in energy: $\Delta E = -W$, such that Q=0 in this situation.)

It is true that for an ideal gaz, the internal energy of a system is a function of T only.

$Q=nc_V\Delta T$ is valid only for an ideal gaz undergoing a quasi-static process during which its volume is held constant. If all these conditions are met, then the equation is indeed saying that if the temperature of an ideal gaz varies by some amount, then the heat absorbed by the gaz as a result of this interaction is proportional to the variation of T.

An isothermic process is a process in which the temperature is constant. If the temperature is constant, and U depends only on T, then during that process, $\Delta U=0$. Then what does the first law says about Q?

$$Q=0+W=W$$

The equation $Q=nc_V\Delta T$ is only valid when the volume of the system is held constant. In general, it is not, and we can't use that.

Last edited: Oct 25, 2006
3. Oct 26, 2006

### student85

ooh.
I love you.

Ok I just appreciate ur explanation a lot. THANKS!

4. Oct 26, 2006

### student85

oh oh, one more thing...
So if U depends on temperature, what does Q depend on. When you add heat (Q) to a gas, u r not necessarily raising its temperature right?

5. Oct 26, 2006

### quasar987

Do you see the equations in my post? Because I don't. Here's the code again...

6. Oct 26, 2006

### quasar987

Not only is the energy a function of temperature only (for an ideal gas), but this relation is invertible, so that we can say that the temperature of an ideal gas is a function of its internal energy only: T=T(U).

Now, when you add heat to a gas, you are raising its internal energy by an amount $\Delta U= Q$, thus raising its temperature by an amount $\Delta T = T(U+\Delta U)-T(U)$.

In the above, I have assumed for simplicity that no work is done by the system at the same time as the heat is transfered, so that we could see exactly how the temperature responds to a heat dump on the system. But suppose that work is done on the system at the same time as we transfer heat to it. In fact, suppose that we let the gas's volume expand in just a way such that a quantity of work W=Q is done by the system as a result of the interaction. In that case, the first law writes

$\Delta U=Q-W=Q-Q=0 \Rightarrow \Delta T =0$

You've already encountered such a procees in which heat is transfered but no variation in temperature ensues. It is what we call an isothermic process. (Though I think the term isothermic process can only be coined when the process is quasi-static, such that the temperature is well defined at all time and constant throughout the process)

Last edited: Oct 26, 2006
7. Oct 26, 2006

### student85

Ok thanks a lot man. It all makes sense now...when you add heat to a gas you are raising its internal energy, but if the gas does work, it will decrease the amount it raises...perfect perfect.
What you said about the sum of all Kinetic and potencial energy is very interesting... so u can also calculte internal energy if you know the mass of the gas, it's potencial energy and what about the speed? You would have to assume all particles move at a same speed I suppose. And I believe speed is very related with pressure so maybe you could use the equation PV=nrT or something?

BTW, I didnt see the equations either, but by clicking in the error notice a window pops up and they show.

8. Oct 26, 2006

### quasar987

The energy of an isolated (no external force is acting on it) system of N particles (such as a gas) is by definition, the sum of all the kinetic energies and the potential energy of the system:

$$U=\sum_{i=1}^N\frac{1}{2}m_iv_i^2+V(\vec{r}_1,...,\vec{r}_N)$$

But it's absolutely impossible to know at a given time the exact velocity and position of all the particles (especially if the system is a gas in which case 1) the particles are moving extremely rapidly, 2) they are extremely small, and 3) their number is humongous: N~10^24 (!))

So this way of talking about energy is not really interesting from an experimental point of vue.

Last edited: Oct 26, 2006