1. Mar 17, 2007

### mindboggling

We are currently learning about isothermal processes in class and there's something i don't really get.

i can visualize what happens when the piston is pulled or pushed very slowly, PV = constant and hence produces an isotherm on a PV diagram.

And if since the temperature is constant, there isn't any change in internal energy and Q = W.

However, i can't visualize this process.

To my understanding, if heat is added to the system, internal energy increase, then the system does work and the volume expands. isn't this more like an isobaric process. How can the temperature keep constant, (also given that the walls of the piston are thermally insulted)?

In other words, what i don't get is how does q = w, reduce the temperature to its original when q is heat added to the system. is q converted directly into w? How? really can't visualize.

2. Mar 17, 2007

### G01

In an isothermal process, q is converted directly into work. Think about it. As heat is added, at the same moment, the volume is increasing, and the pressure of the gas is decreasing. The increase in volume corresponds with an amount of work. If you consider all of these effects, heat added, volume expanding, pressure dropping, occurring at once, does it help you visualize how the process is isothermal? If the volume increases exactly at the same time as the heat is added, it must be the case that the heat is converted directly into work, since it has no time to go anywhere else.

3. Mar 17, 2007

### mindboggling

I can see how q = w from the equations by can't quite grasp this conceptually.

doesn't adding heat increase the internal energy? and thus when volume expands it only does so so that the pressure equalizes atmospheric pressure.

i can visualize volume increasing and pressure dropping when someone pulls the piston slowly, but can't visualize how this happens when say heat is added.

4. Mar 17, 2007

### G01

When heat is added, it causes the molecular energy to increase.

Now, this causes the molecules to move faster, and by doing so, they exert a higher pressure, and thus a higher force on the piston. This force then pushes the piston up, until the pressure is not high enough to raise the piston any more.

Now even though the heat went into the gas, and then into the work, the result is the same as if the heat was turned directly into work. Also, this whole process happens, for the most part, instantaneously, so we say the heat was converted directly into work. So, even though the heat did go into the gas, the gas immediately lost the energy by doing work, so the temperature didn't change(why we call it an isothermal process), because the heat went into the gas, but then right back out as work.

If we fixed the piston, so it couldn't move, the temperature would then increase because the gas would not be able to do work, so it would have to "keep" the heat energy.

This help?

5. Mar 17, 2007

### rbj

in the ideal isothermal expansion/compression. there's a big heat sink on the "cylinder" that conducts the heat generated by the compression away. so the gas in the cylinder remains at the same temperature as ambient. when expansion happens, the temp of the gas would drop but for the big heat sink that draws heat from the ambient to the inside.

the internal energy of the gas remains constant since the temperature does, but if the pressure is greater inside, there will be a force that will push the piston out. that does work and would drop the internal energy of the gas except that it gets replaced by the heat conducted inside by the heat sink.

6. Mar 17, 2007

### mindboggling

thanks g01 and rbj for the replies

i grasp what you mean now

one more question: change in internal energy = - work done by system

please explain how this occurs on a molecular level?

7. Mar 17, 2007

### mindboggling

hmm so is this how an isotherm work

say when the gas is heated, the pressure increases and thus the molecules transfers its energy to the wall and work is done till the pressure equalizes the atmospheric pressure.

at this point, since the molecules transfered its energy to the walls , temperature returns to initial temperature. therefore if we do not allow the piston to move anymore, pressure decreases and volume increases thus representing an isothermal process?

but if heat is added to molecules to do work, and when work is done their temperature drops won't the piston keep rising and dropping only?

Last edited: Mar 17, 2007
8. Mar 18, 2007

### rbj

in an adiabatic process, yes, but not an isothermal because it draws energy from the ambient body. an adiabatic process is sorta the opposite: instead of putting a heat sink on the cylinder and piston, the cylinder is perfectly insulated so that any energy done by compressing is left in the gas inside and it heats up proportionately. in expansion, the system does work on the piston moving out and the temp inside drops.

oh, that's kinda hard. you need a college freshman or sophmore level general physics text. you start out with the ideal gas law and a rectangular box with a single monatomic molecule bouncing around off the walls. you can show that the kinetic energy of the molecule is

$$U = \frac{3}{2}P V = \frac{3}{2} n R T = \frac{3}{2} N_A^{-1} R T = \frac{3}{2} k_B T$$

here, n is the number of moles in the gas law. $N_A$ (Avogadro number) is the number of molecules in a mole and because there in one molecule, $N_A^{-1}$ is the number of moles. then, with more of these monatomic molecules added, there is some handwaving to show that the formula is the same (except now $n = j \cdot N_A^{-1}$ where j is the number of molecules). for a diatomic molecule there are more degrees of freedom and the 3/2 becomes 5/2. this needs a textbook with illustrations. i dunno how to effectively explain it without.

Last edited: Mar 18, 2007
9. Mar 18, 2007

### mindboggling

thanks rbj, this stuff seems to be clearer now :)