Thermodynamic noob questions.

[Anonymous No.11730143]

>In the adiabatic process, since the heat is 0, how can it be possibile that heat capacity times ΔT is 0?
>In the adiabatic process, what's equal to 0? The heat or the change of heat?
I read somewhere that Q is 0, and somewhere that ΔQ is 0. Which is 0?
>If I compress a gas I have a negative Work, so the equation becomes Q = ΔU-W . But shouldn't the heat increase? If the Work is negative, how can the heat increase?

[Anonymous No.11730391]

Negative work = negative heat = heat goes away from the gas elsewhere

[Anonymous No.11732400]

>>11730143
Hi OP, I'm glad you ask these questions, I think most thermodynamics courses aren't very clear with what everything mens, but with a lot of thought it does all make sense. To answer:

1. The heat capacity depends on the path between the initial and final state. Most commonly the heat capacity at constant volume is what is calculated, sometimes also the heat capacity at constant pressure. For an adiabatic process you would be considering the heat capacity at constant entropy and this would in fact be 0. Since it's sort of trivial you don't usually talk about heat capacity at constant entropy.

2. In an adiabatic process, the heat is 0. Heat is a type of transfer of energy from one system to another (the other being work) so Q=0 means there is no transfer of heat energy. If Q=0 and W=0 then the change in internal energy is zero.

3. It doesn't make sense to say the heat increases or decreases after compressing a gas - heat is not a state function so a value of heat cannot be assigned to equilibrium states. Change of heat only makes sense if you mean a change in the rate of heat transfer while the system is not in equilibrium (but this is irrelevant to equilibrium thermodynamics). In this process, you do work W on the gas which increases it's internal energy, and you may or may not transfer heat to the gas as well depending on the manner in which you compress it (e.g. adiabatically, isothermally, etc). But it is the internal energy which will increase or decrease during a process; the work and heat are simply the quantities which determine what this change is.

[Anonymous No.11732410]

>>11732400
And to add, I think the notation where they say Q=0 is clearer, for reasons I stated above. ΔQ=0 is technically wrong but they usually just take it to mean the same thing.

[Anonymous No.11732493]

>>11730143
>If I compress a gas I have a negative Work, so the equation becomes Q = ΔU-W . But shouldn't the heat increase? If the Work is negative, how can the heat increase?

I think you might have a sign error. If you do work on a gas you are increasing its internal energy, and rearranging your equation
>ΔU = Q+W
you see that doing work on a gas is associated with positive W with this convention. As long as you remember that physically doing work on a system or adding heat to a system should be associated to an increase in internal energy you won't ever mix up the signs.

Now if you are wondering why given Q = ΔU-W as W becomes more positive then Q becomes more negative, the answer is that in general ΔU is changing too. If we fix ΔU=0, then of course Q must be negative since if you are adding energy to the system through work, for the internal energy to stay the same then heat must leave the system.