Is there any process adiabatic and isothermal?

1. Jan 24, 2010

1. The problem statement, all variables and given/known data

Is it possible to have a process that is both adiabatic and isothermal.

2. Relevant equations

3. The attempt at a solution

I think the answer is no. Because when a system is doing an adiabatic process, the total energy of this system must be changed. For an adiabatic process, there is no heat transfer to or from the system. So the temperature of the system must be changerd.

But, an adiabatic process only means that there is no heat transfer to or from out side. So when the out side environment is doing work on the system and the system transfer whole work out of itself. the temperature of the system will be same. but i am not sure is this ture.

could somebody explane this for me?

Thanks.

2. Jan 24, 2010

kuruman

If a process is both adiabatic and isothermal, what form does the first law of thermodynamics take?

3. Jan 24, 2010

Mapes

Why?

4. Jan 25, 2010

thanks. I am not sure that I understand completely what u said. But I think u mean that there is impossible a process is both adiabatic and isothermal. That is because when outside environment doing work to the system, the internal energy must be changed. And work can be change to heat 100% but heat can not change to work 100%. Right?

but, if there is a cylinder (cacuum, use a spring to replace gas ), when the work is done by outside. The energy of spring must be changed. and, in a ideal environment, this spring can do 100% work to outside when release. In this process, there is no heat, only work. So is this an adiabatic and isothermal process?

I am confused. pls give me some advice. thanks.

5. Jan 25, 2010

thanks.
when a system is doing an adiabatic process, there is no heat can trasfer from or to outside. so the total energy of this system must be changed.

6. Jan 25, 2010

Mapes

What about letting two gases mix, or a gas expand into an empty chamber? These processes can be made adiabatic and have no associated energy change.

7. Jan 25, 2010

D H

Staff Emeritus

Wrong. What is the first law of thermodynamics?

8. Jan 25, 2010

9. Jan 25, 2010

so? still confuesd...

10. Jan 25, 2010

kuruman

Remember that the change in internal energy ΔU is proportional to the change in temperature ΔT. What is the change in temperature, what is the change in internal energy?

11. Jan 25, 2010

D H

Staff Emeritus
Correct.

Correct.

Whoa! How do you draw that conclusion?

Hint: delta T=0 => delta U=0.

What does this say about work?

12. Jan 26, 2010

delta U=Q+W, and Q=0 => delta U=W

isothermal process

delta U=Q+W, and delta U=0, => Q=-W

combine these two factors, so conclusion is impossible a process adiabatic and isothermal. Right?

13. Jan 26, 2010

Mapes

$\Delta U=0$ only implies $\Delta T=0$ for an ideal gas. Also, processes exist where the work and heat transfer are both zero, as I described above. The processes are both adiabatic and isothermal.

14. Jan 26, 2010

D H

Staff Emeritus
Wrong!

There is one number that is equal to its additive inverse. What is it? (i.e., solve $W=-W$ for $W$.)

Last edited: Jan 26, 2010
15. Jan 26, 2010

sorry, i dont know what u meant

delta U=Q+W, and Q=0 => delta U=W
isothermal process
delta U=0
so if a process is adiabatic and isothermal, when the work done by system or at system, it is impossible to satisfy U=W and U=0

16. Jan 26, 2010

D H

Staff Emeritus
Who said work was being done on the system? Your original question was
The answer to that question is definitely "yes".

17. Jan 26, 2010

quasar_4

Mapes was right.

In a free expansion problem, no heat is exchanged nor is any work done. Thus, dU = 0 for the system ...and so is dT... but it is a real system, and it does create new entropy. So, there is at least one example of such a process.

18. Jan 27, 2010

yes! got it

thanks very much to all

19. Jan 27, 2010

athithi

check this...