# Why are isothermal process values higher than adiabatic ones?

• Ana Mido
So, you can use P, V, and T to solve for dP/dT. But how do you know that the dP/dT for the adiabatic process is more negative than the dP/dT for the isothermal process?That's right !f

#### Ana Mido

why are isothermal process values higher than adiabatic process ones?
I know that the volume is powered by gama in adiabatic process ones, and this has an effect.
but how can I explain it ?!
http://www.popsolving.com/Thermodynamics/Problem2.4_Freebody.jpg [Broken]

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why is isothermal process values higher than adiabatic process ones?
What is the defining quality of an isothermal process? What do you need to maintain that quality?

Ana Mido
What is the defining quality of an isothermal process? What do you need to maintain that quality?
I want to know why when I draw this relation, the isothermal is above & adiabatic is below. I know this is because of the power "gama"
Are there any other reasons ?

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You're looking at two expansions, an isothermal, and an adiabatic. What's going on in the isothermal expansion that is not going on in the adiabatic expansion?

Ana Mido
You're looking at two expansions, an isothermal, and an adiabatic. What's going on in the isothermal expansion that is not going on in the adiabatic expansion?
ok, sorry I've put the question in wrong meaning.

ok
That means you see the difference?

That means you see the difference?
OK, I want to know why is the adiabatic curve steeper than the isothermal one?

What do you have to add to the isothermal process to keep it isothermal?

What do you have to add to the isothermal process to keep it isothermal?
I have to make the temperature still constant.

I have to make the temperature still constant.
Yes. Excellent. And how do you do that?

Yes. Excellent. And how do you do that?
I don't know really.
but may be by closing the system or isolating it ? right ?

or isolating it ? right ?
Wrong. The adiabatic system is closed (no exchange of matter) and insulated, exchanging only work with its surroundings. The isothermal system is closed and not insulated, so it can exchange work and what else with its surroundings?

Wrong. The adiabatic system is closed (no exchange of matter) and insulated, exchanging only work with its surroundings. The isothermal system is closed and not insulated, so it can exchange work and what else with its surroundings?
The closed system exchange energy with surroundings & the mass is still constant

exchange energy
What kind of energy?

What kind of energy?
heat or work

"Or?" Are you certain it's only one or the other?

"Or?" Are you certain it's only one or the other?
no, the both

Okay. Now, compare this to the adiabatic process that can only exchange work with the surroundings.

Okay. Now, compare this to the adiabatic process that can only exchange work with the surroundings.
That's right !
In isothermal: Q=W & T is constant , exchange both
In Adiabatic: Q=0 & W=-ΔU , exchange only work
Is that right ?!

Close enough. The isothermal process picks up extra heat from some external reservoir that maintains the temperature of the working fluid, and that can be converted to work.

Ana Mido
You're right !
Then, how can I prove that Adiabatic curve is more steeper than isothermal curve , using mathematics ?

You're looking at dP/dT for the two processes, and you want to show that it is more negative for the adiabatic process. From the original problem statement, it appears you can use the ideal gas equation of state. You know that T is constant for the isothermal process, and that T is a function of P, and V for the adiabatic process.