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Topic: Ideal Gasses in Isothermic Processes  (Read 6811 times)

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Offline gingi85

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Ideal Gasses in Isothermic Processes
« on: November 15, 2007, 02:29:08 PM »
I'm somewhat new to physical chemistry. I hope you'll bare with me:

I'm having trouble understanding (mathematically and intuitively) why U, the energy of a system, stays constant in an ideal gas in an isothermic process. In other words, Q = -W.


Offline enahs

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Re: Ideal Gasses in Isothermic Processes
« Reply #1 on: November 15, 2007, 07:18:08 PM »
We define an ideal gas as such that the temperature depends only on the internal energy (ΔU=nRΔT).

An isothermal process is one where the Initial temperature and final temperature are the same, therefor there is no change in temperature. If there is no change in temperature, there can be no change in the internal energy (ΔT becomes 0 is the equation above).

In order for this to occur, any energy in the form of heat accepted by the system must be converted entirely to work performed on the surroundings (or vice versa).

 


Offline Yggdrasil

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Re: Ideal Gasses in Isothermic Processes
« Reply #2 on: November 16, 2007, 07:17:29 PM »
(ΔU=nRΔT).

ΔU=nCvΔT

but everything else is absolutely correct.

Offline gingi85

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Re: Ideal Gasses in Isothermic Processes
« Reply #3 on: November 17, 2007, 02:12:48 PM »
Yggdrasil,

What does that mean? Please explain.

Offline Yggdrasil

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Re: Ideal Gasses in Isothermic Processes
« Reply #4 on: November 17, 2007, 02:21:34 PM »
The formula for the change of the internal energy of an ideal gas is given by:

ΔU=nCvΔT

where n is the number of moles of your ideal gas, Cv is the heat capacity of your gas, and ΔT is the change in temperature.  As enahs mentioned, this formula shows that the change in internal energy is directly proportional to the change in temperature.  If you have no change in temperature (ΔT = 0), then you have no change in internal energy (ΔU = 0).

Offline gingi85

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Re: Ideal Gasses in Isothermic Processes
« Reply #5 on: November 17, 2007, 02:31:06 PM »
Thank you!

That part I understood. I didn't I understand why you corrected enah's responce. The equation he used is the one I'm familiar with.

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