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# Internal Energy of an Ideal Gas

We will show that the internal energy of an ideal gas is a function of temperature only. This makes physical sense because there is an assumption in ideal gas behavior that there is no interaction between the molecules when we write

 (11-1)

Consider two processes: one occurring at constant volume, the other occurring at constant pressure.

For process 1: ; This can be integrated because is the only thing that is changing on the righthandside ( is assumed to be independent of and ).

For process 2: ; is constant (i.e., not a function of or ) so it can be integrated directly. Using the ideal gas law:

 (11-2)

So for process 2,

 (11-3)

Since we can make up any quasi-static curve with segments of processes and processes

Evidently, the sum of any such processes is a function only of . Therefore, for an ideal gas

 (11-4)

Comparing the two equations:

 (11-5)

for the constant volume process, and

 (11-6)

for the constant pressure process:

 (11-7)

 (11-8)

 (11-9)

 (11-10)

or

 (11-11)

Next: A New Thermodynamic State Up: Lecture_11_web Previous: Lecture_11_web
W. Craig Carter 2002-09-28