Does internal energy return to zero in a closed cycle on a P-V diagram?

[jaidon]

I have a question concerning internal energy and the First Law of Thermodynamics. If you plotted P vs V and the shape of the graph is an enclosed shape ie) square or triangle, would the total internal energy in one cycle be zero? (one cycle is for example-start at top left of square and finish at top left of square). I think this is right, but am a little uncertain being a Physics moron and all. I would appreciate anyone's input. Thanks

 

[jaidon]

If anyone has any idea on this I would love to hear from you. This thought is what I have based several questions on, and if I am incorrect I need to figure out something else. Thanks.

 

[Clausius2]

Internal Energy is a variable of state. Given some thermodynamic trajectory, for instance from 1 to 2, the change of internal energy is given by:

$$\int_1^2 du=u_2-u_1$$

Due to the fact in ideal gases and perfect liquids the internal energy is only a function of the temperature:

$$u=c_vT$$

then if 1 is the same thermodynamic state than 2 the change of internal energy is zero, no matter which trajectory it was.