Is this against Kelvin-Planck statement?

[pitbull]

Homework Statement

An ideal gas expands isothermally in contact with a heat source. ∆U is zero in this case because it is an ideal gas and T=constant. Is this against Kelvin-Planck statement?

Homework Equations

pv=nRT
dW=-pdV
Kelvin-Planck statement: There is no process whose only result is the full transformation of heat into work.

The Attempt at a Solution

My guess is, ∆U=Q+W=0, therefore
Q=integral(pdv)=nRTlog(Vfinal/Vinitial)>0
W=-nRTlog(Vfinal/Vinitial)
So all the heat that the gas absorbs turns into work. Therefore, this is against K-P statement, but I have a feeling that I am wrong. Can you guys help me?

 

[TSny]

In your wording of the Kelvin-Planck statement you have the phrase "whose only result". This needs to be clarified. Look at a couple of more carefully worded versions:

https://en.wikipedia.org/wiki/Kelvin–Planck_statement

http://www.phy.duke.edu/~rgb/Class/phy51/phy51/node63.html

Specifically, any device used in the process must return to its initial state.