2nd law of thermodynamics violation? What if I assume an ideal gas in a cylinder having frictionless piston and perfectly insulated walls with a perfectly conducting base is kept on a reservoir at temperature $T$ which is greater than the temperature of gas.
The gas inside the cylinder would expand due to heat gained and the piston would rise in proportion to keep the temperature constant.
All I essentially did was describe an isothermal process which is mathematically,
$dU = dQ - dW$
Which in our case would be,
$dQ = dW$
Since the temperature stayed constant and thus the internal energy as it is an ideal gas.
Does it not mean that all of the heat that I'm supplying to the system is being used to do work which is a direct violation of the statements:

All of the heat can never be converted to work.
Or
100% of heat can never be converted to work.
(Which is just one of the variations of second law of thermodynamics I suppose, nevermind if not)

Or is there something else I need to know.
 A: 
The gas inside the cylinder would expand due to heat gained and the
  piston would rise in proportion to keep the temperature constant.

The process you are describing is not an isothermal process because it appears that the external pressure is constant. For an isothermal expansion process the external pressure must be gradually reduced while the volume expands so that the product of the gas pressure and volume is constant throughout. That keeps the temperature constant. Per the ideal gas law
$$Pv=nRT$$
If $T$ is constant throughout the process, then given $n$ and $R$ are constants,  the equation for an isothermal process is $Pv$ = constant.

All I essentially did was describe an isothermal process which is
  mathematically,
=−

That's the first law. Not the equation for an isothermal process, which is $Pv$ = constant.

Does it not mean that all of the heat that I'm supplying to the system
  is being used to do work which is a direct violation of the
  statements:

There is no violation because your statement of the second law is incomplete. All of the heat in a process can be converted to work, but not in a cycle, per the Kelvin-Planck statement of the second law: 
No heat engine can operate in a cycle while transferring heat with a single heat reservoir
The key phrase, missing from your highlighted statement, is "can operate in a cycle". The reversible isothermal expansion process in a  Carnot cycle completely converts heat from the high temperature reservoir into work. That is not a violation because, as @GeorgioP has tried to point out to you, its a process and not a cycle. 
In order to compete the cycle Kelvin-Planck requires that some heat $Q_L$ must be rejected to a lower temperature reservoir. For the Carnot Cycle this occurs in the reversible isothermal compression. That leaves the heat available for work as $W_{net}=Q_{H}-Q_{L}$, where $Q_L$ is the heat rejected to the low temperature reservoir.
An underlying reason for the Kelvin-Planck statement requiring a second low temperature reservoir is that when heat is taken in from the surroundings during the isothermal expansion, there is an increase in the entropy of the system of
$$\Delta S_{system}=+\frac {Q_H}{T_H}$$
In order to complete the cycle, all system properties, including entropy, must return to their original values. Therefore this entropy acquired during the reversible isothermal expansion has to be transferred back to the surroundings in the form of heat $Q_L$. That occurs during the reversible isothermal compression and is 
$$\Delta S_{system}=-\frac{Q_L}{T_L}$$
For the Carnot Cycle
$$\frac{Q_L}{T_L}=\frac {Q_H}{T_H}$$
Therefore the total entropy change for the cycle is.
$$\Delta S_{system}=+\frac{Q_H}{T_H}-\frac {Q_L}{T_L}=0$$
Since the reversible heat transfers also occur isothermally with the surroundings, the change in entropy of the surroundings is also zero.
Hope this helps.
