On a piston cylinder system, when I transfer heat to it, can it be an isobaric and isothermal process? Consider it is not a phase change process, because it is possible there.
Don't rely your answer on ideal gas equation, since it is very fast and easy to know the answer to my question using this equation.
What happens in particles that they can't have constant pressure and constant temperature troughout the process? How do they react?
Some people (myself included) regard an isobaric process as one in which the external pressure $P_{ext}$ is held constant throughout the change until the final volume is attained, even if, during the gas deformation, $P_{ext}$ does not match the initial- or final thermodynamic equilibrium pressure of the gas. This is an irreversible change. This is just a definition that is sometimes used for a constant pressure process.
Some people (myself included) regard an isothermal process as one in which the cylinder is in contact with a constant temperature reservoir throughout the process, so that the gas temperature at the boundary is constant at $T_{res}$ during the change until the final volume is attained, even if, during the gas deformation, $T_{res}$ does not match the initial- or final thermodynamic equilibrium temperature of the gas. This is an irreversible change. This is just a definition that is sometimes used for a constant temperature process.
So, based on these definitions, for irreversible changes like these, it is possible to have an isothermal and an isobaric process at the same time. In these processes, however, the gas temperature and pressure vary with spatial position within the cylinder, and so they are not even spatially constant.
For a reversible process, since the gas must satisfy the equation of state at all times during the process, if the volume is changing, the temperature and pressure cannot both be constant.
So you want constant $P,T$ while $V$ presumably changes due to the moving piston. But if $V$ is increasing (and if external $P\neq0$, i.e., not a so-called "throttling process"), then the gas is doing work on its surroundings. So if the process is adiabatic, then that work energy has to come from within the gas itself, whereby $P,T$ has to change (a la first law) correspondingly. Otherwise, if you want the gas's $P,T$ constant while it's expanding, you'd have to supply outside energy to make up for the energy lost to the piston head, and the process won't be adiabatic.
edit Re your "what happens in particles?" part of the question, just consider a gas molecule colliding with the piston. If the piston's not moving, then, ideally, the collision's elastic, and macroscopic $P,V,T$ aren't affected. But in the non-equilibrium situation where the piston's expanding, some momentum's transferred from the molecule to the piston, resulting in corresponding changes to $P,V,T$.
On a PV graph, an isothermal process is hyperbolic curve because $PV=constant$. On a TS graph, it is horizontal line having constant temperature and entropy is given by, $S=nR\ln V$ with volume.
While an isobaric process on a PV graph, pressure is constant as shown by horizontal line while volume is changing. On a TS graph, an isobaric process is shown as both temperature and entropy increasing while, temperature increasing exponentially, $S=nR\ln T$. In an isothermal process, internal energy, temperature and enthalpy remain constant. While in an isobaric process, all of the above not remain constant, $dE=dH-pdV$ and $Q=dH=H_2-H_1$.
So considering above things, an isothermal and isobaric process are not happen at same, not any of thermodynamic process. Yes some of properties can be same because they are state function but one of any two process has to go through other process to reach there. An isothermal process can be equated with sum of an isobaric and an isochoric process on PV graph, or an isobaric can be equated with sum of an isothermal and an isochoric process on TS graph.
