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?
Best Answer
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.