No, not necessarily. The definition you gave for a quasi-static process is correct, and it clearly states "close to equilibrium", meaning that all three $P, V, T$ parameters should undergo infinitesimal changes, for the process to be considered quasi-static.
Whereas in either isobaric, isochoric and isothermal processes, each time only one of the 3 quantities are held constant, and neither of them will dictate whether the remaining quantities should evolve infinitesimally.
Of course it becomes very simple to understand when you just consider a set of examples where iso-processes occur far from equilibrium:
In Otto cycle: for example after the adiabatic compression, the follow-up combustion is considered isochoric wherein the pressure P shoots up, cannot be quasi-static.
Stirling engine: cycle: an isothermal expansion followed by an isochoric heat-removal, then an isothermal compression followed by an isochoric heat-addition, and all these transformations that the gas undergoes are non-quasistatic, meaning the energy changes of the system either due to heat exchange or work, do not take place in infinitesimal amounts.
EDIT: Further explanations:
Quasi-static processes:
- Along a quasi-static path all intermediate states are equilibrium states;
- if a system progressing along a quasi-static path is “isolated” from its environment, then the values of all properties will remain constant and equal to those just before the isolation.
- Quasi-static processes occur at finite rates but not so rapidly that the system is not able to adjust on a molecular level, in order to reach equilibrium. (perturbation relaxation times $>$ system's relaxation time to reach equilibrium, necessary condition for quasi-static) There would not be in general gradients of any intensive properties such as pressure, density, temperature, etc.
Yes quite! The ideal gas law (and in general any state equation) holds only on equillibrium while the 1st law (and all the rest) hold in general. So, your anslysis mr James Hoyland is inacurrate.
Mr Steven, the post includes the word "expansion" so volume changes by assumption.
And mr or ms PhysC, the first two cases are correct. About the third, expansion would occur by pulling the piston thus removing energy from the system and causing decrease on temperature
Best Answer
The main point: Don't equate heat with temperature. There are more types of energy than thermal and more types of energy transfer than heat.
Isothermal: A process that happens at constant temperature.
Adiabatic: A process that happens without heat transfer help from the surroundings.
If you pop a champagne bottle, then the gas outflow is so fast that it practically has no time to exchange heat with the surrounding air. It just cools down rapidly, and that's why you see a white mist which is small ice crystals forming. This is an adiabatic but non-isothermal process.
In general, remember that heat does not imply temperature. And vice versa. Heat could easily be converted into something else than thermal energy, and thermal energy can be gained in other ways than from heat (for example from work).