# Reversibility of ideal gas processes (isobaric and adiabatic processes)

In an isobaric process, how is it possible to be reversible. In the case that volume is decreased, lets say $2m^3$ to $1m^3$ (at infinitesimally small increments) at a constant $1$ MPa of pressure.

Heat must be extracted such that pressure does not increase in the system. There is a net transfer of heat, so why is it that this process is reversible?

Also in an adiabatic process, with a compression of $2m^3$ to $1m^3$ at a varying pressure. From the idea of adiabatic processes there is no transfer of heat.

So why can an irreversible adiabatic process exist(in an ideal system)?

• "Heat must be extracted such that pressure does not increase in the system." what if the temperature just increased – pentane May 13 '18 at 7:40
• Who says that an isobaric reversible adiabatic compression of an ideal gas is possible? Let me guss. You are trying to determine the entropy change for an adiabatic irreversible compression of an ideal gas that took place at constant externally applied pressure, correct? – Chet Miller May 13 '18 at 12:06

## 1 Answer

"There is a net transfer of heat, so why is it that this process is reversible?"

I assume that you're worried that the transfer of heat from the system in the isobaric case requires the system to be hotter than the surroundings, so heat travels down a temperature gradient, an irreversible process (involving an entropy gain).

The answer is simply that we imagine the process to be so slow that heat will still leave the system even if its temperature is only infinitesimally above the temperature of its surroundings, so the heat flow is as close as we like to being reversible.

I haven't understood what you are concerned about in the case of the ideal adiabatic change. Perhaps you could explain.