![]() ![]() ![]() Irreversible processes are more like sudden jerks or a quick change in the system and therefore the $V$. ![]() In this case, our only alternative (aside from solving the complicated gas dynamics equations for the gas, which include the viscous stresses) is the explicitly specify the external force per unit area imposed by the surroundings. But, for an irreversible process, even though, by Newton's 3rd law, the force per unit area exerted by the gas on its surroundings is equal to the force per unit area exerted by the surroundings on the gas, the force per unit area exerted by the gas on its surroundings includes viscous stresses thus, the force can cannot be determined by simply applying the ideal gas law. The adiabatic process can be either reversible or irreversible. So, you are correct in saying that, for a reversible process, the internal pressure is equal to the external pressure. The thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression. In this case the ideal gas law is recovered. Of course, at thermodynamic equilibrium, the rate of deformation of the gas is zero, and the force per unit area reduces to the pressure. The latter depend, not on the amount that the gas has been deformed, but on its rate of deformation. The force per unit area exerted by the gas on the piston is comprised of two parts in an irreversible process: the local pressure and viscous stresses. In an irreversible process, the gas is not at thermodynamic equilibrium, so the ideal gas law will not apply. The ideal gas law (or any other equation of state) can only be applied to a gas at thermodynamic equilibrium. ![]()
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