Pressure changes at equilibrium

[Ernie Lee 3F]

Hello, when we talk about a reversible process, does the Pressure change in the same way we would expect products and reactants to change in an equilibrium situation? Indeed, does the pressure change very minimally and quickly compensates, hence making P not constant? Therefore, we would not be able to use w=-P(delta)V for the work of expansion, rather -nRT ln (v2/v1)?

[Lauren Kim]

Hi, w = -P * delta V is a formula that can be used only for an irreversible expansion of a gas. For a volume change that happens slowly with external pressure remaining the same, you would use the equation for work with the integral. A reversible expansion means that it can be undone with infinitesimal changes. It follows that in a reversible expansion, there can't be too big of a pressure difference between the system and the environment. Because the changes take place in infinitesimal steps, you can't use w = -P * delta V as is; you have to take its integral, which is where -nRT ln (v2/v1) comes from. If it helps, you can remember than in a reversible expansion, the external pressure and the pressure of the gas are the same, so it does not make sense to use w = -P * delta V.

[BeauBrown]

A reversible reaction happens slowly so it takes a different equation to apply. The -P Delta V is for irreversible situations.