Question

A Joule expansion refers to the expansion of a gas from volume V₁ to volume V₂ against no applied pressure, and is sometimes also called a free expansion. There is no work done, because the P of -PdV is zero. By insulating the system, this process can be done adiabatically, so there is no change in heat. For an ideal gas, the adiabatic process is also isothermal, so there is no change in thermodynamic energy, ∆U = 0 (which is not surprising, since both q and w are 0). Because entropy is a state variable, the change of the system's entropy can be calculated using the most convenient means possible, which would be through a reversible process, and is just given by the usual expression for a reversible isothermal expansion, ∆S = nR ln(V₂/V₁). Thus, even though there is no heat flow, there is certainly an increase in entropy of the system.

Explain how this overall result is consistent with the 2nd Law of Thermodynamics. Is this process reversible or irreversible? Explain your reasoning.

Answer 1

As per the 2nd Law of Thermodynamics, in a cyclic process , the entropy will either increase or will remain the same

In a free expansion there is certainly an increase in volume of the gas and thus , the randomness of the gas molecules increases and so does the entropy.

Similarly, in isothermal/adiabatic expansion, altough ∆E = 0, there is increase in volume of the gas due to work done and thus, the randomness of the gas particles increases and so does the entropy.