Note that the two terms on the right hand side of the equation are exact differentials. So, all one needs to do is to integrate this equation from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state determined obtained with the irreversible process. This is the same thing as a reversible path. Regarding the original question of why the equation for dS in terms of dT and dV for an ideal gas gives the correct answer, it is because the equation applies to the changes in dS, dT, and dV over a continuous sequence of thermodynamic equilibrium states.

I don't know why it isn't taught properly to students. You then calculate the integral of dq/T for the reversible process path. Therefore, even if the irreversible process path is adiabatic, the reversible path does not have to be. The reversible process path does not necessarily have to bear any resemblance to the actual irreversible process path. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states: S S f S i. You then need to identify a reversible process path between the exact same pair of equilibrium states. In calculating the change in entropy of a closed system that has undergone an irreversible process, you need to first focus exclusively on the initial and final equilibrium states. If the divider between the filled and unfilled halves is suddenly removed, that would be a case of adiabatic, isoergic expansion. Think of gas confined to one half of a cylinder, and the other half of the cylinder is empty (i.e. $dS = 0 = racdV > 0$, i.e., the entropy of the gas increases, which makes sense because the process was irreversible. But keep in mind that your equation for $dS$ is only true for ideal gases. Now from S Q / T, the entropy change of the ice is S 16.8 kJ 273 K 61.5 J/K when it melts to water at 0 C. In this reversible process, the temperature of the ice-water mixture is fixed at 0 C or 273 K. In this case you can apply both of your equations. Solution The ice is melted by the addition of heat: Q m L f 50 g × 335 J/g 16.8 kJ. Without heat flow, the only way the entropy of the universe can stay constant is if both the entropy of the universe and of the gas stay constant. If the expansion is done reversibly, then we know entropy of the universe can't be changing (because reversible), and that heat flow from the gas to the surroundings is zero (because adiabatic). The question doesn't have sufficient information for a solution, because we don't know if the adiabatic expansion is reversible or irreversible. 'One mole of ideal gas initially at a pressure of 1 atmosphere and T = 298 K, is expanded into a volume 50% larger adiabatically.'

Entropy change equation