1. The problem statement, all variables and given/known data Consider an ideal monatomic gas that undergoes an adiabatic free expansion starting from equilibrium state A with volume 500 cm3, pressure 40 kPa and temperature 300K to state B, which has a final equilibrium volume of 1000 cm3. Construct an reversible isothermal path that joins the initial and final states and show it on a schematic p-V diagram. Calculate the total entropy change between the two states using the reversible path. Would you expect this to be the same as the entropy change for the irreversible free expansion? 2. Relevant equations 1) P1*V1 = P2*V2 ΔT for free adiabatic expansion = 0 2) ΔS = nRln(Vf/Vi) 3) Ideal Gas Law pV=nRt 3. The attempt at a solution 1) 40*500 = P2 * 1000 P2 = 20 kPa I've attached a P-V diagram. What I'm failing to understand is how i draw a reversible isothermal path that joins the states. If ΔT in an adiabatic free expansion = 0, and then I'm asked to draw a reversible isothermal path that joins the initial and final states together then wouldn't this be the exact same path because ΔT for isothermal path = 0? 2) P1 = 40,000Pa V1 = 0.0005m3 T = 300 K R (gas constant) = 8.3144 ∴ = PV / RT = n = 0.008018217 ΔS = nRln(Vf/Vi) ΔS = (0.008018217)*(8.3144)* ln(0.0005/0.001) = - 0.046209810 J/K 3) The entropy change in the adiabatic free expansion is the same as in question two, only it is positive because entropy is increasing.