Consider a two-compartment system enclosed by rigid, adiabatic walls, and let th
ID: 3164226 • Letter: C
Question
Consider a two-compartment system enclosed by rigid, adiabatic walls, and let the two compartments be separated by a rigid, heat-conducting wall. Assume each compartment is in equilibrium but that they are not in equilibrium with each other. a. Show that for a spontaneous process, dU_1 (1/T_1 - 1/T_2) > 0 b. Use this result to discuss the direction of the flow of energy as heat from one temperature to another. c. Modify this argument to consider the case where the two compartments are separated by a non-rigid, insulating wall. Derive the result dS = (P_1/T_1 - P_2/T_2) dV_1. d. Discuss the direction of a volume change under an isothermal pressure difference. e. Now modify this argument to have the compartments separated by a rigid, insulating, permeable wall. Show that dS = dU/T + P/T dV - mu/T dn. Show that dS = (mu_1/T_2 - mu_2/T_1) dn_1. Discuss this result in the context of the direction of a material flow under a chemical potential difference.Explanation / Answer
If an amount of heat dU flows from partition 1 to partition 2, then entropy of 1 decreases by dU/T1 and that of 2 increases by dU/T2. Because entropy of the universe can only increase in any process, it is necessary that the net change of entropy dU/T2 - dU/T1 > 0. =) T1 > T2. so heat flows from 1 to 2 if T1 > T2. On the same lines heat fliesfrom 2 to 1if T2 > T1.
For a non rigid wall under isothermal conditions dQ = dW = PdV. Therefore Delta S= p1dV/T1 - p2dV/T2 considering expansion of compartment1. As entropy of universe should increase p1/T1 should be greater than p2/T2 of compartment 1 should expand.