Change in Entropy Entropy as a State Function Chapter 20 Second Law of Thermodynamics Change in Entropy Entropy as a State Function We have assumed that entropy, like pressure, energy, and temperature, is a property of the state of a system and is independent of how that state is reached. That entropy is indeed a state function can be deduced only by experiment. To make the process reversible, it is done slowly in a series of small steps, with the gas in an equilibrium state at the end of each step. For each small step, the energy transferred as heat to or from the gas is dQ, the work done by the gas is dW and the change in internal energy is dEint. These are related by the first law of thermodynamics in differential form:

Chapter 20 Second Law of Thermodynamics Change in Entropy Because the steps are reversible, with the gas in equilibrium states, we can use dW = pΔV and dEint = nCVdT. Solving for dQ leads to: Using the ideal gas law, we replace p = nRT/V. After some rearrangements, Integrating each term of the above equation between an arbitrary initial state i and an arbitrary final state f,

Chapter 20 Second Law of Thermodynamics Change in Entropy The left term is the entropy change ΔS (= Sf – Si) as the definition. Substituting this and integrating the quantities on the right:

Chapter 20 Second Law of Thermodynamics Checkpoint An ideal gas has temperature T1 at the initial state i shown in the p-V diagram here. The gas has a higher temperature T2 at final states a and b, which it can reach along the paths shown. Is the entropy change along the path to state a larger than, smaller than, or the same as that along the path to state b? path to a < path to b Path to b requires more Q

Chapter 20 Second Law of Thermodynamics Problem Suppose 1.0 mol of nitrogen gas is confined to the left side of the container as shown here. You open the stop-cock, and the volume of he gas doubles. What is the entropy change of the gas for this irreversible process? Treat the gas as ideal.

Chapter 20 Second Law of Thermodynamics Problem The figure shows two identical copper blocks of mass m = 1.5 kg; block L at temperature TiL = 60°C and block R at temperature TiR = 20°C. The blocks are in a thermally insulated box and are separated by an insulating shutter. When we lift the shutter, the blocks eventually come to the equilibrium temperature Tf = 40°C What is the net entropy change of the two block system during this irreversible process? The specific heat of copper is 386 J/kg·K. Irreversible process from initial state to final state Reversible process from initial state to final state Using thermal reservoir, gradual temperature change dT and gradual heat transfer dQ

Chapter 20 Second Law of Thermodynamics Problem Step 1 Step 2

The Second Law of Thermodynamics Chapter 20 Second Law of Thermodynamics The Second Law of Thermodynamics A closed system may consists of a number of subsystems, which may interact between one and another. Between the subsystems, there can be a number of processes. If the processes in a closed system are irreversible, the entropy of the system will always increase. If the processes are reversible, the entropy will remain constant. Closed System Gas Reservoir Surrounding air

The Second Law of Thermodynamics Chapter 20 Second Law of Thermodynamics The Second Law of Thermodynamics As depicted below, isothermal expansion of an ideal gas brings the gas from a to b, from initial state i to final state f. The heat is transferred at constant temperature from the reservoir to the gas. Now, if we want to reverse the direction of the process, the heat must be extracted from the gas and thus its entropy is decreased. However, if we have to include the reservoir also as part of the system, to have a closed system. The entropy of the reservoir will be: The entropy change of the closed system is the sum of these two quantities = 0.

The Second Law of Thermodynamics Chapter 20 Second Law of Thermodynamics The Second Law of Thermodynamics If a process occurs in a closed system, the entropy of the system increases for irreversible process and remains constant for reversible processes. It never decreases. Although entropy may decrease in part of a closed system, there will always be an equal or lager entropy increase in another part of the system. Thus, the entropy of the system as a whole never decreases. This fact is one form of the second law of thermodynamics: Second Law of Thermodynamics In the real world almost all processes are irreversible due to friction, turbulence, and other factors. So, the entropy of real closed systems undergoing real process always increases. Process in which the system’s entropy remains constant are always idealizations.

Class Group Assignments Chapter 20 Second Law of Thermodynamics Class Group Assignments A 10-g ice dice at –10°C is placed in a huge water tank whose temperature is 15°C. Calculate the change in entropy of the ice cube as it becomes water and reach the water tank temperature. (a) 0.8 J/K (b) 2.0 J/K (c) 3.1 J/K (d) 3.8 J/K (e) 15.2 J/K 1 mol of a certain ideal gas with initial temperature of 25°C and pressure of 2 atm is heated at constant pressure until the volume is increased by 200%. Assume reversible process. Given CP = 25.895 + 32.999×10–3T – 30.46×10–7T2 J/mol·K, determine ΔS: (a) 0 (d) 26 J/K (b) –26 J/K (e) 47 J/K (c) –68 J/K A copper stick conducts 7.50 cal/s from a heat source maintained at 240°C to a large body of water at 27°C. Determine the rate at which entropy changes per unit time in a closed system that covers this process. (a) +1.03 J/K·s (d) +0.065 J/K·s (b)+0.043 J/K·s (e) +0.031 J/K·s (c) +0.017 J/K·s

Homework 8 (20-17) (20-18) Deadline: Thursday, 16 June 2016. Chapter 20 Second Law of Thermodynamics Homework 8 (20-17) (20-18) Deadline: Thursday, 16 June 2016. Validation test will be conducted in the beginning of the class.