2nd Law Thermodynamics

2nd Law Thermodynamics - BME100L:Topic2 SecondLawof...

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1 BME 100L: Topic 2 Second Law of Thermodynamics: Entropy Always Increase Entropy Always Increases From 1 st law to 2 nd law y E = q + w y If I can somehow reduce the energy of a system by E, I can then extract work. I can also do this continuously through a cyclic process (for each cycle E = 0) y Even if I can reduce heat loss (q) to zero, the work I extract will never exceed what’s reduced in the energy of the system (First law: I can’t win ). y Actually, even this cannot be achieved: the reduced energy can’t be 100% converted into useful work in a cyclic process – there is always gonna be some heat loss (Second law: I can’t break even).
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2 From 1 st law to 2 nd law y First Law: Energy can change from one form to another as a result of changes in system states y Second Law: y Will a system change states? OR y Under what conditions will the system change states? T hot Consider an engine trying to extract engine w q total q cold work by operating between a hot reservoir and a cold reservoir. Two important points: 1) In order to do any work through a cycle, we must have T hot > T cold. T cold 2) w < q total; But, what’s the maximum work that can be extracted?
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3 Step I, isothermal expansion : 22 hot 1 1h o t ln 0 VV dw PdV nRT V w PdV dV nRT =− = < ∫∫ 11 2 For ideal gas, Δ E = 0 along an isotherm. Thus, according to 1 st law: 2 111 1 1 h o t 1 0l n 0 V Eqw q wn R T V = Δ=+ ⇒= −= > The system acquires heat and converts it to an equal amount of work during the first step. Step II, adiabatic expansion : 2 c o l d h o t 0; () V q wE qE n C TT = = Thus, the system does work without acquiring any heat, resulting in a decrease in its temperature.
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4 Step III, isothermal compression : 4 3 3 3c o l d 4 ln 0 V V V wP d V n R T V =− = > 33 0 qw < The system will release heat while receiving equal amount of work from the surroundings (cold reservoir). Step IV, adiabatic compression : 0; 4 4 4 4 4 hot cold () V q wE qE n C T T = = The system recovers its temperature while acquiring work from surroundings 4 3 1 hot cold hot cold hot cold 1 24 ln ( ) ln ( ) VV i i V V w w nRT nC T T nRT nC T T = == + + + Total extracted work: 3 1 hot cold ln ln 0 (consistent with 1st law) V V nRT nRT E =+ Δ= 1. Overall, a Carnot cycle clearly obeys the 1 st law. The surroundings (hot reservoir and cold reservoir) supplies the system with heat (q), which is converted into equal amount of work (w).
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5 2. However, there’s something permanent that has happened. Consider the amount of heat extracted from the hot reservoir (from step 1): 2 1h o t 1 ln V qn R T V = Total heat extracted from hot reservoir: Not all of this is converted to w. Instead, there’s a fraction that goes into the cold reservoir: V 4 3c o l d 3 ln R T V = Total heat “wasted” on the cold reservoir: Carnot’s discovery: We know the sum of q over the process is not 0 Consider the sum of q/T over the process.
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This note was uploaded on 04/21/2011 for the course BME 100 taught by Professor Yuan during the Spring '07 term at Duke.

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2nd Law Thermodynamics - BME100L:Topic2 SecondLawof...

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