Cp and Cv for an Ideal Gas Two different ways to change dT=T2 -T1 : Process a: Constant V a a b b a dQa nC v dT Cp and Cv for an Ideal Gas For the same dT : Process b: Constant P a a b b b dQb nC P dT Which is bigger: Qa or Q b , Cv or C P ? Cp and Cv for an Ideal Gas First let consider the constant volume process (a): No work done by/on gas W = 0 Then, 1st law gives, dUa dQ a nC v dT Now, for the constant pressure process (b) with the same dT: We have dW = PdV and dQb = nCpdT (by definition) Substitute them into the 1st law again gives, dUb dQ b dW nC p dT PdV Cp and Cv for an Ideal Gas Taking the differential on both sides of the Ideal Gas Law, we have, d( PV ) nRdT nR is constant with n fixed With P constant, this gives, d( PV ) PdV nRdT Substitute this into the last Eq on the previous page, we have: dUb nC p dT nRdT nC(p RdT ) Cp and Cv for an Ideal Gas Important point: Since U for an ideal gas is a function of T only and process a and process b have the same dT = T2 –T1, dUa dU b (Note: This fact is very useful. Basically, one can use dU = nCvdT to calculate the internal energy change in an ideal gas for any given dT whether V is constant or not). nCv dT n( C p R ) dT Finally, this gives, Cp C v R (True for all Ideal Gases) The Ratio of Heat Capacities A useful ratio g can be defined: C g p Cv 3 Recall that for a monoatomic Ideal Gas, C R v 2 5 5 so that, C C R R and g 1.67 p v 2 3 5 For a diatomic Ideal Gas, Cv R 2 7 7 C C R R and g 1.40 and, p v 2 5 Adiabatic Processes for an Ideal Gas 1 Isothermal P T 0 V 1 Adiabatic P Q 0 V g g 1 Start from definition, dQ=0 for any adiabatic process, Then, from 1st Law, we have, dU dW PdV adiabatic expansion T drops Adiabatic Processes for an Ideal Gas Now, we use the “trick” that for an ideal gas, dU is the same for all processes with the same dT = Tf –Ti. As stated previously, we can calculate dU using, dU nCv dT Then 1st law dU dW gives, nRT nC dT PdV dV v V (In the last step, we used Ideal Gas Law: PV=nRT) Adiabatic Processes for an Ideal Gas Rearrange terms, we have, dT R dV 0 T CVv Using the relations for the molar specific heats, R C C C p v p 1g 1 Cv C v C v Then, we have, dT dV (g 1) 0 T V Adiabatic Processes for an Ideal Gas Integrating this equation, we have, dT dV (g 1) constant T V lnT (g 1)ln V constant lnTV g 1 constant TV g 1 constant Note: T has to be in K. PV Using the Ideal Gas Law again, we can replace T with , nR PV V g 1 constant PV g constant nR (alternate form) Work in an Adiabatic Process (Ideal Gas) We know that, dW dU( dQ 0) Using the same “trick” on dU, we can calculate the work done in an adiabatic process if we know the changes in temperature. dW dU nCv dT W nCv ( T2 T 1 ) C or Wv PV PV R 2 2 1 1 Examples Fire Piston History Fire Piston (demo) http://en.wikipedia.org/wiki/Fire_piston Example 19.68 (Comparison of processes) Fire piston calculations http://complex.gmu.edu/www-phys/phys262/soln/fire_piston.pdf Example 19.68 calculations http://complex.gmu.edu/www-phys/phys262/soln/ex19.66.pdf Given initial state P1, V 1 final V2 3 diff ways: a) Isothermal isotherms b) Adiabatic c) Isobaric Adiabatic Expansion (reversible & nonreversible) insulation Reversible adiabatic expansion (quasi-static) : Expanding gas push piston up work is done by gas W > 0 U < 0 (energy flows out of gas) For an Ideal Gas, U is a function of T only, So, U < 0 also implies T < 0 (temperature drops!) Adiabatic free expansion (non-quasi-static /nonreversible): Gas expands into vacuum no work done W=0 Adiabatic Q = 0 1st law gives U = 0 U remains unchanged and T is a constant! Physics 262/266 George Mason University Prof. Paul So Chapter 20: The 2nd Law of Thermodynamics Preferential Direction in Thermodynamic Processes Heat Engine and Efficiency The 2nd Law of Thermodynamics The Carnot Cycle (the most efficient heat engine) Entropy Entropy and Disorder Preferred Direction of Natural Processes Example 1: Q cold hot Two objects are in thermal contact and heat flows from the cold object to the hot object. Again, energy is conserved (1st Law is NOT violated): Qhot( absorbed ) Q cold ( release ) 0 BUT, we don’t observe this process in nature while the reverse occurs! Preferred Direction of Natural Processes Processes not observed in nature: Example 2: Ball absorbing heat energy Then, converts it into mechanical from surrounding energy and starts to bounce and roll Note: energy is conserved (1st Law is NOT violated): heat mechanical eng. BUT, we don’t observe this process in nature while the reverse occurs! Disorder and Thermodynamic Processes The 2nd Law of Thermodynamics is the physical principle which will delineate the preferred direction of natural processes. We will later see that The preferred The degree of direction of randomness (disorder) of natural processes the resulting state All natural processes in isolation will tend toward a state with a larger degree of disorder ! The 2nd Law of Thermodynamics Historically, there are more than one but equivalent way to state the 2nd Law: To address the condition in example #1, here is the Clausius Statement on the 2nd Law: Q cold hot “It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body.” The 2nd Law of Thermodynamics There is also the Kelvin-Planck’s Statement: “It is impossible for any system to undergo a cyclic process in which it absorbs heat from a reservoir at a given temperature and converts the heat completely into mechanical work.” This implies that all heat engines have limited efficiency ! (efficiency of real mechanical engines ~ 15 to 40%) To understand this form of the 2nd Law, we need to look at a toy model: heat engine Heat Engines Definition: A device that converts a given amount of heat into mechanical energy. All heat engines carry some working substance thru a cyclic process: Engine absorbs heat from hot reservoir at TH Mechanical work is done by engine Engine releases residual heat to cold reservoir at TC Dstering Work Done by an Heat Engine The heat engine works in a cyclic process, U 0 1st Law gives, UQnet W 0 Qnet W where, Qnet QQQ H C H Q C explicit signs for heats.