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Ch19_2_young_freedman.pdf

- StudyBlue
- Virginia
- George Mason University
- Physics
- Physics 262
- Paul So
- Ch19_2_young_freedman.pdf

John R.

File Size:
19
Example: A Thermodynamical Cycle Example 19.4 Given: Q ab = 150 J Q bd = 600 J Find: U =? note ab = ? U abd = ? Q acd = ? Special Processes 3. Adiabatic (no heat exchange, Q = 0): 0Q UQW UW Note: The compression stroke in an internal bi iicombustion eng ne s quick and it can be well approximated by an adiabatic process . (will come back to this example later) Adiabatic Processes: 2 examples Quasi-static adiabatic expansion: 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, S U 0l i li T 0 o, < 0 also mplies < (temperature drops!) Adiabatic free e pansion (non qasistatic): insulation D fire piston x (non-quasi- Gas expands into vacuum no work done W=0 Adiabatic Q = 0 1 st law gives U =0 gives = 0 U remains unchanged and T is a constant! Special Processes 4. Isochoric (constant V): NO volume change NO work (W=0) change 0W UQW UQ 5. Isobaric (constant P) ()WPVV Since P is a constant, dW = PdV can be integrated easily and gives, 21 Special Processes Isothermal (constant T) For an Ideal Gas, U(T) depends only on T U = 0 ! 0U UQW QW Recall work done by an isothermal process in an Ideal Gas:, work in an Ideal 22 2 ln VV VnRT W PdV dV nRT VV 11 1 VV Summary 1. Adiabatic (Q=0) 2. Isochroic (V=0) 3. Isobaric (P=0) 4. Isothermal (T=0) 1 st Law for infinitesimal changes: dU dQ dW dQ PdV C p and C v for an Ideal Gas Two different ways to change dT=T 2 -T 1 : Process a: Constant V av dQ nC dT a a b b a C p and C v for an Ideal Gas Process b: Constant P For the same dT : bP dQ nC dT a a b b b Which is bigger, dQ a or dQ b ? C p and C v for an Ideal Gas First let consider the constant volume process (a): No work done by/on gas W = 0 Then, 1 st law gives, dU dQ nC dT aav Now, for the other constant pressure process (b) with the same dT: We have dW = PdV and dQ b = nC p dT (by definition) Substitute these into the 1 st law gives, bb p dU dQ dW nC dT PdV C p and C v for an Ideal Gas With P constant, we can consider the following differential using the Ideal Gas Law: ()d PV PdV nRdT Substitute this into the previous equation, we have: b p dU nC dT nRdT () p nC RdT C p and C v 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 = T 2 –T 1 , ab dU dU (Note: This fact is very useful. Basically, one can use dU = nC v dT to calculate the internal energy change for a given dT whether V is constant or not). () vp nC dT n C R dT Finally, this gives, pv CCR (True for all Ideal Gases) The Ratio of Heat Capacities A useful ratio can be defined: p C v C Recall that for a monoatomic Ideal Gas, 3 2 v CR so that, 55 1.67 23 pv CCR Rand For a diatomic Ideal Gas, 5 2 v CR 77 1.40 25 pv C C R R and and, Questions Some end of chapter questions Adiabatic Processes (more) Isothermal 1 P V 1 Adiabatic process P V By definition, we have dQ=0 for any adiabatic process, Then, from 1 st 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 = T f –T i . As stated previously, we can ll dU icalculate us ng, v dU nC dT Then 1 st law gives, nRT nC dT PdV dV v V (In the last step, we used (p, Ideal Gas Law: PV=nRT) Adiabatic Processes for an Ideal Gas Rearrange terms, we have, 0 dT R dV v TCV Using the relations for the molar specific heats, 11 pv p vvv CC C R CCC (1) 0 dT dV Then, we have, TV Adiabatic Processes for an Ideal Gas Integrating this equation, we have, ( 1) constant dT dV 1 () ln ( 1) ln constant ln constant TV TV 1 constantTV Note: T has to be in K. Using the Ideal Gas Law again, we can replace T with , PV nR 1 constant PV V nR constantPV (alternate form) Work in an Adiabatic Process (Ideal Gas) (0)dU dW dQ We know that, Using the same “trick” on dU we can calculate the work done the , work in an adiabatic process if we know the changes in state variables. v dW dU nC dT 21 () v WnCTT 22 11 v C WPVPVor R Examples Fire Piston (demo) Example 19.66 (Comparison of processes) Fire Piston History p( Fire piston calculations Example 19.66 calculations Administrator Microsoft PowerPoint - ch19_young_freedman.ppt [Compatibility Mode]