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- Paul So
- ch19_2_young_freedman.pdf

John R.

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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 WPVPV???or 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]

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