physics 52 - heat and optics dr. joseph f. becker physics department san jose state university ©...
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Physics 52 - Heat and Optics
Dr. Joseph F. BeckerPhysics Department
San Jose State University© 2005 J. F. Becker
Ch. 19 First Law of Thermodynamics
1. Thermodynamic systems2. Work done during volume changes3. Paths between thermodyn. states4. Internal energy and the First Law 5. Kinds of thermodynamic processes6. Internal energy of an ideal gas7. Heat capacities of an ideal gas8. Adiabatic processes for ideal gas
Sign conventions:a) Q is positiveb) Q is negativec) W is positived) W is negative
e) In a steam engine the heat in and the work out (done) are both defined to be POSITIVE.
(a) Molecule hits a piston moving away from it; speed and KE of molecule decrease. (expan. -> cooling)
(b) Molecule hits a piston moving toward it; speed and KE of molecule increase. (comp. -> heating)
When a molecule hits a wall moving away from it, the molecule does work on the wall; the molecule’s speed and kinetic
energy decrease. The gas does positive work (on the piston);
and the internal energy of the gas decreases.Molecule hits a wall moving toward it, the
wall does work on the molecule; the molecule’s speed and kinetic energy
increase. The gas does negative work (on the
piston); and the internal energy of the gas
increases.
WORK-ENERGY THEOREM
The infinitesimal work done by the system (gas) during the small expansion dx is dW = p A dx = F dx so W = p dV
The work done equals the area under the curve on a p-V diagram. (a) Volume increases: work and area are positive. (b) Volume decreases: work and area are negative. (c) A constant-pressure: Volume increases, W > 0.
(a) Three different paths between state 1 and state 2.
(b)-(d) The work done by the system during a transition between two states depends on
the PATH (or PROCESSES) chosen.
The final states are the same. The intermediate states (p and V) during the transition from state 1 to 2 are entirely different. Heat, like work, depends on the initial and final states AND the path between the states. (a) Heat is added slowly so as to keep the temperature constant. (b) Rapid, free expansion does no work and there is no heat transfer.
HEAT ADDED
In a thermodynamic process, the internal energy of a system may increase, decrease, or remain the same:(a) U > 0(b) U < 0 (c) U = 0
The First Law of Thermodynamics
U = Q - W
KINDS OF THERMODYNAMIC PROCESSES
ADIABATIC – NO HEAT TRANSFER; Q=0; U=-W
ISOCHORIC – CONSTANT VOLUME; W=0; U = Q
ISOBARIC – CONSTANT PRESSURE; W=p (V2 – V1)
ISOTHERMAL – CONSTANT TEMPERATURE; only for ideal gas U = 0, Q =
W
The First Law of Thermodynamics
U = Q - W or Q = U + W
Raising the temperature of an ideal gas by various
processes:Q = U + W and U = f(T)Isochoric W = 0 so Q = UIsobaric Q = U + p(V2-V1)
A p-V diagram of an adiabatic process
for an ideal gas from state a to state b.
pV = constant (isotherm)
pV = constant (adiabat)
ADIABATIC PROCESSES for IDEAL GAS
Q = 0 so U = Q – W = 0 – p V and U = n Cv dT for an ideal gas
U = n Cv dT = – p V = -(n R T/V) dV dT/T + (R/ Cv) dV/V = 0dT/T + ( - 1) dV/V = 0 where – 1 = (Cp/Cv
) –1 dT/T + ( - 1) dV/V = 0 ; lnT+(-1) lnV = const
ln (TV -1 ) = constant and T1V1 -1 = T2V2 -1
And since pV = nRT, (pV/nR) V -1 = constant’
(p/nR) V = constant’ p1V1 = p2V2
A quantity of air is taken from state a to state b along the
straight line.
Is Tb > Ta?
Given values for all p’s and V’s,
calculate the work.
Wadc =120 J
Wabc = 450 J
Ua = 150 J.
Ub = 240 J. Uc = 680 J.
Ud = 330 J.
Qab = ?
Qbc = ?
Qdc = ?
Qad = ?