Lecture 1: Zeroth and First Law Flashcards
(28 cards)
zeroth law of thermodynamics
if two systems are separately in thermal equilibrium with a third, then they must also be in thermal equilibrium with each other
zeroth law leads to the concept of
temperature as a measurable quantity
the zeroth law legitimises the use of
thermometers and calibration of temp
what is thermal equilibrium
bulk physical properties are uniform and invariant
in particular, the state variables are unchanged
example: PV curves for 2 gases in thermal equilibrium
get isotherms for each
hence shows property of temp can be identified
the empirical temperature
a function θB(PB, VB) - an equation of state - whose value is the same for two thermally equilibrated systems.
makes sense to define θ(P,V) as
linearly increasing with ‘hotness’ but it makes sense to do so by defining a scale based on a thermometric property ‘X’ and two arbitrary fixed points θ1 and θ2
first law in differential form
dU=δQ+δW
dU is
the infinitesimal change in internal energy (in J) for a process
δQ is
the infinitesimal heat (thermal energy in J) transferred in the process.
δW is
the infinitesimal work (in J) done by the surroundings on a system during the process
what kind of differential is dU
exact differential
U is a function of state so ΔU is path independent
what type of differentials are δQ and δW
inexact differentials
neither are uniquely defined for a process and depend on path taken between initial and final states
inexact differentials often relate to
irreversible processes
the first law tells us that although the distribution between heat (thermo) and work (dynamics) can’t always be determined,
their combination (thermodynamics) is calculable: U is state function
what can we do to calculate alternative paths with well defined δQ and δW and hence explore ΔU
choose quasistatic, reversible processes
1st law calculation - typical situation
frictionless piston compressing a fluid
we consider reversible changes from state (P1,V1) to state (P2, V2) by
applying a force to the piston
we treat applying a force to the piston as
a quasistatic process (do it slowly, one dx then find eqbm, then another dx, another eqbm etc)
make applying force to a piston a reversible process by
neglecting frictional dissipation
for the reversible process, the incremental work done by the surroundings on the gas is given by
δW=-PdV
why is there a negative sign in δW=-PdV
by the surroundings on gas
gas compressed (surroundings doing work)
volume decreases
(+ve if gas doing work on surroundings)
δW=-PdV for the complete process
ΔW=- ∫ PdV
ΔW is
path dependent (area under curve so depends on process for path taken)