Thermodynamics Flashcards
change in entropy
and for a heat reservoir
dS = dQ/T
= integral from Ti to Tf (Cp dT/T)
heat reservoir:
object losing heat to reservoir:
dS = integral from Ti to TR (CT/T dT)
resevoir itself:
dS = integral (dQR/TR)
= (change in QR)/TR
= -(change in Qobject)/TR
entropy
statistical definition
S = kB ln(ohm)
where ohm = N!/(capital pi)j nj !
ideal gas
pV = nRT
pV^(gamma) = c
TV^(gamma - 1) = c2
dU = 0 so, dQ = -dw
mono-atomic gas defining quality
gamma = 5/3
expected Cv = 3R/2
dU from 1st law
dU = TdS - pdV
dU = TdS - mdB
defining characteristic of a paramagnetic material
-m = -kB/T
1st law of thermodynamics
energy can neither be created nor destroyed, only altered in form
zeroth law of thermodynamics
if two systems, seperatly, are in thermal equilibrium with a third, then the original two are in thermal equilibrium with eachother
2nd law of thermodynamics
the entropy of any natural & spontaneous process either increases or remains constant
reversible, change in S =0
irreversible, change in S > 0
3rd law of thermodynamics
entropy of a system approaches a constant value as the temperature approaches absolute zero
irreversibility equation
I = t0 (change in Suniverse)
boltzman distribution
population nr for level r
nr = e^(-Er/kT)/z
where z = partititon function
= sumi (e^(-Ei/kT))
diathermal
heat can be exchanged
adiathermal
no heat exchange
adiabatic
no heat exchange & reversible
adiabtic expansion: dQ = 0
constant what?
isotherm, adiabat, isochore, isobar
isotherm, constant temperature & pV
adiabat, constant pV^(gamma), gamma>1
isochore, constant volume
isobar, constant pressure
expression for work done
dw = f dx
= -p dV
statistical defininition of boltzmann distribution
nj = Aexp(-𝜷𝜺j)
where 𝜷 = 1/(kbT)
carnot cycle relation
|QL|/QH = TL/TH
applying 1st law to carnot cycle to get an equation for work done
QH - |QL| = -W
clausius inequality
closed integral (dQ/t) </= 0
describe the adiabatic process for cooling a gas from room temperature until it is liquified
begins with isothermal compression, removing heat from the gas.
As the gas adiabatically expands it comes back to the starting pressure but because adiabats are steeper than isotherms this new point is on a lower temperature isotherm.
process repeats, as T –> 0 the isotherms get closer together so cooling becomes more difficult.
describe the joule-kelvin process for cooling a gas from room temperature until it is liquified
constant enthapy
in an expansion of pi –> pf < pi, curve of T against p decreases below inversion point so tf < Ti , leading to cooling
