Thermodynamics

Question 1

change in entropy

and for a heat reservoir

Answer 1

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

Question 2

entropy

Answer 2

statistical definition
S = kB ln(ohm)
where ohm = N!/(capital pi)j nj !

Question 3

ideal gas

Answer 3

pV = nRT
pV^(gamma) = c
TV^(gamma - 1) = c2
dU = 0 so, dQ = -dw

Question 4

mono-atomic gas defining quality

Answer 4

gamma = 5/3
expected Cv = 3R/2

Question 5

dU from 1st law

Answer 5

dU = TdS - pdV
dU = TdS - mdB

Question 6

defining characteristic of a paramagnetic material

Answer 6

-m = -kB/T

Question 7

1st law of thermodynamics

Answer 7

energy can neither be created nor destroyed, only altered in form

Question 8

zeroth law of thermodynamics

Answer 8

if two systems, seperatly, are in thermal equilibrium with a third, then the original two are in thermal equilibrium with eachother

Question 9

2nd law of thermodynamics

Answer 9

the entropy of any natural & spontaneous process either increases or remains constant

reversible, change in S =0
irreversible, change in S > 0

Question 10

3rd law of thermodynamics

Answer 10

entropy of a system approaches a constant value as the temperature approaches absolute zero

Question 11

irreversibility equation

Answer 11

I = t0 (change in Suniverse)

Question 12

boltzman distribution

Answer 12

population nr for level r
nr = e^(-Er/kT)/z
where z = partititon function
= sumi (e^(-Ei/kT))

Question 13

diathermal

Answer 13

heat can be exchanged

Question 14

adiathermal

Answer 14

no heat exchange

Question 15

adiabatic

Answer 15

no heat exchange & reversible

adiabtic expansion: dQ = 0

Question 16

constant what?
isotherm, adiabat, isochore, isobar

Answer 16

isotherm, constant temperature & pV
adiabat, constant pV^(gamma), gamma>1
isochore, constant volume
isobar, constant pressure

Question 17

expression for work done

Answer 17

dw = f dx
= -p dV

Question 18

statistical defininition of boltzmann distribution

Answer 18

nj = Aexp(-𝜷𝜺j)
where 𝜷 = 1/(kbT)

Question 19

carnot cycle relation

Answer 19

|QL|/QH = TL/TH

Question 20

applying 1st law to carnot cycle to get an equation for work done

Answer 20

QH - |QL| = -W

Question 21

clausius inequality

Answer 21

closed integral (dQ/t) </= 0

Question 22

describe the adiabatic process for cooling a gas from room temperature until it is liquified

Answer 22

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.

Question 23

describe the joule-kelvin process for cooling a gas from room temperature until it is liquified

Answer 23

constant enthapy
in an expansion of pi –> pf < pi, curve of T against p decreases below inversion point so tf < Ti , leading to cooling