Thermal Flashcards
energy, specific heat capacity
ΔQ = mcθ
where Q is heat
c is shc
θ is temp
find specific heat capacity of iron
heater and thermometer in iron block
circuit with voltmeter and ammeter connected to heater
since E = ItV
and c = E/mθ
so c = ItV/mθ
latent heat of fusion
energy needed to melt 1kg of substance
latent heat of vaporization
energy needed to vaporize 1kg of substance
general equation for specific latent heat
l = E/m
energy needed for change in temp of substance when there is a change of state involved
E = mcΔT + ml
ice in juice equation
energy as they go equilibrium temp
juice is 20, ice is -5
E = mc(20-T) = mc(T-(-5)) + ml
where T is final temp
20-T is temp taken
T-(-5) is temp given
ml is melting of ice
Boyles law
P ∝ 1/V
at constant temp
Charles’s law
V ∝ T
at constant pressure
(KELVIN)
Gay Lussac law
P ∝ T
at constant vol
(KELVIN)
Ideal gas law
(2 forms)
pV = NkT
where N is number of molecules
k is Boltzmann’s constant
pV = nRT
where n is number of moles
R = 8.31 (gas constant)
KELVIN
pressure, vol, moles, temp before and after
P1V1/n1T1 = P2V2/n2T2
isothermic
isobaric
isothermic - T constant
isobaric - P constant
Ideal gas rules
- motion is random
- volume of particles negligible
- no internal forces
- collisions elastic
- duration of collisions negligible
Kinetic theory of gasses equation
P = 1/3 (Nm(Crms)^2/V)
where N is number of molecules
Crms^2 is root mean squared speed
average KE of particle
Ek = 3/2 kT
where k is boltzmans constant
KELVIN
1st law of thermodynamics
Q = ΔU + W
where Q is heat given/taken
ΔU is change in internal energy
W is work done on or by
Isothermic process
1st law of T.d
- internal energy of gas not changing
Q = W
P1V1 = P2V2
adiobatic process
1st law of T.d
- no heat lost or gained
Q = 0
ΔU = -W
work done by/on gas result
Volume changes
area under PV graph
energy
gas compressing vs expanding work done
gas compressing - work done on gas
gas expanding - work done by gas
PV graphs
straight line up: no work done, Q = ΔU
straight horizontal line: W = pΔV (area under graph)
arrow going towards y axis: work done ON
2nd law of thermodynamics
for a spontaneous process, entropy of the universe increases
3rd law of thermodynamics
a perfect crystal at 0K has 0 entropy
0th law
if 2 systems are in equilibrium with a third system, the two systems are in equilibrium with each other
