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Specific heat capacity
Q=mcΔT
atm to Pa
1atm=1.01×10^5 Pa
Ideal gas law
pV=nRT, pV=NkT
(On formula sheet)
What is an isochoric process
Volume is constant
What is an isobaric process
Constant pressure
What is an isothermal process
Constant temperature
Work done on a system
∫V1V2pdV
work is negative if the gas is compressed
work is positive is the gas expands
Change in internal energy
dU=dQ-dW
Where dQ is the change in heat
is path independent, constant in an isolated system
(On formula sheet)
work done in an isothermal process
dW=dQ so dU=0
W=nRTlnV1V2
Work done in an isochoric process
no work is done
dU=nCvdT
where C is molar heat capacity at a constant volume (C=Molar mass x c)
Work done in isobaric process
dW=p(V2−V1)
dU=ndT(Cp−R)
relationship between molar heat capacity at a constant pressure and at constant volume
Cp=CV+R
Adiabatic constant
γ=CVCp
Adiabatic process
no heat flow into or out of system (dQ=0)
dU=-W
TVγ−1=const
pVγ=const (On formula sheet)
W=γ−1p1V1−p2V2
Internal energy in a cycle
No net change in internal energy
Heat in a heat engine
Q=W=∣QH∣−∣QC∣
thermal efficiency of a heat engine
η=1−QHQC
the Carnot cycle
Contains two isothermal and two Adiabatic processes
Efficiency for the carnot cycle
η=1−THTC
Energy in a refrigerator
∣QH∣=∣QC∣+∣W∣
Coefficient of performance in refrigerators
Kr=∣QH−QC∣∣QC∣=TH−TCTC
coefficient of performance in a heat pump
Kr=∣QH−QC∣∣QH∣=TH−TCTH
Zeroth law
When two bodies are each separately in thermodynamic equilibrium with a third body they will be in thermodynamic equilibrium which each other
First law
For a closed system, the change in internal energy is the difference between the heat accumulated by the system and the work done by it
Second law
It is impossible to construct a cyclic engine that only converts thermal energy from a reservoir at a given temperature into mechanical heat. A portion of the thermal energy must be transferred to a second reservoir at a lower temperature.
what is entropy
Increasing entropy is an increase in disorder
Adiabatic and reversible processes have no change in entropy
Increased by:
Add heat
Increase volume
Change in entropy
ΔS=SB−SA=∫ABT1dQ (JK−1)
CONVERT dQ TO dT USING SHC FORMULA
Positive value=decrease in order
Entropy for microstates
S=kBlnw
w is number of microstates
change in s: kB ln(w2-w1)
Third law
The entropy of a perfect crystal at absolute zero is equal to zero
(Absolute zero is impossible to reach)
Kinetic theory of gases
molecules move randomly
total number of molecules is large
volume occupied by molecules is small compared to that of the gas
no forces between molecules unless they collide
collisions are elastic and occur instantaneously
kinetic pressure
p=31ρvrms2
(On formula sheet)
root mean squared speed
vrms2=ρ3P
P is pressure
Average kinetic energy
KE=23kT
each degree of freedom the molecule has contributes 1/2kT
Eg translational x,y,z =3/2kT
translational and two rotational=5/2kT
Average velocity
v=πm8kT
rms velocity
vrms=m3kT (On formula sheet)
most probable velocity
vmp=m2kT
How are the velocities related
vmp<v<vrms
mp<avg<rms
van der waals equation
ideal gas law corrected
(p+a(Vn)2)(nV−b)=RT
a- corrects pressure/ intermolecular forces
b- corrects volume/ volume of molecules