Thermodynamics Revision Sheet
Temperature
TF=9/5 TC + 32°
TC = TK - 273.15 K
The pressure,volume and temperature in kelvin of such gases obey the equation
PV= nRT ----(1)
Thermal expansion
ΔL=αLΔT
Specific Heat Capacity=ΔQ/mΔT
Gas Laws
Boyles Law: PV=constant
Charles Law : V/T=constant
Dalton Law of Partial Pressure: P=P1 + P2 +P3
Root mean Square Velocity
Vrms=√3RT/M
Mean Velocity
Vm=√8RT/πM
Most Probable Velocity
V=√2RT/M
Also
Vrms > Vm > V
Average kinetic Energy of Gas=3/2nRT
First law of Thermodynamics
ΔU=Q-W
Gas Processes
Isothermal Process : PV=constant ,ΔU=0,Q=W,Molar Specific Heat=infinity
Adaibatic Process : PVy=constant,Q=0,ΔU=-W,Molar Specific Heat=zero
Polytropic Process : PVn=constant,Molar Specific Heat=[R/(y-1) + R/(1-n)]
Volume Constant : P/T=constant W=0,ΔU=Q,Molar Specific Heat=Cv
Pressure Constant : V/T=constant ΔU=Q-W ,Molar Specific Heat=CP
Internal energy depends on Temperature.
So for same temperature change ΔT
nCvΔT=Q1-W1=Q2-W2=Q3-W3
Molar Specfic Heat Capacity of any process is given by
C=Cv + Pdv/ndT where n is no of moles of the gas
Workdone by Gas= ∫PdV
Heat Conduction
Q=-KAdT/dx
Wein displacement lawλT4=Constant
Stefan's Law
Q=eσT4
Newton law of Cooling
dT/dx=b(T-Ts)
Problem Solving tips
1. Net Heat flow at a junction is Zero
2. Internal Energy does not depend on the path
3. Internal Energy in a cyclic Process is zero
4. Heat is positive when given to the system and negative when taken out of the system
5. Workdone by the system on the surrounding is positive while the workdone on the system by the surrounding is negative
Temperature
TF=9/5 TC + 32°
TC = TK - 273.15 K
The pressure,volume and temperature in kelvin of such gases obey the equation
PV= nRT ----(1)
Thermal expansion
ΔL=αLΔT
Specific Heat Capacity=ΔQ/mΔT
Gas Laws
Boyles Law: PV=constant
Charles Law : V/T=constant
Dalton Law of Partial Pressure: P=P1 + P2 +P3
Root mean Square Velocity
Vrms=√3RT/M
Mean Velocity
Vm=√8RT/πM
Most Probable Velocity
V=√2RT/M
Also
Vrms > Vm > V
Average kinetic Energy of Gas=3/2nRT
First law of Thermodynamics
ΔU=Q-W
Gas Processes
Isothermal Process : PV=constant ,ΔU=0,Q=W,Molar Specific Heat=infinity
Adaibatic Process : PVy=constant,Q=0,ΔU=-W,Molar Specific Heat=zero
Polytropic Process : PVn=constant,Molar Specific Heat=[R/(y-1) + R/(1-n)]
Volume Constant : P/T=constant W=0,ΔU=Q,Molar Specific Heat=Cv
Pressure Constant : V/T=constant ΔU=Q-W ,Molar Specific Heat=CP
Internal energy depends on Temperature.
So for same temperature change ΔT
nCvΔT=Q1-W1=Q2-W2=Q3-W3
Molar Specfic Heat Capacity of any process is given by
C=Cv + Pdv/ndT where n is no of moles of the gas
Workdone by Gas= ∫PdV
Heat Conduction
Q=-KAdT/dx
Wein displacement lawλT4=Constant
Stefan's Law
Q=eσT4
Newton law of Cooling
dT/dx=b(T-Ts)
Problem Solving tips
1. Net Heat flow at a junction is Zero
2. Internal Energy does not depend on the path
3. Internal Energy in a cyclic Process is zero
4. Heat is positive when given to the system and negative when taken out of the system
5. Workdone by the system on the surrounding is positive while the workdone on the system by the surrounding is negative