Thermodynamics revision sheet for IIT-JEE and AIEEE

Temperature
T_F=9/5 T_C + 32°
T_C = T_K - 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
V_rms=√3RT/M
Mean  Velocity
V_m=√8RT/&#960M
Most Probable Velocity
V=√2RT/M
Also
V_rms > V_m > 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 : PV^y=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=C_v
Pressure Constant : V/T=constant ΔU=Q-W ,Molar Specific Heat=C_P

Internal energy depends on Temperature.
So for same temperature change ΔT
nC_vΔT=Q₁-W₁=Q₂-W₂=Q₃-W₃

Molar Specfic Heat Capacity of any process is given by

C=C_v + 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-T_s)

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 megative