Kinetic Theory of Gases

Adiabatic Processes

for an Ideal GasAn

adiabatic processis one in whichno heatis transferred. You may think of it as a "fully insulated" process. We want to considerreversibleadiabatic processes. "reversible" means the system is near thermal equilibrium.If the process is reversible, we can always count on the ideal gas law,

P V = n R T A reversible adiabatic process can be described by

P V ^{}= constantNow,

whyis that true? Havingstatedthatis true for an adiabatic process, can we

provethat orderivethat ordemonstratethat?Remember, for an adiabatic process, there is

noheat flow; that is,Q = 0 For an infinitesimal change in volume, we can still use

dU = n C _{V}dTdU = - dW = - P dV

n C

_{V}dT = - P dVdT = - (P/n C

_{V}) dV(We will use that in a moment).

From the ideal gas law,

P V = n R T we can write the differential

P dV + dP V = n R dT P dV + V dP = n R [ - (P/n C

_{V}) dV ]P dV + V dP = R [ - (P/C

_{V}) dV ]P dV + V dP = - [R/C

_{V}] P dVRemember that

R = C _{P}- C_{V}P dV + V dP = - [(C

_{P}- C_{V})/C_{V}] P dVdV/V + dP/P = - [(C

_{P}- C_{V})/C_{V}] dV/VdV/V + dP/P =[(C

_{V}- C_{P})/C_{V}] dV/VdV/V + dP/P =[1 - C

_{P}/C_{V}] dV/VdV/V + dP/P =[1 - ] dV/V

dV/V + dP/P =dV/V - dV/V

dP/P = - dV/V

dP/P + dV/V = 0

ln P + ln V = constant

An isotherm is a graph of

P V = constant Since > 1,

the graph of an adiabat is

steeperthan the graph of an isotherm.During an adiabatic expansion, the gas

coolsso T < 0.

Return to Ch21 ToC(c) Doug Davis, 2002; all rights reserved