Kinetic Theory of Gases

Adiabatic Processes

for an Ideal Gas

An adiabatic process is one in which no heat is transferred. You may think of it as a "fully insulated" process. We want to consider reversible adiabatic 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 = constant

Now, why is that true? Having stated that

is true for an adiabatic process, can we prove that or derive that or demonstrate that?

Remember, for an adiabatic process, there is no heat flow; that is,

Q = 0

For an infinitesimal change in volume, we can still use

dU = n CV dT

dU = - dW = - P dV

n CV dT = - P dV

dT = - (P/n CV) 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 CV) dV ]

P dV + V dP = R [ - (P/CV) dV ]

P dV + V dP = - [R/CV] P dV

Remember that

R = CP - CV

P dV + V dP = - [(CP - CV)/CV] P dV

dV/V + dP/P = - [(CP - CV)/CV] dV/V

dV/V + dP/P =[(CV - CP)/CV] dV/V

dV/V + dP/P =[1 - CP/CV] dV/V

dV/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 steeper than the graph of an isotherm.

During an adiabatic expansion, the gas cools so T < 0.

Specific Heat

Equipartition of Energy

Return to Ch21 ToC

(c) Doug Davis, 2002; all rights reserved