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.
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