8
1 Basic Background Material
To conform with tradition, we shall write this factor in terms of the heat capacity
ratio γ as γ − 1, to obtain the expression
P V = (γ − 1)U .
(1.2.6)
For a monatomic gas (which is relatively well represented by our hard-sphere
model), γ ≡ C P /C V , with C P and C V , the heat capacities at constant pressure
P and volume V (defined formally in Sects. 2.2 and 2.3), has the value 5/3. The
differential work, dW , done on the gas in compressing it by moving the piston by
−dx is
dW = F (−dx) = −P Adx = −P dV .
Now assume that the compression is adiabatic, i.e., there is no energy added or
removed during the process: this implies that work can be converted into internal
energy, so that P dV = −dU . Thus, if we take the differential of expression (1.2.6),
assuming that γ is a constant, we obtain
P dV + V dp = (γ − 1) dU .
If we now combine these last two results, we obtain the differential equation
γ
dV
V
+
dP
P
= 0 ,
which has the solution
γ ln V + ln P = ln C,
or
P V
γ
= C .
(1.2.7)
Under adiabatic conditions, when the compression of a gas can only increase its
temperature, because the work done cannot go anywhere else but into internal
energy, we thus predict that the product of pressure and volume to the power γ
is a constant. For a monatomic gas (such as He, Ar, etc.), P V 5/3 = const., and this
works. Of course, you may already have seen this result previously as an application
of the First Law of Thermodynamics: however, you should note that we have now
obtained this same result from basic microscopic considerations and from Newton’s
laws of motion.
Our simple model gives Eq. (1.2.5) as the ideal gas equation of state. This is
also one of the forms obtained in thermodynamics. For a gas, the thermodynamic
internal energy U is the sum of all kinetic and potential energies for the gas particles
that, in the present case, are hard-spheres, which possess no energies other than
1 Basic Background Material
To conform with tradition, we shall write this factor in terms of the heat capacity
ratio γ as γ − 1, to obtain the expression
P V = (γ − 1)U .
(1.2.6)
For a monatomic gas (which is relatively well represented by our hard-sphere
model), γ ≡ C P /C V , with C P and C V , the heat capacities at constant pressure
P and volume V (defined formally in Sects. 2.2 and 2.3), has the value 5/3. The
differential work, dW , done on the gas in compressing it by moving the piston by
−dx is
dW = F (−dx) = −P Adx = −P dV .
Now assume that the compression is adiabatic, i.e., there is no energy added or
removed during the process: this implies that work can be converted into internal
energy, so that P dV = −dU . Thus, if we take the differential of expression (1.2.6),
assuming that γ is a constant, we obtain
P dV + V dp = (γ − 1) dU .
If we now combine these last two results, we obtain the differential equation
γ
dV
V
+
dP
P
= 0 ,
which has the solution
γ ln V + ln P = ln C,
or
P V
γ
= C .
(1.2.7)
Under adiabatic conditions, when the compression of a gas can only increase its
temperature, because the work done cannot go anywhere else but into internal
energy, we thus predict that the product of pressure and volume to the power γ
is a constant. For a monatomic gas (such as He, Ar, etc.), P V 5/3 = const., and this
works. Of course, you may already have seen this result previously as an application
of the First Law of Thermodynamics: however, you should note that we have now
obtained this same result from basic microscopic considerations and from Newton’s
laws of motion.
Our simple model gives Eq. (1.2.5) as the ideal gas equation of state. This is
also one of the forms obtained in thermodynamics. For a gas, the thermodynamic
internal energy U is the sum of all kinetic and potential energies for the gas particles
that, in the present case, are hard-spheres, which possess no energies other than