Problem 5 - Chapter 21 problem 68
Consider a cylinder of cross sectional area A and mass M . At equilibrium
the cylinder contains n moles of an ideal gas at pressure P0 and temperature
T0 : If the cylinder is displaced from equilibrium by a small distance x show
that the angular frequency of oscillations of the cylinder head is found from
! 2 = P02 A2 =M nRT0 :
The change in volume in the cylinder is V = A x: Assuming no heat
transfer during this displacement then the rate of change in pressure in the
cylinder is found from
dP
dV
d
(P V ) =
V + PV 1
=0
dx
dx
dx
P dV
P
dP
=
=
A:
dx
V dx
V
With the restriction x small we have
P = P0 +
dP
dx
P0
A x
V
x = P0
P =P0
P02
A x
nRT0
P ' P0
The net force on the piston is found from
PA
P0 A =
dP
dx
A x=
P =P0
P02 2
A x
nRT0
From Newton’s third law
d2
P02 2
A x=M 2 x
nRT0
dt
and the angular frequency is
P02
! =
A2 ! ! =
M nRT0
2
r
M nRT0
P0 A
Problem 6 - Adiabatic Free Expansion
(a) Consider the adiabatic free expansion of n1 moles of an ideal gas at
pressure P1i from a volume V1i into a volume V = V1i + V2i in which the
1
volume V2i contains n2 moles of another ideal gas at pressure P2i < P1i :
Find the change in entropy due to this free expansion in terms of the initial
and …nal pressures and the number of moles of the gases. Hint, consider
two reversible isothermal processes. In the …rst process the gas in volume
V2i is compressed isothermally into a volume V2f consistent with the …nal
equilibrium pressure, Pf . Then consider an isothermal expansion of the gas
in volume V1i into a volume V1f consistent with this same …nal equilibrium
pressure. (b) From the result obtained in part (a) show that the change in
entropy reduces to the change in entropy for an adiabatic free expansion of
an ideal gas from volume Vi into a …nal volume Vf : Also show from the result
obtained in part (a) that s = 0 if the pressure of the gas in the second
chamber is equal to the pressure of the gas in the …rst chamber.
(a) The …nal equilibrium pressure is found from
Pf (V1f + V2f ) = P1i V1i + P2i V2i
P1i V1i + P2i V2i
P1i V1i + P2i V2i
:
=
Pf =
V1f + V2f:
V
To …nd the change in entropy we consider two reversible isothermal processes.
In the …rst we compress the gas in the volume V2i until it reaches …nal
equilibrium pressure. This volume is given by
P2i V2i = Pf V2f ! V2f =
P2i V2i
n2
P2i V2i
=
V =
V;
Pf
P1i V1i + P2i V2i
n1 + n2
where n1;2 are the number of moles of each gas. Then we follow that with a
second reversible isothermal process. In this process the gas that was initially
in the volume V1i expands isothermally into a volume
V1f = V1i + (V2i
V2f ) = V
n1
n2
V =
V:
n1 + n2
n1 + n2
It is useful to note that the total volume remains unchanged, V1f + V2f =
V1i + V2i = V:
For an isothermal expansion dQ=T = dW=T = P dV =T = nRdV =V; hence
the total change in entropy is simply
V1f
V2f
+ n2 R ln
or
V1i
V2i
P1i
P2i
s = n1 R ln
+ n2 R ln
:
Pf
Pf
s = n1 R ln
2
(1a)
(1b)
It is useful to note that in an isothermal expansion the P V product is constant. Since we know the …nal equilibrium pressure, Pf , we can write equation
(1b) immediately.
(b) First we consider the limit as n2 ! 0; i.e. the …rst gas expands into
a vacuum. For that case
V1f
s = n1 R ln
V1i
which is the result obtained in class for a free adiabatic expansion. Now
consider the limit in which the gas in both chambers are initially at the same
pressure P . Then P1i = P2i = P = Pf : Since ln 1 = 0; the second expression
for s vanishes and s = 0:
3