Physics 105 Problem Set 11 Solutions
Problem 1
(3 pts)
(a) When bouncing off a wall perpendicular to the
x
axis, only the
x
component of the atom’s velocity
changes its sign. Collisions with the other walls do not affect the
x
component of the atom’s velocity.
Hence the atom transfers to the wall orthogonal to the
x
axis the amount of momentum equal to
Δ
p
= 2
mv
x
during each collision; and it happens exactly once during the time interval equal to
Δ
t
=
2
L
v
x
.
Consequently, the time averaged force is
F
=
Δ
p
Δ
t
=
mv
2
x
L
.
(b) Using the result from the previous part, the pressure of the gas on the wall orthogonal to the
x
axis is
p
=
∑
N
i
=1
F
i
L
2
=
1
L
3
N
X
i
=1
m
i
v
2
ix
.
Given that the direction of the atom’s velocity is equally likely to point in any direction, we find:
1
2
N
X
i
=1
m
i
v
2
ix
=
1
2
N
X
i
=1
m
i
v
2
iy
=
1
2
N
X
i
=1
m
i
v
2
iz
=
N
2
k
B
T .
This yields for the pressure
pV
=
N
N
A
(
k
B
N
A
)
T .
Recalling that the ideal gas constant is
R
=
k
B
N
A
, we recover the ideal gas law.
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 '08
 LYMANA.PAGE
 Physics, mechanics

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