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Quiz 5 Solutions
Part (a)
At room temperature, there are 5 modes/molecule for diatomic gas (3KE
translational
and 2 KE
rotational
,
vibrational modes are frozen out) and 6 modes/molecule for solid Au/liquid Hg (3 KE and 3 PE).
Part (b)
7:30am lecture:
The total number of modes for each substance: 7N
A
*5=35N
A
for N
2
, 6N
A
*6=36N
A
for Hg, or,
alternatively, C
v
=(5modes per particle)/2*(7moles)*R=145.4J/K for N
2
, and C
v
=(6modes per
particle)/2*(6moles)*R=149.8J/K for Hg.
Then the substance with the smaller total number of modes or
smaller C
v
(N
2
) will have greater temperature change when the same amount of heat is added to each,
because Q=
Δ
E
th
=(total #modes)/2*k
B
*
Δ
T or Q=
Δ
E
th
=C
v
*
Δ
T.
9:00am lecture:
The total number of modes for each substance: 5N
A
*5=25N
A
for O
2
, 3N
A
*6=18N
A
for Au, or,
alternatively, C
v
=(5modes per particle)/2*(5moles)*R=103.9J/K for O
2
, and C
v
=(6modes per
particle)/2*(3moles)*R=74.8J/K.
Then the substance with the smaller total number of modes or smaller
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This note was uploaded on 01/05/2011 for the course PHY 7A taught by Professor Pardini during the Winter '08 term at UC Davis.
 Winter '08
 PARDINI

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