P19_003.PDF
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Chapter 19 - 19.3
3. Let
T
L
be the temperature and
p
L
be the pressure in the left-hand thermometer. Similarly, let
T
R
be the temperature and
p
R
be the pressure in the right-hand thermometer. According to the problem
statement, the pressure is the same in the two thermometers when they are both at the triple point of
water. We take this pressure to be
p
3
. Writing Eq. 19-5 for each thermometer,
T
L
= (273
.
16K)
p
L
p
3
and
T
R
= (273
.
16K)
p
R
p
3
,
we subtract the second equation from the first to obtain
T
L
−
T
R
= (273
.
16K)
p
L
−
p
R
.
p
3
First, we take
T
L
= 373
.
125 K (the boiling point of water) and
T
R
= 273
.
16 K (the triple point of water).
Then,
p
L
−
373
.
125 K
−
273
.
16 K = (273
.
16K)
120 torr
p
3
for
p
3
. The result is
p
3
= 328 torr. Now, we let
T
L
= 273
.
16 K (the triple point of water) and
T
R
be the
unknown temperature. The pressure difference is
p
L
−
p
R
=90
.
0 torr. Solving
273
.
16K
−
T
R
= (273
.
16K)
90
.
0torr
328 torr
for the unknown temperature, we obtain
T
R
= 348 K.
p
R
= 120 torr. We solve
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p19_102.pdf
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P19_002.PDF
(57 KB)
P19_001.PDF
(63 KB)
P19_003.PDF
(70 KB)
P19_005.PDF
(65 KB)
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