The temperature in Part (d) is the absolute zero of temperature. Lord Kelvin suggested that we set up a scale on which that temperature is OK. On that scale, T=t+ k. Show that, on the Kelvin scale, your equation reduces to P= mil.
The temperature in Part (d) is the absolute zero of temperature. Lord Kelvin suggested that we set up a scale on which that temperature is OK. On that scale, T=t+ k. Show that, on the Kelvin scale, your equation reduces to P= mil.
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Chapter9: Liquids, Solids, And Materials
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Can some help me, work through the problem in section (e) using my answers in section (a) trough (d)? I think the wording in section (e) has me a bit confused.
![2. In this experiment a student found that when she increased the temperature of a 544 mL sample of air
from 22.8°C to 33.6°C, the pressure of the air went from 1012 cm H₂O up to 1049 cm H₂O. Since the air
expands linearly with temperature, the equation relating P to t is of the form:
V=5414mL T₁ = 22.8°C 7+2=33.6°C
P₁ =1012 an H₂0 Pf=1049 cm H ₂0
P=mmit b
where m is the slope of the line and b is a constant.
a. What is the slope of the line? (Find the change in P divided by the change in t.)
AD 1049-1012 cm H ₂0 = 3.41,3 cm H ₂0/2
At 33.6-22.8
3.41259
b. Find the value of b. (Substitute known values of P and t into Equation 9 and solve for b.)
1012
= (3.415)(228) + b
78.204
-78.204
=
933,796=6
7 = 3.43
933.796
-934
b=
cm H₂O/°C
cm H₂O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a2e06cd-89b1-4949-854a-ccc1923a28fc%2Fc4566808-ecfe-4ca3-8d61-296b8e23bae7%2Fm82et3u_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2. In this experiment a student found that when she increased the temperature of a 544 mL sample of air
from 22.8°C to 33.6°C, the pressure of the air went from 1012 cm H₂O up to 1049 cm H₂O. Since the air
expands linearly with temperature, the equation relating P to t is of the form:
V=5414mL T₁ = 22.8°C 7+2=33.6°C
P₁ =1012 an H₂0 Pf=1049 cm H ₂0
P=mmit b
where m is the slope of the line and b is a constant.
a. What is the slope of the line? (Find the change in P divided by the change in t.)
AD 1049-1012 cm H ₂0 = 3.41,3 cm H ₂0/2
At 33.6-22.8
3.41259
b. Find the value of b. (Substitute known values of P and t into Equation 9 and solve for b.)
1012
= (3.415)(228) + b
78.204
-78.204
=
933,796=6
7 = 3.43
933.796
-934
b=
cm H₂O/°C
cm H₂O
![c. Express Equation 9 in terms of the values of mandi b.
d. At what temperature t will P become zero?
0=(3.43) 1934
9301
-9341= 13.436
P= (3.43) € + 934
-273
P = 0 at t =
°C = t₁ = -k
3.43
(Please note that you are unlikely to get results anywhere mear this good when actually carrying outt
this experiment, as it involves a very large extrapolation.)
e. The temperature in Part (d) is the absolute zero of temperature. Lord Kelvin suggested that we set up
a scale on which that temperature is OK. On that scale, T=t+ k. Show that, on the Kelvin scale, your
equation reduces to P= mil.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a2e06cd-89b1-4949-854a-ccc1923a28fc%2Fc4566808-ecfe-4ca3-8d61-296b8e23bae7%2Fmgdlh3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:c. Express Equation 9 in terms of the values of mandi b.
d. At what temperature t will P become zero?
0=(3.43) 1934
9301
-9341= 13.436
P= (3.43) € + 934
-273
P = 0 at t =
°C = t₁ = -k
3.43
(Please note that you are unlikely to get results anywhere mear this good when actually carrying outt
this experiment, as it involves a very large extrapolation.)
e. The temperature in Part (d) is the absolute zero of temperature. Lord Kelvin suggested that we set up
a scale on which that temperature is OK. On that scale, T=t+ k. Show that, on the Kelvin scale, your
equation reduces to P= mil.
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