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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.

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter9: Liquids, Solids, And Materials
Section: Chapter Questions
Problem 45QRT: At the critical point for carbon dioxide, the substance is very far from being an ideal gas. Prove...
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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
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.
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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