Calculate the pressure (in atm) at which the freezing temperature of water is -20°C = 253 K. H₂O(s)→ H₂O(l) AH°f (= AHM) V solid 1.963x10-5 m³/mole Vliquid 1.800x10-5 m³/mole = = = Caution: signs and units 6010 J/mole at TM = 273 K, (fusion = melting) Assume AH°f, Av₁ = V₁ - Vs are constant · · 1 atm = 1.013×105 Pa = 1 atm
Calculate the pressure (in atm) at which the freezing temperature of water is -20°C = 253 K. H₂O(s)→ H₂O(l) AH°f (= AHM) V solid 1.963x10-5 m³/mole Vliquid 1.800x10-5 m³/mole = = = Caution: signs and units 6010 J/mole at TM = 273 K, (fusion = melting) Assume AH°f, Av₁ = V₁ - Vs are constant · · 1 atm = 1.013×105 Pa = 1 atm
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter8: Thermochemistry
Section: Chapter Questions
Problem 12QAP: The heat of neutralization, Hneut, can be defined as the amount of heat released (or absorbed), q,...
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Can We use Clausius-Clapeyron Equation Here?
Since Enthalpy Of Vaporisation Not given,But enthalpy of Fusion Given.
![Calculate the pressure (in atm) at which the freezing temperature of water is -20°C = 253 K.
H₂O(s)→ H₂O(l) AH°f (= AHM)
V solid
1.963x10-5 m³/mole
Vliquid 1.800x10-5 m³/mole
=
=
=
Caution: signs and units
6010 J/mole at TM = 273 K,
(fusion = melting)
Assume AH°f, Av₁ = V₁ - Vs are constant · ·
1 atm
=
1.013x105 Pa
= 1 atm](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8d5954fb-d2b2-496d-a418-a5c577f57b02%2F8943f594-d77b-4116-8a0e-5225459ed11a%2F0gygpi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Calculate the pressure (in atm) at which the freezing temperature of water is -20°C = 253 K.
H₂O(s)→ H₂O(l) AH°f (= AHM)
V solid
1.963x10-5 m³/mole
Vliquid 1.800x10-5 m³/mole
=
=
=
Caution: signs and units
6010 J/mole at TM = 273 K,
(fusion = melting)
Assume AH°f, Av₁ = V₁ - Vs are constant · ·
1 atm
=
1.013x105 Pa
= 1 atm
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