SOLUTION: 22. Vapor Pressure Problem: The vapor pressure of water is the pressure that water vapor (steam) would exert if it, alone, occupied the space above the water in a closed container.

Question 962262: 22. Vapor Pressure Problem: The vapor pressure of water is the pressure that water vapor (steam) would exert if it, alone, occupied the space above the water in a closed container. About 200 years ago French scientists Clausius and Clapeyron found that the vapor pressure varies exponentially with the reciprocal of the absolute temperature of the water. That is, if P is the vapor pressure and T is the absolute temperature, then P = a x 10^((k x (1/T))
a. Find the particular equation expressing vapor pressure in terms of absolute temperature if water at 0 degrees Celsius has a vapor pressure of 4.6 millimeters of mercury (mm), and at 100 degrees Celsius, has 760 mm. (The absolute temperature is 273 plus the Celsius temperature.) You must be very clever to figure a way to evaluate the two constants!
b. Use your equation to predict the vapor pressure of water on a hot summer day, 40 degrees Celsius.
c. Water boils when it's vapor pressure reaches the pressure if its surrounding atmosphere. At what Celsius temperature would water boil
i. atop Mt. Everest, where the air is 220 nm?
ii. in the reactor of a nuclear power plant, where the pressure is kept at 100,000 mm?
iii. in the deepest part of the ocean, where the pressure is 800,000 mm?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
That is, if P is the vapor pressure and T is the absolute temperature, then
P = a x 10^((k x (1/T))
a. Find the particular equation expressing vapor pressure in terms of absolute temperature if water at 0 degrees Celsius has a vapor pressure of 4.6 millimeters of mercury (mm), and at 100 degrees Celsius, has 760 mm. (The absolute temperature is 273 plus the Celsius temperature.) You must be very clever to figure a way to evaluate the two constants!
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You have 2 points (T,P)::
Using (273+0,4.6) you get 4.6 = a*10^(k*(1/273))
Using (273+100,760) you get 760 = a*10^(k(1/373))
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Dividing the lower equation by the upper you can solve for "k"::
165.22 = 10^((k/373)-(k/273))
165.22 = 10^[k(273-373)/(373*273)
165.22 = 10 ^[-0.0000982k)
-0.0000982k = log165.22
-0.000982k = 2.218
Finally:: k = -22587.2
---
Now, solve for "a":
4.6 = a*10(-22587.2/273)
4.6 = a*1.83^-83
a = 2.51x10^83
----
Equation:
P = 2.51x10^83*10^(-22587.2(1/(273+t)) where t is the temp in Celcius.
-------------------------------------
I'll let you check the arithmetic above and apply the
equation below.
Cheers,
Stan H.
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b. Use your equation to predict the vapor pressure of water on a hot summer day, 40 degrees Celsius.
c. Water boils when it's vapor pressure reaches the pressure if its surrounding atmosphere. At what Celsius temperature would water boil
i. atop Mt. Everest, where the air is 220 nm?
ii. in the reactor of a nuclear power plant, where the pressure is kept at 100,000 mm?
iii. in the deepest part of the ocean, where the pressure is 800,000 mm?

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