Chemistry ideal gas law question..
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Chemistry ideal gas law question..

[From: ] [author: ] [Date: 11-07-09] [Hit: ]
and the partial pressure due to escaped water vapor would be 0.198 atm.Next, lets write the significant variables in the same units as they appear in the ideal gas constant were using:P (escaped H2O) = 0.198 atmV = 0.072 LT = 287.......
P = Pressure
V = Volume available to the gas
n = Number of moles of gas
R = ideal gas constant
T = temperature in kelvin

Now, the total pressure in the problem is from a mixture of gasses, and we're only interested in the water that escaped from a liquid solution into the vapor phase. To solve this based on given information, we're going to have to make some assumptions. First, let's assume that no other molecules collected, except for water, were due to them evaporating (i.e. they didn't escape into the vapor phase). Second, let's assume that the atmospheric pressure where the gas sample was collected is exactly 1 atm. Thus, the partial pressure due to atmospheric gasses would be 1 atm, and the partial pressure due to escaped water vapor would be 0.198 atm.

Next, let's write the significant variables in the same units as they appear in the ideal gas constant we're using:
P (escaped H2O) = 0.198 atm
V = 0.072 L
T = 287.15 K

Now, we solve the ideal gas equation for n, number of moles, then plug in our values to get the number of moles:

n = P*V/(R*T) = 6.050 * 10^(-4) mol

Now, we just convert mols of water to grams using the molar mass of water, which gives us 0.0109 grams of water.

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I cannot tell unless I have the partial pressure of the other gasses. The total pressure refers to the gasses and the water together, right? How do you know the pressure of the water without the pressure of the other gasses?

Edit based on your info:
You want to use the equation PV = nRT, but solve the equation for n first.
n = (PV) / (RT)
P = 12.0 torr = 12.0/760 atm
V = 72mL = .072 L
R= .0821 L-atm/ mol-K
T = 14C = 14 + 273K= 287K
Plug in the values into the calculator to figure out the moles of H2O
n= [ (12.0/760 atm) * (.072 L)] / [ .0821 L-atm/ mol-K * 287K]
n = 4.82 x 10⁻⁵ mole of water

Multiply the moles by the molar mass of water, which is about 18.0 g/mol
4.82 x 10⁻⁵ mole * 18.0 g/mole = 8.68 x 10⁻⁴ grams of water escaped into the vapor

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bottom line, youre not a chemist. give it up kid.
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