GASES (Part 2)
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Transcript of GASES (Part 2)
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Equal volumes of gases at the Equal volumes of gases at the same T and P have the same T and P have the same number of molecules.same number of molecules.
V = knV = kn
V and n are directly related.V and n are directly related.
twice as many twice as many moleculesmolecules
MOLEY…MOLEY…MOLEY…MOLEY…MOLEY!!!MOLEY!!!
V1 = V2n1 n2
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STP Values
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PV=kPV=k
V/T=kV/T=k
P/T=kP/T=k
V/n=kV/n=k
=k=kPVPV
nTnTthenthen
k is given the variable R, and is called the Ideal Gas Constant
So…
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Brings together gas properties.
Can be derived from experiment and theory.
BE SURE YOU KNOW THIS EQUATION!
P V = n R TP V = n R T
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P = PressureV = VolumeT = TemperatureN = number of moles
R is a constant, called the Ideal Gas Ideal Gas ConstantConstant
Instead of learning a different value for R for all the possible unit combinations, we can just memorizememorize oneone value and convert the units to match R.convert the units to match R.
R = 0.08206R = 0.08206
L • atm mol • K
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How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 250 C?
Solution1. Get all data into proper units V = 27,000 L T = 250 C + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg)
= 0.980 atmAnd we always know R, 0.08206 L atm /
mol K
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How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 250 C?
SolutionSolution2. Now plug in those values and solve for
the unknown.
PV = PV = nnRTRT
n = 1.1 x 10 -3 mol
RT RTRT RT
n = (0.980 atm) (27,000 L) _ (0.08206) (298 K)
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Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?
P(20.0 L) = (2.86moles)(0.08206)(296K)P = 3.47 atmP = 3.47 x 760 mmHg 1 atmP = 2640 mmHg
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A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?
(.97atm)(5.0L)=n(0.08206)(293K)4.85 = 24.04n.20 moles = n.20 moles O2 x 32.00 g O2
1 mole= 6.4 grams O2