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Chapter 9 - Gases Cheat Sheet by

chem200 gases cheat sheet to use when studying

Gas Laws

Boyle's Law
Charles's Law
Vi÷Ti = Vf÷Tf
Combined Gas Law
PiVi÷Ti = PfVf÷Tf
Avogadro's Law
Ideal Gas Law
Dalton's Law of Partial Pressure

Dalton's Law of Partial Pressure

Partial Pressure
Pgas(atm)=­(total pressure x molesgas)÷total moles
PP when volumes are different
Mole fraction
moles of gas ÷ total moles
Wet Gas
Pwet gas=Ptotal-PH2O
then use PV=nRT to solve for variables

Real Gases

Van der waal's equation
P=[(nR­T)÷­(V-nb)] - [(a*n2)÷(V2)]
When comparing real gases
a gas with a larger "­a" value will require the largest correction to account for interm­ole­cular forces
a gas with a smaller "­b" value will behave most ideally at high pressures
If Vdw's pressure is lower than the ideal pressure, attractive forces dominate
If Vdw's pressure is higher than ideal pressure, repulsive forces dominate
Real Gas Behavior
attractive forces between molecules cause a decrease in pressure
As molecules increase in size deviations from ideal behavior become apparent at relatively HIGH temps
In general, most gases behave most ideally at HIGH temps and LOW pressures

Pressure Units and Conver­sions

1 atm=
1 atm (R= .08206)
760 mmHg (R= 62.364)
760 torr
1.013x105 Pa
101.3 kPa
29.92 inches Hg
14.69 psi
1.01325 bar

Stoich­iometry and Gases

Mole ratio = Volume ratio
2mL A:3mL B

Kinetic Molecular Theory

If temper­ature is increased, Pressure and KE increase by a factor of Tf÷Ti and rms increases by a factor of √Tf÷Ti
If volume is increased, Pressure increases by a factor of Vi÷Vf while KE and rms increase by a factor of 1 (because they are not affected))
If moles are increased, pressure increases by a factor of nf÷ni, while KE and rms increase by a factor of 1 (no change)

Using Ideal Gas Law to Calculate Gas Properties

Ideal Gas Law
0 degrees celcius, 273 degrees Kelvin, 1 atm, 22.4 L/mol
d=MP÷RT where M is molar mass
When not given volume, but told to assume ideal gas behavior, use V=1L

Diffusion and Effusion

G1=gas 1
G2=gas 2
Average Kinetic Energy
KEG1 = KEG2 when TG1=TG2
Molecular Speed
√u2 = √3RT÷M where M is the molar mass
√u2G1 ÷ √u2G2 = √MG2 ÷ √MG1
d/dx G1 ÷ d/dx G2 = √MG2 ÷ √MG1
tG2 ÷ tG1 = √MG2 ÷ √MG1


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