CHEMISTRY GAS LAWS REVIEW

TASK 1

APPLICATION OF GAS LAWS

• IDEAL GAS LAW
• PV=nRT
• Diffusion:even distribution of particles to fill a space

VOLUME PRESSURE AND TEMPERATURE

• As volume increases, temperature increases
• As volume increases, pressure decreases
• As temperature increases, pressure increases

TASK 2

GAS LAW EXAMPLE
Hot air balloons fly when the air inside the hot air balloon is less dense than the air surrounding it. Hot air is less dense than cool air; the heated air causes the balloon to rise simply because it is lighter than an equal volume of cold air.

TASK 3

DIFFUSION
Diffusion is the process of a substance spreading out to evenly fill its container or environment. In a solution, a concentrated solute diffuses to spread evenly in its solvent. ... Diffusion occurs because the gas molecules are in continuous random motion.

EFFUSION
Effusion refers to the movement of gas particles through a small hole. Graham's Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.

DIFFERENCE
DIFFUSION is the process by which molecules move and travel from one place to another without requiring bulk motion. EFFUSION is the process by which molecules travel through a pinhole from a place of high concentration to low concentration. EFFUSION refers to the ability of gas to travel through a small pin-hole.

TASK 4

REFERENCE SHEET

• Ideal gas law equation
• Combined gas law equation
• Pressure units
• Pressure conversions
• Kelvin equation
• Celsius equation
• Universal gas constant

TASK 5

GAS LAW NOTES

• P1=1.05atm
• V1=5.0 L
• T1= 293 K
• Combined gas law((p1)(v1)/t1)=(p2)(v2)/(t2))
• 20.0 degrees C +273=293K
• -15 degrees C+273=258
• (1.05atm)(5.0 L)/(293K)=(0.65atm)(x)(258 K)
• X=7.11L

TASK 6

Example:
A 47.948 L sample of CO experiences a change in pressure from 1.345 atm to 4.2409 atm. Identify the initial volume.
The initial volume is 47.948 L

Example
A gas has a volume of 4 liters at 50 Celsius. What will its be (in Liters) at 100 degrees Celsius? Which variable(s)do not remain constant in this problem?

TASK 7

IDEAL GAS LAW
(relations to airbags)
Airbags work with the ideal gas law every time they explode. When airbags react with sodium azide, it heats up very quickly, and a large amount of nitrogen gas is made. Before this reaction, the sodium azide was a solid, so there was no gas. By reacting these two chemicals to create nitrogen gas, moles of gas are added to the system.

Chemistry of airbags
Once the sensor has turned on the electrical circuit, a pellet of sodium azide (NaN3) is ignited. A rapid reaction occurs, generating nitrogen gas (N2). This gas fills a nylon or polyamide bag such that the front face of the bag travels at a velocity of 150 to 250 miles per hour. This process, from the initial impact of the crash to full inflation of the airbag, takes only about 40 milliseconds (Movie 1). Ideally, the body of the driver or passenger should not hit the airbag while it is still inflating. In order for the airbag to cushion the head and torso with air for maximum protection, the airbag must begin to deflate (i.e., decrease its internal pressure) by the time the body hits it.

IDEAL GAS LAW

• P=pressure
• V=volume
• n=number of moles of gas
• R=gas constant
• T=temperature

TASK 8

EXAMPLE
at a certain temperature, 3.24 moles of CO2 gas at a 2.15 atm take up a volume of 35.28 L. What is this temperature in Celsius?

SOLUTION

• P=2.15 atm
• V=35.28
• n=3.24 mol
• R=(0.0821 L)(atm)(K)(mol)
• T next

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• T=(2.15 atm)(35.28 L)/3.24 mol x (K)(mol)/(0.0821 K(atm)
• =285.1536K
• (3 sig figs)=285K
• Convert to Celsius=285K-273=12 degrees Celsius

TASK 9

EXAMPLE: If I have 17 moles of gas at a temperature of 67 degrees C, and a volume of 88.89 L, what is the pressure of the gas?

Untitled Slide

• P=
• V=88.89 L
• n=17 moles
• R=(0.0821 L)
• T=67 degrees Celsius

Convert temp to kelvin:
67 degrees Celsius +273=340K

P=(17 moles)(340K)/88.89L x(0.0821L)(atm)/(K)(moles)
=
474.538/88.89=5.338

2 sig figs=5.338=5.3
P=5.3atm

N represents moles in the ideal gas equation.
R represents the universal gas constant in the ideal gas law equation convert to kelvin

TASK 11

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LAB REPORT

• Slope of Boyle’s Law is indirect
• As volume increases, pressure decreases

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LAB REPORT

• Slope of Charles Law is indirect
• As volume increases, temperature increases

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LAB REPORT

• Slope of lussacs law is direct
• As temperature increases, pressure increases

STP=standard temperature and pressure=0degrees Celsius and 1 atm
At STP, 1 mol of gas takes up 22.4 L of volume

EXAMPLE
What volume would 0.735 moles of O2 gas occupy at 1 atm of pressure and 0 degrees Celsius
(0.735 moles)(22.4 L)/(1 mol)= 16.5 L

TASK 13

1mol of gas contains 22.4L