Gas Laws

Variables influencing gases Pressure (P) Three units (mm Hg, atm, kPa) Volume (V) Units (L or mL) Temperature (T) Units (K) Moles (n) mol = n

Pressure Conversion 760 mm Hg = 1 atm = 101.3kPa Also called standard pressure Example: The atmospheric pressure in MI is around 750mm Hg. Determine the pressure in atmospheres and kPa. 750 mm Hg = 0.987 atm = 100 kPa

Temperature Conversion °C + 273 = Kelvin Standard temperature is at 273 K Example: If the temperature is 47°C, what is the temperature in Kelvin? 47°C + 273 = 320K

Boyle’s Law Robert Boyle described the relationship between pressure (P) and volume (V) of a gas. As V ↓, P ↑. As V ↑, P ↓. This is an inversely proportional relationship. When one variable goes up, the other goes down.

Boyle’s Law P1V1 = P2V2 Boyle Visual

Boyle’s Law - Example The gas in a balloon has a volume of 7.5L at 100kPa. The balloon is released into the atmosphere, and the gas expands to a volume of 11L. Assuming temperature is constant, what is the new pressure? Answer: 68 kPa

Charles’ Law Jacques Charles observed the relationship between volume (V) and temperature (T). As T ↑, V also ↑. As T ↓, V ↓. This is a directly proportional relationship. When one variable goes up, the other goes up too.

Charles’s Law

Charles’s Law V1/T1 = V2/T2 Charles Visual

Charles’s Law - Example A gas sample occupies 2.5L at 300.0K. What volume will the gas occupy at 80.0K? Answer: 0.67L

Guy-Lussac’s Law Joseph Guy-Lussac studied the relationship between pressure (P) and temperature (T). As T ↑, P also ↑. As T ↓, so did P. This is a directly proportional relationship. As one variable goes one way, so does the other.

Guy-Lussac’s Law P1/T1 = P2/T2

Guy-Lussac’s Law - Example A gas has a pressure of 6.58kPa at 539K. If the temperature is decreased to 211K at a constant volume, what will the resulting pressure be? Answer: 2.58 kPa

Sources http://www.nasa.gov/images/content/122625main_k_sun2.jpg http://www.unit5.org/christjs/Gas_La1.gif http://boomeria.org/chemtextbook/fig18-6.jpg&imgrefurl=http://boomeria.org/chemtextbook/cch18.html&h=325&w=449&sz=23&hl=en&start=20&sig2=piNW_iTpdOE8TcWP8JVBMw&um=1&tbnid=XNrhJ7VDxtnPZM:&tbnh=92&tbnw=127&ei=8ry4R47DNYrKiAGi1aGoCQ&prev=/images%3Fq%3Dcharles%2Blaw%26gbv%3D2%26um%3D1%26hl%3Den%26sa%3DG http://www.algebralab.org/practice/practice.aspx?file=Reading_BoylesLaw.xml http://www.nasa.gov/images/content/122625main_k_sun2.jpg http://www.coolest-gadgets.com/wp-content/uploads/3.jpg http://www.aerostar.com/jpeg_historical_photos/Dr-Pepper-Swirl-Bottles_Lg.jpg http://bp3.blogger.com/_ovJS1Em-6dg/RlL7ARCWRUI/AAAAAAAAIkY/yy1k3BTJtDQ/s1600-h/balloon1.jpg http://jholmes.net/wordpress/wp-content/uploads/2007/11/faa_flights.jpg http://www.balloonatics.biz/_wp_generated/wp11bae3c4.jpg http://epicself.com/wp-content/uploads/2007/10/high_heel_xray_300x510shkl.jpg

Combined Gas Law Putting them together

Combined Gas Law Often volume, temperature, and pressure all change at the same time. Deals with a fixed amount of gas. T represents temperature in Kelvin Units must match on both sides

Example A gas at 155 kPa and 25°C occupies a container with initial volume of 1.00 L. By changing the volume, the pressure of the gas increases to 605 kPa and the temperature is raised to 125°C. What is the new volume? Answer: 0.342 L

STP Standard pressure is at 1atm Standard temperature is 273 K STP stands for standard temperature and pressure.

Sources http://www.infoorganizers.com/fileadmin/template/main/images/puzzle-pieces-2.jpg http://www.cardsunlimited.com/largeimage/Balloons.jpg http://imagecache2.allposters.com/images/pic/NIM/AF577~Balloons-Posters.jpg

Partial Pressures

Dalton’s Law of Partial Pressures The total pressure of a gas mixture is the sum of the partial pressures of the component gases. Ptotal = P1+P2+P3+P4+…

Example A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.9 kPa. If PO2=6.6 kPa and PN2=23.0 kPa, what is PCO2? Answer: 3.3 kPa

Sources http://www.kinneybrothers.com/FLASH%20CARDS/confused.jpg http://www.learner.org/channel/courses/essential/physicalsci/images/s4.dalton.jpg http://beconfused.com/blog/2008/01/19/julian-beevers-chalk-optical-illusion-part-2/

Ideal Gas Law The Culmination

R R is the ideal gas constant Its value depends on the units used for pressure. If P is in atm:

Ideal Gas Law PV=nRT

Example You fill a rigid steel container (volume = 20.0 L) with nitrogen gas to a final pressure of 2.00 * 104 kPa at 28°C. How many moles of nitrogen gas does the cylinder contain? Answer: 1.60 * 102 mol N2(g)

Example – Round Two What pressure will be exerted by 0.450 mol of a gas at 25°C if it is contained in a 0.650 L vessel? Answer: 16.9atm

Ideal Gas Law Practice 

There are 8g H2 gas in a balloon There are 8g H2 gas in a balloon. What is the volume at 25°C and 758mmHg?

If the total air pressure is 0.99atm, the partial pressure of CO2 is 0.05atm, and the partial pressure of hydrogen sulfide is 0.02atm, what is the partial pressure of the remaining air?

30g of a gas occupy a volume of 82. 0mL at 3. 00atm and 27°C 0.30g of a gas occupy a volume of 82.0mL at 3.00atm and 27°C. Calculate the molar mass of the gas.

2H2(g) + O2(g)  2H2O(g) How many liters of water can be made from 55g of O2(g) at 12.4atm and 85°C?

At what pressure would 0.150mol N2(g) at 23.0°C occupy 8.90L?

How many grams of carbon dioxide gas are present in a closed 3 How many grams of carbon dioxide gas are present in a closed 3.0L container if the gas is exerting a pressure of 131.69kPa at 28°C?

Graham’s Law of Effusion

Diffusion Movement of particles from areas of higher concentration to lower concentration

Kinetic Energy The average kinetic energy of molecules in a gas depends only on temperature. KE = 1/2mv2 The velocity of the gas depends on its mass.

Effusion The process where molecules of a gas in a container randomly pass through a hole in the container.

Graham’s Law Equation

Larger GFM = slower rate of effusion. Graham’s Law Larger GFM = slower rate of effusion.

Answer: CO2 effuses 0.9X as fast as HCl. Example Compare the rate of effusion of carbon dioxide with that of hydrogen chloride at the same temperature and pressure. Answer: CO2 effuses 0.9X as fast as HCl.

Example - Again If a molecule of neon gas travels at an average of 400m/s at a given temp., estimate the average speed of a molecule of butane gas, C4H10, at the same temp. Answer: 235m/s

Sources http://www.coletechnologies.us/files/effusion_box.JPG http://www.moleplace.com/images/mole02.gif