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1 Unit 3 The 3 rd planet in the solar system, Earth Ch. 10: Boyle s Law This law shows the relation between pressure and volume of a gas in a closed system at a constant temperature. Volume and pressure are inversely proportional The equation is constant temperature where k is a constant for the gas in a closed system at a This equation can be extended to when the gas is under the same temperature in a closed system, as k would be the same. Given a gas that occupies 2.00 L with a pressure of 1.58 atm, find the pressure if the volume is increased 4.00 L. First, it is easiest to solve for the unknown variable, P 2. Then, we plug in the numbers L is the first volume and the first pressure is 1.58 atm. The second volume is given as 4.00 L and the second pressure is unknown. Simplify the equation next, getting atm. Charles Law This law shows the relationship between volume and temperature in a closed system at a constant pressure Temperature and volume are directly proportional When using this equation, temperature must be in Kelvin

2 The equation is where k is the constant for the gas in a closed system at a constant pressure Like Boyle s Law, this law can be extended to At 20 C, a gas occupies a volume of 10.0 L. The temperature is raised and the volume increases to 15.0 L. Find the new temperature in degrees Celsius. K= =293K C= = 166 C First, you must convert the temperature to Kelvin. Next, it is helpful to solve for the unknown value. Then plug in the values; 10.0 L for the first volume, 15.0 L for the second volume, and the temperature found for the first temperature. Finally, simplify the equation to get 439.5K, then convert to degrees Celsius to get 166 C. Gay-Lussac s Law This law shows the relationship between pressure and temperature in a closed system at a constant volume Pressure and temperature are directly proportional Temperature must be in Kelvin The equation is in a closed system at a constant volume

3 Like Boyle s Law and Charles Law, this law can be extended so that Given a gas at 30 C, the pressure of this gas is measured to be.50 atm. Find the pressure at 10 C. K=C K 1 = =303K K 2 = =283K First, convert the temperatures to kelvin. Solve for the unknown value, and then plug the values into the equation. Finally, simplify, finding the final pressure to be 0.47 atm. Combined Gas Law This law combines the laws of Boyle, Charles and Gay-Lussac into one law, so long as it is in a closed system The relationships of the variables as previously mention hold true in this law (temperature directly proportional to volume and pressure, volume and pressure inversely proportional) Temperature must be in Kelvin The equation is in a closed system Like the previous laws, this law can be extended so that

4 Given a gas at 20 C that in a 5.00 L container has a pressure of 2.00 atm, find the volume of the gas at STP conditions. STP is 1 atm and 0 C K=C K 1 = =293.15K K 2 = =273.15K First, as with previous examples, you must put the temperature into the correct units of Kelvin. Then solve the equation for the unknown value, and plug in values. Simplify the equation to get a volume of 9.01 L. Dalton s Law To find the total pressure, add up all the partial pressures The duh law Avogadro s Hypothesis: Equal volumes of all gases under the same temperature and pressure have the same number of moles Therefore, as long as it is the same temperature and pressure Given.0500mol of O 2 occupies 2.00L at a certain temperature and pressure, find the volume if the number of moles was increased to.100mol.

5 Solve for the unknown value, which is the second volume. Then plug in all the values into the equation and simplify. Ideal Gas Law This law combines all the gas laws so that if given 3 conditions of the 4 (pressure, volume, temperature, and number of moles) you could find the last variable. Temperature must be in Kelvin The equation is R can equal kpa L/mol K, or atm L/mol K depending on which unit of pressure is used Given a gas at 20 C has a pressure of.500 atm and occupies a volume of 1.00 L, find the number of moles of gas are present. K=C K= = Convert to correct units first; in this case, convert Celsius to kelvin. Next, solve the equation for the unknown value which is n. Then plug in the numbers and simplify.

6 Effusion- escape of gas molecules through a tiny hole Diffusion- spread of one substance through a space In this unit, we will apply diffusion to gases. To find the comparative speed of different molecules of a gas, Graham s Law is used which relates rate and molar mass. The equation is where r is the rate and M is the molar mass of each molecule or atom. A mixture of CO 2 and N 2 diffuse through a room. What is the rate of diffusion if the rate of diffusion for N 2 is 35.7mol/s? Molar mass of CO 2 is Molar mass of N 2 is First you must calculate the molar mass of each molecule using the molar mass of each element in the formula. Then solve for the unknown value and plug in numbers and simplify. Application Find the molar volume of Propane at STP using experimental data provided: Mass of gas Volume of gas Room temperature Barometric pressure Water vapor pressure.0500 g.0277 L 20.0 C 760. mmhg 17.5 mmhg

7 P i = PV=nRT n= mol First, find the pressure inside by subtracting water vapor pressure from the barometric pressure. Then convert this to atmospheres. Using this pressure, and the volume and temperature recorded, find the number of moles I used the lab we did, but changed the numbers. Determine the molar mass of a gas given the information below: 2HCl + CaCO 3 CaCl 2 CO 2 H 2 O Mass of CaCO 3 ribbon 0.206g Volume of CO 2 gas.154l Height of water tower Room temperature 20.0 C Water vapor pressure 17.5mmHg Barometric pressure 762mmHg

8 22.4 L/mol First, find the number of moles of calcium carbonate from the mass used and divide by the molar mass. Then, using the BCE, you can find the moles of CO 2 produced. Next, you need to find the pressure inside using the equation, and use this information to find the volume at STP (1 atm and 0 C). Then, divide this volume by the moles found previously. I used the lab we did and changed the numbers and used a different reaction found here: arbonate_and_hydrochloric_acid

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