Gas Law1.docx

Andrea Nicole S. Ramchand

Gas Laws 

table of contents 1.

Kinematic Molecular Theorem 2. Boyle’s Law 3. Charles’ Law 4. Gay-Lussac’s Law 5. Combined Gas Law 6. Avogadro’s Law 7. Ideal Gas Law

Kinematic Molecular Theorem Submitted To: Maam Elena B. Mohammad Submitted By: Jade Danielle S. Mihammad 

states that :

-gas particles are in constant motion and exhibit perfectly elastic collisions -gas particles are in constant, random, straightline motion -gas particles don’t attract or repel each other -can both explain Charles’ and Boyle’s Law -the average kinetic energy of a collection of gas particles is directly proportional to absolute temperature only

Boyle’s Law -states that pressure and volume are related at constant temperature -as one goes up, the other goes down - PV

= K-P1V1= P2V2

Sample Problems: 1. A balloon with a volume of 2.0 L is filled with a gas at 3 atmospheres. If the pressure is reduced to 0.5 atmospheres without a change in temperature, what would be the volume of the balloon? Solution: PiVi = PfVf where Pi = initial pressure Pf = final pressure Vf = final volume Vi = initial volume To find the final volume, solve the equation for Vf: Vf = PiVi/Pf Vi = 2.0 L

Pi = 3 atm Pf = 0.5 atm Vf = (2.0 L) (3 atm) / (0.5 atm) Vf = 6 L / 0.5 atm Vf = 12 L Answer: The volume of the balloon

will expand to 12 L.

2. An ideal gas occupying a 2.0 L flask at

760 torr is allowed to expand to a volume of 6,000 mL. Calculate the final pressure in atm.

Charles’ Law -states that volume of a gas varies directly with the absolute temperature at constant pressure -V = KT - V1/T1 = V2/T2 Sample Problems:

Answer:

1. Calculate the new volume, if in a container there is a mass of gas that occupies a volume of 1.3 liters, at a temperature of 280 K. Calculate the volume when reaching a temperature of 303 K.

V 1 = 1.3 l. T 1 = 280 K V 2 =? T 2 = 303 K

Answer: 1.41 liters.

2. If we have a gas that at 10 degrees Celsius occupies 2.4 liters, calculate the final temperature, if at the end it occupies 2.15 liters. V1 T1 V2 T2

= 2.4 l = 10 ° C = 283 K = 2.15 l =?

-states that at constant volume, pressure and absolute temperature are directly related.

-P = KT -P1T1/P2/T2 Sample Problem:

The new temperature is 253 K, which is equal to -20 ° C.

0.370 atm 1. A gas has a pressure of 0.370x atm at = ––––– 50.0 °C. What is–––––––– the pressure at standard Avogadro’s 273 K Law temperature? 323 K Solution: -states that at constant temperature and

Gay-Lussac’s Law

pressure, the volume of a gas is directly xrelated = 0.313 atm to the number of moles

-V=Kn

-P1V1/T1 = P2V2/T2

-V1 = n1/V2 = n2

Sample Problems: 1) A sample of sulfur dioxide occupies a volume of 652 mL at 40.° C and 720 mm Hg. What volume will the sulfur dioxide occupy at STP?

Sample Problem:

Combined Gas Law -the combination of2)Boyle’s, Charles’ Use Avogadro's Law: and Gay Lussac’s V1 / n1 = V2laws. / n2

1. A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.) Solution:

/ 0.500product mol = 2.70 Lof /x -states that the ratio2.00ofL the x = and 0.675 mol pressure and volume the absolute 3) Compute grams ofto He added: temperature of a gas is equal a 0.675 mol - 0.500 mol = 0.175 mol constant

1) Convert grams of He to moles: 0.175 mol x 4.00 g/mol 2.00 g / 4.00 g/mol = 0.500 mol

2) A sample of argon has a volume of 5.0 dm3 and the pressure is 0.92 atm. If the final 1) P1 = 720 mm P2 = 760 mm V1 = 652 mL temperature is 30.° C, the final volume is 5.7 V2 = ? L, and the final pressure is 800. mm Hg, what was the initial temperature of the argon?  T1 = 40.° C + 273 = 313 K  T2 = 0° C + 273 = 273 K  P1V1/T1 = P2V2/T2  V2 = P1V1/T1 x T2/P2  V2 = 720 mm x 652 mL x 273 K/(313 K x 760 mm) Answer: = 540 mL SO2

2) P1 = 0.92 atm P2 = 800. mm V1 = 5.0 dm3, V2 = 5.7 L, T2 = 30.° C + 273 = 303 K

Answer = 0.7 grams of He added T1 = ?  P1V1/T1 = P2V2/T2 T1 = P1V1/P2 x T2/V2 T1

1.) P = 1.00 x 10-6 mm Hg 273 K V = 985 mL L·atm/mol·K

T = 0.0° C + 273 = R = 0.0821

 PV = nRT n = PV/RT  n = 1.00 x 10-6 mm x 1 atm/760 mm x 985 mL x 1 L/103 mL/ (0.0821 L·atm/mol·K x 273 K) = 5.78 x 10-11 moles N2  nmolecules = 5.78 x 10-11 moles N2 x 6.02 x 1023 N2 molecules/1 mol N2

-the equation of state of a hypothetical ideal gas -a good approximation of the behavior of many gases under many conditions -a combination of Boyle’s, Charles’, GayLussac’s, and Avogadro’s laws

Answer:

-PV = nRT

= 3.48 x 1013 N2 molecules

Sample Problems:

1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?

Ideal Gas Law

2) Calculate the mass of 15.0 L of NH3 at 27° C and 900. mm Hg.

2) P = 900. mm Hg

T = 27° C + 273 = 300 K

V = 15.0 L

R = 0.0821 L·atm/mol·K

 PV = nRT n = PV/RT  n = 900. mm x 1 atm/760 mm x 15.0 L/(0.0821 L·atm/mol·K x 300 K)  n = 0.721 moles NH3 x 17.04 g NH3/1 mol NH3 Answer: = 12.3 g NH3