logging in or signing up ideal-gas-law ankush85 Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: Embed: Flash iPad Copy Does not support media & animations WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 2372 Category: Education License: All Rights Reserved Like it (1) Dislike it (0) Added: June 15, 2009 This Presentation is Public Favorites: 1 Presentation Description For Online Photography Course, Join: http://www.wiziq.com/course/5247-photography-tips-for-beginners Comments Posting comment... Premium member Presentation Transcript The Ideal Gas Law : The Ideal Gas Law PV = nRT Ideal Gases : Ideal Gases An “ideal” gas exhibits certain theoretical properties. Specifically, an ideal gas … Obeys all of the gas laws under all conditions. Does not condense into a liquid when cooled. Shows perfectly straight lines when its V and T & P and T relationships are plotted on a graph. In reality, there are no gases that fit this definition perfectly. We assume that gases are ideal to simplify our calculations. We have done calculations using several gas laws (Boyle’s Law, Charles’s Law, Combined Gas Law). There is one more to know… The Ideal Gas Law : The Ideal Gas Law PV = nRT P = Pressure (in kPa) V = Volume (in L) T = Temperature (in K) n = moles R is constant. If we are given three of P, V, n, or T, we can solve for the unknown value. Recall, From Boyle’s Law: P1V1 = P2V2 or PV = constant From combined gas law: P1V1/T1 = P2V2/T2 or PV/T = constant Developing the ideal gas law equation : Developing the ideal gas law equation PV/T = constant. What is the constant? At STP: T= 273K, P= 101.3 kPa, V= 22.4 L/mol Mol is represented by n, constant by R: Rearranging, we get: PV = nRT Because V depends on mol, we can change equation to: At STP: (101.3 kPa)(22.4 L) = (1 mol)(R)(273K) Sample problems : Sample problems How many moles of H2 is in a 3.1 L sample of H2 measured at 300 kPa and 20°C? PV = nRT (300 kPa)(3.1 L) = n (8.31 kPa•L/K•mol)(293 K) = n = 0.38 mol How many grams of O2 are in a 315 mL container that has a pressure of 12 atm at 25°C? P = 300 kPa, V = 3.1 L, T = 293 K PV = nRT = n = 0.1547 mol P= 1215.9 kPa, V= 0.315 L, T= 298 K 0.1547 mol x 32 g/mol = 4.95 g Ideal Gas Law Questions : Ideal Gas Law Questions How many moles of CO2(g) is in a 5.6 L sample of CO2 measured at STP? a) Calculate the volume of 4.50 mol of SO2(g) measured at STP. b) What volume would this occupy at 25°C and 150 kPa? (solve this 2 ways) How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C? At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate its molar mass. 98 mL of an unknown gas weighs 0.087 g at SATP. Calculate the molar mass of the gas. Can you determine the identity of this unknown gas? Slide 7: P=101.325 kPa, V=5.6 L, T=273 K PV = nRT (101.3 kPa)(5.6 L) = n (8.31 kPa•L/K•mol)(273 K) Moles of CO2 is in a 5.6 L at STP? = n = 0.25 mol a) Volume of 4.50 mol of SO2 at STP. P= 101.3 kPa, n= 4.50 mol, T= 273 K PV=nRT (101.3 kPa)(V)=(4.5 mol)(8.31 kPa•L/K•mol)(273 K) (4.50 mol)(8.31 kPa•L/K•mol)(273 K) V = = 100.8 L Slide 8: b) Volume at 25°C and 150 kPa (two ways)? Given: P = 150 kPa, n = 4.50 mol, T = 298 K (4.50 mol)(8.31 kPa•L/K•mol)(298 K) V = = 74.3 L From a): P = 101.3 kPa, V = 100.8 L, T = 273 K Now P = 150 kPa, V = ?, T = 298 K Slide 9: How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C? PV = nRT = n = 3.97 mol P= 1000 kPa, V= 10.0 L, T= 303 K 3.97 mol x 70.9 g/mol = 282 g At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate molar mass. PV = nRT = n = 0.02845 mol P= 100 kPa, V= 1.00 L, T= 423 K g/mol = 2.506 g / 0.02845 mol = 88.1 g/mol Slide 10: 98 mL of an unknown gas weighs 0.081 g at SATP. Calculate the molar mass. PV = nRT = n = 0.00396 mol P= 100 kPa, V= 0.098 L, T= 298 K g/mol = 0.081 g / 0.00396 mol = 20.47 g/mol It’s probably neon (neon has a molar mass of 20.18 g/mol) Determining the molar mass of butane : Determining the molar mass of butane Using a butane lighter, balance, and graduated cylinder determine the molar mass of butane. Determine the mass of butane used by weighing the lighter before and after use. The biggest source of error is the mass of H2O remaining on the lighter. As a precaution, dunk the lighter & dry well before measuring initial mass. After use, dry well before taking final mass. (Be careful not to lose mass when drying). When you collect the gas, ensure no gas escapes & that the volume is 90 – 100 mL. Place used butane directly into fume hood. Submit values for mass, volume, & g/mol. Molar Mass of Butane: Data & Calculations : Molar Mass of Butane: Data & Calculations Atmospheric pressure: Temperature: For more lessons, visit www.chalkbored.com You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.