Chapter 9 - Part 2

Chapter 9 - Part 2 - Chapter 9 Gases Lecture 20 (CHEM 1010)...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 9 – Gases Lecture 20 (CHEM 1010) March 31 Agenda: Ideal Gas Law Applying the Ideal Gas Law Dalton’s Law of Partial Pressures Kinetic Molecular Theory of Gases Graham’s Law (Diffusion and Effusion Real world gas
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Pressure 1 atm = 760 mmHg = 1 torr = 101.325 kPa Boyle’s Law V inversely proportional to P Charles’ Law V directly proportional to T Avogadro’s Law V directly proportional to n Ideal Gas Law PV = nRT R can be 8.314 J/(K mol) or 8.314 kg m2/(s2 K mol) 0.08206 L atm/(K mol) Ideal Gas Law
Background image of page 2
Problem 9.68 challenge Pure oxygen gas was first prepared by heating mercury (II) oxide, HgO: 2HgO( s ) → 2Hg ( l ) + O2( g ) What volume (in liters) of oxygen at STP is released by heating 10.57 g of HgO?
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Problem 9.69 challenge How many grams of HgO would you need to heat if you wanted to prepare 0.0155 mol of O2 according to 2HgO( s ) → 2Hg ( l ) + O2( g ).
Background image of page 4
Propane gas (C3H8) is used as fuel in rural areas. How many liters of CO2 are formed at STP by the complete combustion of the propane in a container with a volume of 15.0L and a pressure of 4.5 atm at 25°C. The unbalance equation is C3H8(g) + O2(g)  CO2(g) + H2O(l) Gas Laws and Chemical Reactions Challenge 9.12 5th
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Propane gas (C3H8) is used as fuel in rural areas. How many liters of CO2 are formed at STP by the complete combustion of the propane in a container with a volume of 15.0L and a pressure of 4.5 atm at 25°C. The unbalance equation is C3H8(g) + O2(g)  CO2(g) + H2O(l) Gas Laws and Chemical Reactions Challenge 9.12 5th
Background image of page 6
Calculate the volume of oxygen in liters at 35°C and 630.0 mmHg, that could be obtained by heating 10.0 g of potassium chlorate (KClO3). Step #2 : # mols O2 = 0.1223 mol of O2 V = ?, R = 0.0821 L atm K-1 mol-1 T = 35 + 273 = 308 K, P = 630 / 760 = 0.82895 atm V = nRT/P V = [(0.1223mol)(0.0821L•atm•K-1•mol-1)(308K)]/0.82895atm = 3.73L Gas Laws and Chemical Reactions
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Dalton’s Law of Partial Pressures Many chemical problems involve mixtures of gases and not pure gases. In the case of gas mixtures the total gas pressure can be related to the pressures of individual gas components ( partial pressures ). Dalton’s Law of Partial pressures states:
Background image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 34

Chapter 9 - Part 2 - Chapter 9 Gases Lecture 20 (CHEM 1010)...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online