Chapter 6 TAVSS

Chapter 6 TAVSS - CHEM 161-2007 CHAPTER 6 - THERMOCHEMISTRY...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHEM 161-2007 CHAPTER 6 - THERMOCHEMISTRY PRACTICE PROBLEMS DR. ED TAVSS • Heat and work • Enthalpy • Calorim. & Heat capacity (Also, see Chapter 10 - “Phase Changes and Diagrams”) • Hess’s Law (Also see Chapter 8 - “Bond Energies & Lengths”) • Enthalpy of form’n HEAT AND WORK 15 Chem 161-2006 Final Exam Chapter 6 – Thermochemistry Heat and Work What volume change will a system undergo if it absorbs 4.55kJ of heat and ΔU is 3.26kJ and the external pressure is 0.833atm? A. The system will be compressed by 15.3 L B. The system will be compressed 1.55L C. The system will expand by 1.55L D . The system will expand by 15.3 L E. The system will expand by 1.29L ∆E = q + w ∆E = q – P∆V ∆V = (∆E –q)/-P ∆V = (3260J – 4550J)/(-0.833 atm) Convert J into Latm 3260J x (1Latm/101.3J) = 32.18 Latm 4550J x (1Latm/101.3J) = 44.92 Latm ∆V = (32.18 Latm – 44.92 Latm)/(-0.833 atm) = +15.29 L Since ∆ means final minus initial, then +15.29 L means an expansion of 15.29 L. Note: I think that the key, which says “B”, is incorrect. D CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 6 THERMODYNAMICS HEAT AND WORK 10. A gas absorbs 125 J of heat and expands from 2.00 L to 5.00 L under a constant pressure of 1.00 atm. Calculat ∆ U for this process. (a) 122 J (b) 429 J (c)-179 J (d) -122 J (e) 347 J ∆E = q – P∆V q = +125J-P∆V = -1.00 atm x (5.00L – 2.00L) = -3.00 Latm Convert to Joules:-3.00 Latm x (101.3 J/Latm) = -304 J ∆E = +125J -304J = -179J 19 Chem 161-2005 Hourly Exam II Chapter 6 – Thermodynamics Heat and work Assume the sign convention of the text, in which w is positive when work is done on the system. In which of the following processes is the sign of “w” negative? X. A gas is compressed under 2.0 atm pressure Y. A substance undergoes combustion in a bomb calorimeter (constant volume). Z. 1 mole of liquid water vaporizes to 1 mole of gaseous water under atm pressure A. X only B. X and Y only C . Z only D. Y only E. X and Z only X. When a gas is compressed this is work done on the system, and the energy of the system increases; w is positive. Y. When a substance undergoes combustion in a bomb calorimeter, there is no change in volume. Since w = -p∆V, and ∆V = 0, then w = 0, which is neither positive nor negative. Z. When 1 mole of liquid water vaporizes to 1 mole of gaseous water under atm pressure, the gas expands from 0 L to 22.4 L. Expansion is positive work on the environment, and resulting increase in energy of the environment, so it is therefore negative work and negative energy on the system. 37 Chem 161-2005 Final exam CHAPTER 6 - THERMODYNAMICS Heat and work Which reaction should have a ∆ E>0 under conditions of constant temperature and pressure?...
View Full Document

This note was uploaded on 07/17/2010 for the course 160 162 taught by Professor Kimmel during the Spring '10 term at Rutgers.

Page1 / 74

Chapter 6 TAVSS - CHEM 161-2007 CHAPTER 6 - THERMOCHEMISTRY...

This preview shows document pages 1 - 5. Sign up to view the full document.

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