APSC 132 MULTIPLE CHOICE REVIEW
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A gas absorbs 0.0 J of heat and then performs 15.2 J of work. The change in internal energy of the gas is: (a) -24.8 J (b) 14.8 J (c) 55.2 J (d) -15.2 J (e) none of these Under conditions of constant pressure, the heat f

APSC 132 PROBLEM SET 1
1. 0.500 mol of an ideal gas expands at a constant temperature of 0oC from an initial pressure of 5.000 atm to a final pressure of 0.500 atm. Calculate q, w, E and H for both a reversible expansion and for an irreversible expansion.

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APSC 132 PROBLEM SET 1 SOLUTIONS
Note: In this problem set, no chemical reactions occurred in the system. Problem 1 Isothermal reversible This means that T = 0. Since H=ncpT and E=ncvT, H = E =0 Therefore, 0 = q + w or q = -w wsystem = -nRTln(P1/P2)= -(

APSC 132 PROBLEM SET 2
1. 1.00 mol of nitrogen gas expands adiabatically and irreversibly from an initial pressure of 5050 kPa and a temperature of 500oC against an outside pressure of 101 kPa. Calculate q, w. E and H. (Cp = 29.3 J/mol.K) 2. At 101kPa and

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APSC 132 PROBLEM SET 2 SOLUTIONS
Problem 1 The expansion is adiabatic and irreversible.
Pgas
Vgas V1 = n = 1.00 mol P1 = 5050 kPa T1 = 773 K Cv = Cp R = 21.0 J/mol.K V2 = n = 1.00 mol P2 = 101 kPa T2 =
V1 = nRT1/P1 = (1.00)(8.31)(773)/5050 L = 1.27 L Be

Information about APSC 131 chemistry
Welcome to each and every one of you. The following information is to assist you in
your transition into first year Queens engineering.
Course Instructors
Professor Bill Newstead will teach weeks 1 4 of the course
Pr

Chemistry 132 Lecture 12 Thermodynamic Description of the Equilibrium State
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Thermodynamics views a chemical reaction as the spontaneous flow of atoms from reactants to products carried out at constant T and P and requires that G<0 to be spontaneous. Whe

Chemistry 132 Lecture 11 Equilibrium Continued
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Thermodynamic Equilibrium Constant We have discussed the equilibrium constants Kc and Kp. However, we often run into the thermodynamic equilibrium constant, K. K is a unit less constant found by using ratio

132 Midterm Examination Equations and Data
R = 8.314 J/mol.K = 8.314 kPa.l/mol.K = 0.0821 atm.L/mol.K STP = standard temperature and pressure = 273.2 K and 101.3 kPa SATP = standard ambient temp. and press. = 298 K and 100.0 kPa Equations for finding E, H

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Queens University Faculty of Applied Science APSC 132 Final Examination April 2001
YOUR NAME: _
STUDENT NUMBER: _
HAND IN TEST PAPER
Time: three hours Only approved gold sticker calculators are admitted. Answer all questions in the space prov

Chemistry 132 Lecture 10 Introduction to Equilibrium (Oxtoby Chapter 9)
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Chemical Equilibrium
Chemical reactions left to themselves ultimately come to a state of equilibrium. Chemical equilibria display all of the following characteristics:
The system is