An Application of Linear Transformations to Thermodynamics Elvin Hughes, Jr. Southeastern Louisiana University, Hammond, LA 70402
Students in physical chemistry ordinarily do not develop experience with linear transformations until they encounter group theory and quantum mechanics. This article describes a problem in thermodynamics that can be simplified by using linear transformations. Consider the triangular reversible cycle diagrammed in the figure. P and Vrefer to the pressure and volume of the working fluid. The objective is to develop a formula for calculating the work done during the execution of the reversible cycle ABCA. The method can be applied to other types of multiple-step cycles. One can factor the cycle into triangular sections and obtain the total work by summing the work obtained in each section. There are manv " wavs " to solve this tme ". of oroblem. but the objective here is to use linear transformations in develoninr!the solution. The strateev is to translate the trianrrle to th;? origin and then rotate g e triangle in a manner scch that its base is collinear with the volume axis. One can then easily calculate the area, that represents the work of the cycle, by multiplying the height by one-half of the base. The first step involves translating the triangle to the origin by subtracting PA and VA from each of the vertices of the triangle. The resulting cycle is shown in the figure, and the points of the triangle are labeled A', B', and C'. The second step involves rotating the triangle to align OC' with the volume axis. In general, if a vector r, with coordinates xl and yl and angle B is rotated clockwise by an angle A, then the new coordinates of the vector are given by the following equations ( I ) :
VOLUME, ARBITRARY A triangular reversible cycle.
AREA = U2 Vc"Ps"= 1R (-VB'Pc'+PB'Vc')
Id this example Vc', VB',PC', and Ps' become upon rotation through the angle C'A'C", Vc", V { PC", and PB", respectively The values of these components are given by the following equations:
zm
VC"= vc'vc'(vc'2 + PCf2
+~
~ , ~ +~PCP2 ' l)" ~ v ~ , ~
VC"= (vc'2 + PCf2)1/2 PC"= - v c ~ ~ c ~ l ( v+cpcc2 ' Z Im
+~
~ ' v ~ +~ l ( v ~ ' ~
PC" = 0 VBn= VBrVCXVC,2+
+~ PB"= - V ~ ' P ~ , I (+Vpcd ~ , ~)112+ ~ 1"
Example
Calculate the work, in liter atmospheres, done during the execution of a cycle that encompasses the points (1L, 5 atm), (2 L, 8 atm), and (5 L,7atm).
~ ~ ~ ~ ~ 1" l ( v ~ Solution , ~ + coordinates are (0,0), (1,3), and (4,Z) with ~ ~ + v pcA ~ I1l2~ l ( v The ~ ~relative ~
where eos (A) = V ~ ' I ( V+~PCC2 '~
sin (A) = P ~ / ( V+~PC" ' ~)m
Pc'=Pc -PA PB'=PB -PA Vc'=Vc
This formula can be arrived a t bv alternative nrocedures. However, this illustration provides an application to a familiar ~roblemand illustrates that certain auantities. such as area, are invariant with respect to rotations.
-v*
V,'= VB - VA
The area of the translated and rotated triangle can be calculated using the followingformula and the appropriate substitutions:
units of liters and atmospheres, respectively The triangle can now be rotated through an angle A, such that The components (1,3) and (4,2) are now transformed into the following: (10/20°.6, 10/Z0°.5)and 0). The area i f the transformed triangle is given by the following formula: AREA= 1/2*20~.~*10/20~.~ L atm = 5 L a h =WORK Literature Cited F.Albert. Chemiml Applicafione of Gmup Throry, 2nd ed.; John Wley: New Yar*,1911, p69.
1. Cotton,
Volume 70 Number 2 February 1993
93
