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of actual problems by graphical methods. This paper is intended 

 to suggest methods for obtaining such solutions. 



Figure I (Boyle's Law). — Draw two lines OX and OY at 

 right angles. Suppose we have a mass of gas whose volume is 

 Vj at the pressure P lt Take OA to represent V x and OB to 

 represent P x . Also take AA t = OA and BB 3 = OB. Join 

 Aj B x and through A and B draw lines parallel to OY and OX 

 meeting each other and A } B x at the point Q. Now AAj represents 

 Vj and BB X represents P lt Take BB 2 to represent any other 

 pressure P 2 . Join B 2 Q and produce it to meet OX in A a . Then, 

 as will be easily seen from the similarity of the triangles, B 2 BQ 

 and QAA 0 , AA., represents V 2j the volume under the new pressure 

 P 2 . 



We can easily obtain from this construction any desired 

 number of points on an isothermal curve. Along OX and OY 

 (Figure II) take OA and OB to represent the. volume and 

 pressure of a mass of gas, say that which occupies unit volume 

 at unit pressure. Draw lines through A and B parallel to the axes 

 and let them intersect at O 1 . Take any number of points B lf B 2 , 



B s along OY, their distances from B representing various 



pressures, and draw lines from them through O 1 cutting OX 



at A 1( A 2) A 3 The distances of these from A will 



represent the corresponding volumes. By drawing vertical lines 

 through each of the " A " points and horizontals through each of 



the " B " ones we obtain a series of intersections Q lt Q 2 , Q 3 



which will be points on a PV curve whose axes are OX 1 and OY 1 . 

 By taking enough of these and drawing a smooth curve through 

 them we obtain the isothermal we want. 



Figure III (Law of Charles). — Take axes as before and let 

 OA t represent the volume of a given mass of gas under any assigned 

 (constant) pressure and at an absolute temperature represented by 

 OB v Join Aj B 1# If OB 2 , 0B 3) etc. represent other temperatures 

 on the absolute scale the corresponding volumes will be found by 

 drawing through B 2 , B 3l etc. lines parallel to By Aj and cutting OX 

 at the points A 2 , A 3 , etc. Then 0A 2r 0A 3 , etc. represent the 

 volumes required. 



Figure IV. (Boyle's Law). —An alternative method for finding 

 the volume corresponding to any assigned pressure or conversely, 

 i.e., for problems in Boyle's Law, is shown in Figure IV. Let 

 OAj be the volume corresponding to the pressure OBj. It is 

 required to find the volume corresponding to the pressure OB 2 . 

 Join Aj B 2 and through B x draw a line parallel to A. x B 2 cutting 

 OX in A 2 . Then 0A 2 will represent the required volume. This is 

 evident from the similarity of the triangles OB x A 2 , 0B 2 A lt in 

 which 



OA 2 : 0A X = 0B 2 : OB 2 



i.e., the volumes are inversely as the pressures. In like manner if 

 OB 3 represent a third pressure and it is required to find the cor- 

 responding volume, we join A 2 B 3 and through Bj draw a line 

 parallel to A-j B 3 cutting OX in A s . Then 0A 3 represents the 



