. A text book of physics, for the use of students of science and engineering . axes parallel to 420 HEAT CHAP. the edges of the cube in Fig. 408. Resolve V into componentsalong these axes ; this may be done by first resolving V into com-ponents OM and OK, the latter being in the plane containing OX andOZ. OK is then resolved into velocities OL and ON along OX and OZrespectively. The components are thus u along OX, ux along OYand u2 along OZ. From the geometry of the figure, OA2 =OK2 + KA2 = OL1 + LK2 +OM2 = OL2+ON2+OM2 ; .-. \f2=u2 + u1z + u22. The velocities of the other molecules have differ
. A text book of physics, for the use of students of science and engineering . axes parallel to 420 HEAT CHAP. the edges of the cube in Fig. 408. Resolve V into componentsalong these axes ; this may be done by first resolving V into com-ponents OM and OK, the latter being in the plane containing OX andOZ. OK is then resolved into velocities OL and ON along OX and OZrespectively. The components are thus u along OX, ux along OYand u2 along OZ. From the geometry of the figure, OA2 =OK2 + KA2 = OL1 + LK2 +OM2 = OL2+ON2+OM2 ; .-. \f2=u2 + u1z + u22. The velocities of the other molecules have different directions, but all can be resolved into com-ponents parallel to the edgesof the cube. If u-f and w22have the same relation to ufand w22 respectively that ft2has to w2, as explained above,and if V2 has the same relationto V2, V2 is the mean ofthe squares of the actual velo-cities, then we may write V2=w2+%2+w22. Since there is no tendencyfor molecules to accumulate in any part of the cube, it is reasonableto suppose that the velocities u, ux and u2 are equal; hence. O ru L X Fig. 409.—Component velocities of a molecule. W- ?U-. ?U0 lv2. 1 -2 3V (2) Hence, in the case of a centimetre cube containing n molecules ofa gas moving in all directions, we have from (1) and (2) p=±nmSf2 dynes per sq. cm (3) If the cube has a volume v, each cubic centimetre containing «molecules, then pv = ~nmv\/2. (4) Now inn is the total mass of the molecules in one cubic centimetre, nm is the density d of the gas, and d x v is the total mass M in a volume v 5 .\ ^ = ^MV2 (5) If the cube contains unit mass of gas. M is unity, and pv = ^\f2 (6) xxxn AVOGADROS LAW 421 Some important relations.—It is known from Boyles law that theproduct of the pressure and volume of a given mass of gas is constant,provided the temperature is constant, and it has just been shownthat this product depends upon V2, since M in (5) above is it may be inferred that V2 is constant if the temperat
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