1) Unit conversions: Express the temperature on the Celsius and Kelvin scales. Express the pressure in atmospheres. (Hint: First convert inches to millimeters, and use that fact that a pressure of one mm mercury is the same as one torr.) Express the velocity in meters per second.
2) Use the formula for the root mean square speed of a gas from the modern kinetic theory of gases to estimate the speed of a molecule of hydrogen at this temperature and pressure. If we take this as the correct value, how close was Joule's estimate?
3) [recommended only for students who have had physics] Let us make an estimate of the speed along the lines of Joule's estimate, but using MKS units. The essential idea was that the pressure of the gas is due to the pounding of the molecules against the walls of its container. How fast would the mass of hydrogen in a vessel have to be moving to impart a pressure of one atmosphere? For simplicity, consider a cube exactly one meter on a side containing H2 at 289 K and one atmosphere (1.013x105 Pa) pressure. The force exerted on a wall of the cube, then, is the pressure times the surface area,
(1.013x105 Pa)(1 m2) = 1.013x105 N.a) What is the mass of H2 in this cube? Now suppose that this mass is concentrated in three particles of equal mass, each bouncing back and forth between a different pair of walls at a constant speed. Each time a particle collides with the wall, its momentum is changed by 2mv, where m is the mass of the particle and v the constant speed. (The factor of 2 results from the fact that the speed changes from v to -v when the particle changes direction, i.e., a change of 2v.) The force on a wall, then, is 2mv times the frequency of collisions. (Recall that force has dimensions of mass x length/(time)2 or momentum/time; thus force is the momentum of each collision times the number of collisions per unit time.) If the constant speed is v meters per second, then the particle will collide with a wall v times per second (because the distance between collisions is 1 meter). Half of these collisions will be with each wall, so the number of collisions per second with each wall is v/2. The force, 1.013x105 N, is numerically equal to mv2.[1]
4) [recommended only for students who have had physics] Joule did not use algebra, just arithmetic, in his calculation, which he expressed as follows:
Let us suppose an envelope of the size and shape of a cubic foot to be filled with hydrogen gas, which, at 60° [Fahrenheit] temperature and 30 inches barometrical pressure, will weigh 36.927 grs. Further, let us suppose the above quantity to be divided into three equal and indefinitely small elastic particles, each weighing 12.309 grs.; and, further, that each of these particles vibrates between opposite sides of the cube, and maintains a uniform velocity except at the instant of impact; it is required to find the velocity at which each particle must move so as to produce the atmospherical pressure of 14,831,712 grs. on each of the sides of the cube. In the first place, it is known that if a body moving with the velocity of 32 1/6 feet per second be opposed, during one second, by a pressure equal to its weight its motion will be stopped, and that, if the pressure be continued one second longer, the particle will acquire the velocity of 32 1/6 feet per second in a contrary direction. At this velocity there will be 32 1/6 collisions of a particle of 12.309 grs. against each side of the cubical vessel in every two seconds of time; and the pressure occasioned thereby will be 12.309 x 32 1/6 = 395.938 grs. Therefore, since it is manifest that the pressure will be proportional to the square of the velocity of the particles, we shall have for the velocity of the particles requisite to produce the pressure of 14,831,712 grs. on each side of the cubical vessel, v = (14,831,712/395.938)1/2 x 32 1/6 = 6225 feet per second.a) Use the ideal gas law to compute the mass of a cubic foot of hydrogen in grains, if 1 gram = 15.43 grains.
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