![]() We can compare this situation to that of a casino, where the outcomes of the bets are random and the casino’s takings fluctuate by the minute and the hour. However, because a huge number of molecules collide with the wall in a short time, the number of collisions on the scales of time and space we measure fluctuates by only a tiny, usually unobservable fraction from the average. ![]() In a sample of gas in a container, the randomness of the molecular motion causes the number of collisions of molecules with any part of the wall in a given time to fluctuate. A force is thus exerted on the wall, creating pressure. Similarly, if the average velocity of the molecules is higher, the gas pressure is higher.įigure 2.9 When a molecule collides with a rigid wall, the component of its momentum perpendicular to the wall is reversed. As the number of molecules increases, the number of collisions, and thus the pressure, increases. These collisions are the source of pressure in a gas. (Later, we discuss the validity of this assumption for real monatomic gases and dispense with it to consider diatomic and polyatomic gases.)įigure 2.9 shows a collision of a gas molecule with the wall of a container, so that it exerts a force on the wall (by Newton’s third law). Then, we can assume that the atoms have no energy except their translational kinetic energy for instance, they have neither rotational nor vibrational energy. Second, we begin by considering monatomic gases, those whose molecules consist of single atoms, such as helium. First, we let the container be a rectangular box. We make still further assumptions that simplify the calculations but do not affect the result. Furthermore, if the velocities of gas molecules in a container are initially not random and isotropic, molecular collisions are what make them random and isotropic. They do not disturb the derivation either, since collisions between molecules moving with random velocities give new random velocities. The collisions between molecules do not appear in the derivation of the ideal gas law. Other forces on them, including gravity and the attractions represented by the Van der Waals constant a, are negligible (as is necessary for the assumption of isotropy). The molecules make perfectly elastic collisions with the walls of the container and with each other.In other words, we take the Van der Waals constant b, the volume of a mole of gas molecules, to be negligible compared to the volume of a mole of gas in the container. The molecules are much smaller than the average distance between them, so their total volume is much less than that of their container (which has volume V).To derive the ideal gas law and the connection between microscopic quantities such as the energy of a typical molecule and macroscopic quantities such as temperature, we analyze a sample of an ideal gas in a rigid container, about which we make two further assumptions: The molecules obey Newton’s laws and are in continuous motion, which is random and isotropic, that is, the same in all directions.There is a very large number N of molecules, all identical and each having mass m.First, we make two assumptions about molecules in an ideal gas. We can gain a better understanding of pressure and temperature from the kinetic theory of gases, the theory that relates the macroscopic properties of gases to the motion of the molecules they consist of. Pressure is the force divided by the area on which the force is exerted, and temperature is measured with a thermometer. We have examined pressure and temperature based on their macroscopic definitions. Solve problems involving the distance and time between a gas molecule’s collisions.Solve problems involving mixtures of gases. ![]()
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