{"id":3748,"date":"2018-08-15T04:14:35","date_gmt":"2018-08-15T11:14:35","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=3748"},"modified":"2018-08-15T04:14:35","modified_gmt":"2018-08-15T11:14:35","slug":"kinetic-theory-of-gases","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/kinetic-theory-of-gases.htm","title":{"rendered":"Kinetic theory of gases","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">The k<\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">inetic theory states that every substance consists of atoms or molecules and that the atom or molecule moves continuously carelessly. This assumption of kinetic theory matches the situation and condition of the atom or molecule of the gas constituent. The force of attraction between the atoms or molecules making up the gas is feeble so that atoms or molecules can move freely.<\/span><\/span><!--more--><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">When moving, atoms or molecules have speed. Atoms or molecules also have mass. Because it has mass (m) and velocity (v), then the atom or molecule has kinetic energy (KE) and momentum (p). Kinetic energy: KE = 1\/2 m v<\/span><\/span><sup><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2<\/span><\/span><\/sup><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">. While momentum: p = m v. In addition to kinetic energy and momentum, there is also force (F). When moving freely, collisions must occur. So, force raises because of changes in momentum when a collision occurs. Kinetic energy, momentum, and impulse forces are the core of our discussion of the dynamics (the laws of motion, impulse, and momentum). We can say that the kinetic theory of gases applies dynamics at the atoms or molecules level.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><b>Ideal Gas Concept (based on macroscopic properties of gas)<\/b><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">In the discussion of the laws of gas, it has been explained about three quantities which state the macroscopic nature of real gas. The three quantities in question are Temperature (T), Volume (V) and pressure (P). The relationship between these three macroscopic quantities is stated in Boyle&#8217;s Law, Charles&#8217;s law, and Gay Lussac&#8217;s law. Kindly note that these three laws only apply to real gas that has a pressure and density (density = mass\/volume) that is not too large. These three laws also only apply to a real gas whose temperature is not near the boiling point.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Boyle&#8217;s law, Charles&#8217;s law, and Gay-Lussac&#8217;s law do not apply to all real gas conditions so that we can make an ideal gas model. Ideal gas does not exist in everyday life; the ideal gas is just the perfect form that is made to help our analysis, much like ideal rigid and fluid bodies. So, we consider Boyle&#8217;s law, Charles law and Gay-Lussac law apply to all ideal gas conditions. The existence of an ideal gas model helps us review the relationship between macroscopic quantities of gas.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">The ideal gas law is expressed in two equations, PV = nRT (ideal gas law in the number of moles) and PV = NkT (ideal gas law in the number of molecules). We assume that the ideal gas meets these two equations. In other words, the ideal gas law applies to all ideal gas conditions, whether the ideal gas pressure or mass is huge or when the ideal gas temperature approaches the boiling point. In contrast, the ideal gas law does not apply to all real gas conditions. The ideal gas law only applies when the pressure and density of the real gas are not too large. The ideal gas law also only applies when the temperature of the real gas is not near the boiling point. Based on this brief description, we can say that real gas has similar properties with ideal gas only when the real density,<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> and <\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">gas pressure are not too large and when the real gas temperature is not near the boiling point.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">The concept of the ideal gas that has been described above is reviewed based on macroscopic properties. Although an ideal gas is only an ideal model, an ideal gas is still considered a gas that consists of atoms or molecules that move freely. Therefore, it would be nice if we also discuss the ideal gas concept from a microscopic perspective.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><b>Ideal Gas Concept (based on microscopic properties of gas)<\/b><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">The following are some brief descriptions that describe the microscopic conditions of the ideal gas, which are based on the Gas Kinetic Theory:<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1. An ideal gas consists of particles, called molecules. The ideal gas molecules can include of one atom or several atoms. Each molecule has mass (m) and moves randomly in any direction at a specific rate (v).<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2. The distance between each molecule is higher than the diameter of each molecule.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">3. These molecules obey the laws of motion and interact with each other when collisions occur.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">4. Collisions between molecules and molecules or between molecules with a container wall are the perfect collision, and each collision happens in a concise time.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">In the perfect collision, the law of conservation of energy applies (energy before collision = energy after collision) and the law of conservation of momentum (momentum before collision = momentum after collision).<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><b>The review of impulse-collision for the kinetic theory of gases<\/b><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Review the quantitative relationship between the macroscopic quantities and the microscopic quantities of the gas. The quantities that state the macroscopic nature of gas are temperature (T), volume (V) and pressure (P). While the quantities that say the microscopic nature of gas are velocity or velocity (v), momentum (p), force (F) and kinetic energy (EK) of atoms or molecules making up the gas.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-3749\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-1.png\" alt=\"Kinetic theory of gases 1\" width=\"202\" height=\"196\" \/>We review some gas molecules in a closed container. The side length of the box = l and the cross-sectional area = A.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Molecules have mass (m) and, when moving, the molecule has velocity (v). Because the container is closed, there is a possibility of collisions between molecules and the wall of the vessel that has a surface area of A.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">To simplify the analysis, we only the collision that occurs on the left wall (the wall that is parallel to the z-axis). First, we discuss the collisions experienced by one molecule. Call it molecules 1. The molecular mass = m1 and the speed of movement = v1. The direction of movement to the left is set to be negative, while the direction of movement to the right is set to be positive.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">We can assume that before colliding the container wall, the molecular motion is parallel to the x-axis and the direction of the movement to the left. Therefore, there is a component of speed on the x-axis which is negative (-v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1x<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">). Because it has mass (m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">) and speed (-v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1x<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">), the molecule has momentum (p<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = -m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1 <\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1x<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">). This is the initial momentum. When colliding with a wall, the molecule provides an action force on the wall. Because there is an action force, the wall provides a reaction force. The reaction force from the wall causes the molecule to be reflected on the right. Because of the direction of movement to the right, the molecular velocity component is positive (v<sub>1x<\/sub>). The molecular momentum after the collision is p<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">2<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1x<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">. This is the final momentum.<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">The magnitude of the change in momentum due to collisions is:<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Total momentum = final momentum\u2014initial momentum<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">p total = p<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">2<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> &#8211; p<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1<\/span><\/sub><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">p total = m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1x<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> &#8211; (-m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1x<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">)<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">p total = 2m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1x<\/span><\/sub><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">2m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">1x<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = total momentum for one collision. Because molecular collisions are the perfect collision, the collisions occur not only once, but repeatedly. In the perfect collision, the law of conservation of energy and the law of conservation of momentum applies. Energy and momentum before collision = energy and momentum after the collision. Therefore, the molecule will never stop moving (endless energy). Molecular speed also never decreases (continued momentum).<\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">After colliding with the left wall, the molecule moves to the right until it hits the right wall. After colliding the right wall, the molecule moves back to the left and hit the left wall again. Because the length of the side of the box = l, then after colliding the left wall for the first time, the molecule will travel a distance of 2l before colliding the left wall a second time. When moving as far as 2l, molecules must require a certain time interval (<\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">\u0394<\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">t). The amount of time interval (<\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">\u0394<\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">t) that the molecule needs to move as far as 2l, is mathematical:<\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3750\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-2.png\" alt=\"Kinetic theory of gases 2\" width=\"112\" height=\"107\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">\u0394t is the time interval between each collision. When colliding a wall, the molecule provides an action force on the wall. Because of the action force, the wall provides a reaction force. The existence of this reaction force makes the molecule move again to the right. In this case, the direction of molecular movement changes. At first, the molecule moves to the left (-v1x), after colliding the wall, the molecule moves to the right (v<sub>1x<\/sub>). Changes in direction of movement cause changes in momentum (final momentum &#8211; initial momentum = m<sub>1<\/sub> v<sub>1x<\/sub> &#8211; (-m<sub>1<\/sub> v<sub>1x<\/sub>) = 2m<sub>1<\/sub> v<sub>1x<\/sub>). We can say that the change in momentum occurs because of the total force is given by the wall. The amount of total force given by the wall, mathematically:<\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-3751\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-3-127x300.png\" alt=\"Kinetic theory of gases 3\" width=\"127\" height=\"300\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/08\/Kinetic-theory-of-gases-3-127x300.png 127w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/08\/Kinetic-theory-of-gases-3.png 157w\" sizes=\"auto, (max-width: 127px) 100vw, 127px\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">In the box above, only one molecule is described. This does not mean that only one gas molecule in the box. In reality, there are many gas molecules. The amount of total force for all gas molecules in the box, mathematically:<\/span><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">F = F<\/span><\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span lang=\"en-US\">1 <\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">+ F<\/span><\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span lang=\"en-US\">2<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"> + F<\/span><\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span lang=\"en-US\">3<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"> + &#8230;.. + F<\/span><\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">n<\/span><\/span><\/span><\/sub><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">F<\/span><\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span lang=\"en-US\">1<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"> = total force for molecule 1<\/span><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">F<\/span><\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span lang=\"en-US\">2 <\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">= total force for molecule 2<\/span><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">F<\/span><\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span lang=\"en-US\">3 <\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">= total force for molecule 3<\/span><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">&#8230;&#8230; = and so on<\/span><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">F<\/span><\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">n <\/span><\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">= total force for molecules 4<\/span><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">n = the last molecule.<\/span><\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-3752\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-4-300x85.png\" alt=\"Kinetic theory of gases 4\" width=\"300\" height=\"85\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/08\/Kinetic-theory-of-gases-4-300x85.png 300w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/08\/Kinetic-theory-of-gases-4.png 317w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = molecular mass 1, m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">2<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = molecular mass 2, m<\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><sub>3<\/sub> <\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">= molecular mass 3, mn = mass of the last molecule. m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">1<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> + m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">2<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> + m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">3<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> + &#8230;.. + m<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">n<\/span><\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = m (the mass of gas in the box). l = length of the side of the box.<\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3753\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-5.png\" alt=\"Kinetic theory of gases 5\" width=\"268\" height=\"64\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">12x <\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">= molecular speed 1, v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">22 x<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = molecular speed 2, v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">33 x<\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"> = speed of molecule 3, v<\/span><\/span><sub><span style=\"font-family: Times new roman, serif\">n2 x <\/span><\/sub><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">= speed of the last molecule. The speed of each molecule varies, so we need to calculate the average velocity of all molecules. To calculate the average velocity of a molecule, we can divide the speed of all molecules by the number of molecules. In the kinetic theory of gas, the number of molecules is usually given the symbol N. Mathematically, the average speed of all molecules is written:<\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3754\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-6.png\" alt=\"Kinetic theory of gases 6\" width=\"265\" height=\"119\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">W<\/span><span lang=\"en-US\">e combine <\/span><span lang=\"en-US\">the <\/span><span lang=\"en-US\">equation b with <\/span><span lang=\"en-US\">the <\/span><span lang=\"en-US\">equation a :<\/span><\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3755\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-7.png\" alt=\"Kinetic theory of gases 7\" width=\"105\" height=\"45\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><i>F = force, m = mass of gases, l = length of the side of the box, N = number of molecules.<\/i><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">In the previous explanation, assumed the movement of molecules parallel to the x-axis. This presupposition is made to easy our analysis. In reality, all gas molecules in a box do not move in any direction at random. Because the motion occurs randomly, besides having an average velocity component on the x-axis, the molecule also has an average velocity component on the y-axis or z-axis. Thus, the average velocity of a gas molecule = the total number of components of the average velocity on the x-axis, y-axis, and z-axis. Mathematically:<\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3756\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-8.png\" alt=\"Kinetic theory of gases 8\" width=\"172\" height=\"58\" \/><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3756\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-8.png\" alt=\"Kinetic theory of gases 8\" width=\"172\" height=\"58\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Because molecules move randomly, the velocity component on the x-axis, y-axis, and z-axis has the same magnitude. Mathematically:<\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3757\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-9.png\" alt=\"Kinetic theory of gases 9\" width=\"118\" height=\"52\" \/><\/p>\n<p class=\"western\" lang=\"en-US\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">We combine <i>the <\/i><i>equation 2<\/i> with <i>the <\/i><i>equation 1<\/i> :<\/span><\/span><\/p>\n<p lang=\"en-US\" align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3758\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-10.png\" alt=\"Kinetic theory of gases 10\" width=\"166\" height=\"200\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">W<\/span><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">e combine <\/span><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"><i>the <\/i><\/span><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"><i>equation <\/i><\/span><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"><i>3<\/i><\/span><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"> with <\/span><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"><i>the <\/i><\/span><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"><i>equatio<\/i><\/span><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\"><i>n c <\/i><\/span><\/span><\/span><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><span lang=\"en-US\">:<\/span><\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3759\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-11.png\" alt=\"Kinetic theory of gases 11\" width=\"126\" height=\"183\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><i>F = the amount of force exerted by gas molecules on the wall of the container, which has a surface area of A.<\/i><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><b>The relationship between pressure (P) and microscopic magnitude<\/b><\/span><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\">Pressure (P) is a quantity that states the macroscopic nature of gas. Review Pressure based on the microscopic properties of the gas. The amount of pressure given by the gas molecule on the wall that has cross-sectional area A is:<\/span><\/span><\/p>\n<p align=\"justify\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-3760\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/08\/Kinetic-theory-of-gases-12-115x300.png\" alt=\"Kinetic theory of gases 12\" width=\"115\" height=\"300\" srcset=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/08\/Kinetic-theory-of-gases-12-115x300.png 115w, https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/sites\/28\/2018\/08\/Kinetic-theory-of-gases-12.png 147w\" sizes=\"auto, (max-width: 115px) 100vw, 115px\" \/><\/p>\n<p class=\"western\" style=\"text-align: justify\" align=\"justify\"><span style=\"font-family: Times new roman, serif\"><span style=\"font-size: medium\"><i>P = pressure, N = number of gas molecules, m = mass, v = Average speed of a molecule, V = volume of container<\/i><\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>The kinetic theory states that every substance consists of atoms or molecules and that the atom or molecule moves continuously carelessly. This assumption of kinetic theory matches the situation and condition of the atom or molecule of the gas constituent. The force of attraction between the atoms or molecules making up the gas is feeble &#8230; <a title=\"Kinetic theory of gases\" class=\"read-more\" href=\"https:\/\/gurumuda.net\/physics\/kinetic-theory-of-gases.htm\" aria-label=\"Read more about Kinetic theory of gases\">Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Kinetic theory of gases","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[2],"tags":[],"class_list":["post-3748","post","type-post","status-publish","format-standard","hentry","category-basic-physics-tutorials"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3748","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=3748"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3748\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=3748"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=3748"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=3748"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}