Lecture (3) Equation of State and Kinetic Theory of Ideal Gases PDF

CHAPTER TWO EQUATION OF STATE AND KINETIC THEORY OF IDEAL GASES ‫املستوي الثالث‬ )‫شعبة علوم أساسي (إجنليزي‬ Dr. Essam Gamal LECTURE THREE Ideal Gas A gas is ideal if it has the following properties: 1. It consists of a huge number of small particles (mol...

CHAPTER TWO EQUATION OF STATE AND KINETIC THEORY OF IDEAL GASES ‫املستوي الثالث‬ )‫شعبة علوم أساسي (إجنليزي‬ Dr. Essam Gamal LECTURE THREE Ideal Gas A gas is ideal if it has the following properties: 1. It consists of a huge number of small particles (molecules) at zero degrees Celsius ( 𝟎 ℃ ) and under normal atmospheric pressure (𝟏 𝒂𝒕𝒎). These molecules may consist of a single atom or a group of atoms. 2. Particles move randomly in straight lines at a uniform speed in all directions, which is called Brownian motion. 3. Collisions between molecules are elastic, meaning that: ❑ The speed (𝒗) of the particle before the collision = its speed (𝒗) after the collision. ❑ The momentum (𝒎𝒗) of the particle before the collision = the momentum (𝒎𝒗) of the particle after the collision. 𝟏 ❑ The kinetic energy ( 𝒎𝒗𝟐 ) of the particle before the collision = its 𝟐 𝟏 kinetic energy ( 𝒎𝒗𝟐 ) after the collision. 𝟐  Gas laws 𝑷 ∝ 𝑻 𝒂𝒕 𝑽 = 𝒄𝒐𝒏𝒔𝒕. 𝟏 𝑽∝ 𝒂𝒕 𝑻 = 𝒄𝒐𝒏𝒔𝒕. 𝑷1 𝑻1 𝑷 = 𝑷1 𝑽2 𝑷2 𝑻2 = 𝑷2 𝑽1 𝑽 ∝ 𝑻 𝒂𝒕 𝑷 = 𝒄𝒐𝒏𝒔𝒕. 𝑽1 𝑻1 = 𝑽2 𝑻2  Coefficients of Gas ❶ Thermal coefficient of volume expansion (𝜷): 𝟏 𝝏𝑽 𝜷= °𝑪−𝟏 ❸ Coefficient of compressibility (𝑲): 𝑽𝒐 𝝏𝑻 𝑷 𝟏 𝝏𝑽 −𝟏 “is the ratio between the rate of change of the 𝑲=− ቀ𝑵Τ𝒎𝟐 ൯ 𝑽𝒐 𝝏𝑷 𝑻 volume of the system by changing its temperature when its pressure is constant to its volume at 𝟎 ℃”. The negative sign is due to the inverse relationship between the volume and ❷ Coefficient of tension (𝜸): pressure. “is the ratio between the rate of change 𝟏 𝝏𝑷 of volume of the system by changing its 𝜸= °𝑪−𝟏 pressure when its temperature is 𝑷𝒐 𝝏𝑻 𝑽 constant to its volume at 𝟎 ℃”. “is the ratio between the rate of change of the pressure of the system by changing its temperature when its volume is constant to its pressure at 𝟎 ℃”.  The relation between the coefficients of gas 𝜷, 𝜸 and 𝑲 𝟏 𝝏𝑽 𝑽 = 𝒇 𝑷, 𝑻 𝜷= 𝜕𝑽 𝜕𝑷 𝜕𝑽 𝑽𝒐 𝝏𝑻 𝑷 =− 𝜕𝑽 𝜕𝑽 𝜕𝑷 𝜕𝑻 𝜕𝑻 𝒅𝑽 = 𝒅𝑷 + 𝒅𝑻 𝟐 𝑻 𝑽 𝑷 𝜕𝑽 𝜕𝑷 𝜕𝑻 = 𝜷𝑽𝒐 𝑻 𝑷 𝜕𝑻 𝑷 −𝑲𝑽𝒐 𝜸𝑷𝒐 = − 𝜷𝑽𝒐 At constant volume 𝟏 𝝏𝑷 𝑽 = 𝒄𝒐𝒏𝒔𝒕. 𝜸= 𝑷𝒐 𝝏𝑻 𝜷 = 𝑲 𝜸 𝑷𝒐 𝑽 𝒅𝑽 = 𝟎 𝜕𝑷 𝝏𝑽 𝝏𝑽 = 𝜸𝑷𝒐 𝟎= 𝒅𝑷 + 𝒅𝑻 𝜕𝑻 𝑽 𝝏𝑷 𝝏𝑻 𝑻 𝑷 𝟏 𝝏𝑽 𝜕𝑽 𝜕𝑽 𝑲=− 𝒅𝑷 = − 𝒅𝑻 𝑽𝒐 𝝏𝑷 𝑻 𝜕𝑷 𝜕𝑻 𝑻 𝑷 𝜕𝑽 𝜕𝑽 𝒅𝑷 𝜕𝑽 = −𝑲𝑽𝒐 =− 𝜕𝑷 𝑻 𝜕𝑷 𝑻 𝒅𝑻 𝜕𝑻 𝑷  𝜕𝑽 𝜕𝑷 𝜕𝑽 At constant volume =− 𝜕𝑷 𝑻 𝜕𝑻 𝑽 𝜕𝑻 𝑷 𝑽 = 𝒄𝒐𝒏𝒔𝒕. 𝒅𝑽 = 𝟎 𝜕𝑷 𝜕𝑽 𝜕𝑽 =− ൘ 𝜷 𝜕𝑻 𝑽 𝜕𝑻 𝑷 𝜕𝑷 𝑻 𝒅𝑷 = 𝒅𝑻 𝑲 𝜕𝑷.‫بالتكامل‬ = − 𝜷𝑽𝒐 Τ −𝒌𝑽𝒐 𝜕𝑻 𝑽 𝑷𝒇 𝑻𝒇 𝜷 න 𝒅𝑷 = න 𝒅𝑻 𝜕𝑷 𝜷 𝑲 = 𝑷𝒊 𝑻𝒊 𝜕𝑻 𝑽 𝑲 𝜷 𝑷 = 𝒇 𝑽, 𝑻 ሺ𝑷𝒇 − 𝑷𝒊 ቁ = 𝑻𝒇 − 𝑻𝒊 𝑲 𝜕𝑷 𝜕𝑷 𝒅𝑷 = 𝒅𝑽 + 𝒅𝑻 𝜷 𝜕𝑽 𝑻 𝜕𝑻 𝑽 𝚫𝑷 = 𝚫𝑻 𝑲 𝜕𝑷 𝜷 Where 𝑷𝒊 , 𝑷𝒇 are the initial and final pressure of the gas and 𝑻𝒊 , 𝒅𝑷 = 𝒅𝑽 + 𝒅𝑻 𝜕𝑽 𝑻 𝑲 𝑻𝒇 are the initial and final temperature of the gas, respectively. The relation between the coefficients of gas 𝜷, 𝜸 and 𝑲 𝜷 = 𝑲 𝜸 𝑷𝒐 𝜷 ሺ𝑷𝒇 − 𝑷𝒊 ቁ = 𝑻𝒇 − 𝑻 𝒊 𝑲 𝜷 𝚫𝑷 = 𝚫𝑻 𝑲 Where 𝑷𝒊 , 𝑷𝒇 are the initial and final pressure of the gas and 𝑻𝒊 , 𝑻𝒇 are the initial and final temperature of the gas, respectively.  Atmospheric pressure conversions ÷ 𝟏𝟎 ÷ 𝟕𝟔 𝟕𝟔𝟎 𝒎𝒎𝑯𝒈 ሺ𝑻𝒐𝒓𝒓) 𝟕𝟔 𝒄𝒎𝑯𝒈 𝟏 𝒂𝒕𝒎 𝑷𝒂 = 𝟏𝟑𝟔𝟎𝟎 × 𝟗. 𝟖 × 𝟎. 𝟕𝟔 × 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 ÷ 𝟏𝟎𝟓 𝟏. 𝟎𝟏𝟑 𝑩𝒂𝒓 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑵Τ𝒎𝟐 (Pascal) 𝟏 𝒂𝒕𝒎 = 𝟕𝟔 𝒄𝒎𝑯𝒈 𝑷𝒂 = 𝑯𝒈𝒈𝒉𝑯𝒈 𝑵Τ𝒎𝟐 ሺ𝐏𝐚𝐬𝐜𝐚𝐥) 𝑷𝒂 = 𝟏𝟑𝟔𝟎𝟎 × 𝟗. 𝟖 × 𝟎. 𝟕𝟔 = 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑵Τ𝒎𝟐 𝑩𝒂𝒓 = 𝟏𝟎𝟓 𝑵Τ𝒎𝟐 ሺ𝐏𝐚𝐬𝐜𝐚𝐥) EXAMPLE A mass of mercury is kept at a pressure (𝟏 𝒂𝒕𝒎) and a temperature of 𝟎 ℃ and at a constant volume, its temperature is raised to 𝟏𝟎 ℃, calculate what the pressure becomes if you know that the density of mercury is 𝟏𝟑𝟔𝟎𝟎 𝒌𝒈/𝒎𝟑 and its acceleration free fall 𝟗. 𝟖 𝒎/𝒔𝟐. Consider that 𝜷 and 𝐊 for mercury are constant over the temperature range used and their values are: −𝟔 −𝟏 −𝟏𝟏 𝟐 −𝟏 𝜷 = 𝟏𝟖𝟏 × 𝟏𝟎 ℃ and 𝐊 = 𝟒 × 𝟏𝟎 ሺ𝑵Τ𝒎 ). Sol. 𝑷𝒂 = 𝑯𝒈𝒈𝒉𝑯𝒈 𝑵Τ𝒎𝟐 𝑷𝒂 = 𝟏 × 𝟏𝟑𝟔𝟎𝟎 × 𝟗. 𝟖 × 𝟎. 𝟕𝟔 = 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑵Τ𝒎𝟐 𝟏 𝒂𝒕𝒎 = 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑵Τ𝒎𝟐  Equation of state for ideal gas A gas is considered ideal if it obeys Boyle’s, and Charles’s laws, pressures law, and most other laws for gases at all pressures and temperatures. Therefore, an ideal gas does not exist in nature and is a gas that follows the general gas law. Deriving the equation of state for ideal gas From Boyle's and Charles's laws we find that: 𝑷1 𝑽1 𝑷2 𝑽2 1 = 𝑽∝ 𝒂𝒕 𝑻 = 𝒄𝒐𝒏𝒔𝒕. 𝑻1 𝑻2 𝑷 𝑽 ∝ 𝑻 𝒂𝒕 𝑷 = 𝒄𝒐𝒏𝒔𝒕. 𝑷𝑽 = 𝒄𝒐𝒏𝒔𝒕. = 𝑹 𝑻 𝑻 𝑽∝ Where 𝑹 is the universal gas constant and equals 𝟖. 𝟑𝟏 𝑱/𝒎𝒐𝒍. 𝑲. 𝑷 𝑻 𝑷𝑽=𝑹𝑻 𝑽 = 𝒄𝒐𝒏𝒔𝒕. × 𝑷 For 𝒏 number of moles. 𝑷𝑽 = 𝒄𝒐𝒏𝒔𝒕. ሺ𝟏) 𝑷𝑽=𝒏𝑹𝑻 2 𝑻  𝑷𝑽=𝒏𝑹𝑻 2 𝒎 𝑵 𝒏= 𝒏= 𝑴 𝑵𝑨 𝒎 𝑷𝑽= 𝑹𝑻 Where 𝑵 is the number of molecules, 𝑵𝑨 is Avogadro's 𝑴 number and equals 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑 atom or molecule. 𝒎 𝑹𝑻 𝑵 𝑷= 𝑷𝑽= 𝑹𝑻 𝑽 𝑴 𝑵𝑨 𝒎 𝑹 𝝆= 𝒌= 𝑽 𝑵𝑨 𝑹𝑻 𝑷𝑽=𝑵𝒌𝑻 𝑷=𝝆 𝑴 Where 𝒌 is the Boltzmann constant and equals 𝟏. 𝟑𝟖 × 𝟏𝟎−𝟐𝟑 𝑱/𝑲. 𝒎 𝑷𝑽=𝒏𝑹𝑻=𝑵𝒌𝑻= 𝑹𝑻 𝑴 Mass of one (single) molecule 𝒎′ ′ 𝑴 𝐌𝐨𝐥𝐞𝐜𝐮𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝒎 = = 𝑵𝑨 Avogadro′s number Standard temperature and pressure STP (standard conditions) 𝑻 = 𝟎 ℃ = 𝟎 ℃ + 𝟐𝟕𝟑 = 𝟐𝟕𝟑 𝑲 𝑷𝒂 = 𝝆𝑯𝒈 × 𝒈 × 𝒉𝑯𝒈 = 𝟏𝟑𝟔𝟎𝟎 × 𝟗. 𝟖 × 𝟎. 𝟕𝟔 = 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑵Τ𝒎𝟐 𝑷 = 𝟏 𝒂𝒕𝒎 = 𝟏 × 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑵Τ𝒎𝟐  Summary 𝒎 𝑻𝒐𝒕𝒂𝒍 𝒎𝒂𝒔𝒔 𝒏= = 𝑴 𝑴𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒎𝒂𝒕𝒕𝒆𝒓 𝑵 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝒏= = 𝑵𝑨 Avogadro’s number 𝑵 = 𝒏 𝑵𝑨 𝑷𝑽=𝒏𝑹𝑻 𝑷𝑽=𝑵𝒌𝑻 𝑹 𝒌= 𝑵𝑨 Mass of one (single) molecule 𝒎′ ′ 𝑴 𝐌𝐨𝐥𝐞𝐜𝐮𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝒎 = = 𝑵𝑨 Avogadro′s number Standard temperature and pressure STP (standard conditions) 𝑻 = 𝟎 ℃ = 𝟎 ℃ + 𝟐𝟕𝟑 = 𝟐𝟕𝟑 𝑲 𝑷𝒂 = 𝝆𝑯𝒈 × 𝒈 × 𝒉𝑯𝒈 = 𝟏𝟑𝟔𝟎𝟎 × 𝟗. 𝟖 × 𝟎. 𝟕𝟔 = 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑵Τ𝒎𝟐 𝑷 = 𝟏 𝒂𝒕𝒎 = 𝟏 × 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑵Τ𝒎𝟐 EXAMPLES ① What is the value of the temperature at which the volume of a given mass of an ideal gas doubles at 𝟐𝟕 ℃ while the pressure remains constant? Sol. ② Calculate the mass of the carbon dioxide molecule, given that Avogadro’s number is 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 and the molecular mass of the gas is 𝟒𝟒 𝒈. Sol. ③ A vessel with capacity 𝟏𝟎𝟎 𝑳 is filled with helium gas at 𝟐𝟕 ℃ and a pressure of 𝟐𝟎 𝒂𝒕𝒎. Assuming that the gas is ideal, calculate the mass and number of molecules of helium gas used, knowing that the molecular mass of the gas is 𝟒 𝒈 , the universal constant for gases is 𝟖. 𝟑𝟏 𝑱/𝒎𝒐𝒍., Avogadro’s number 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆, the density of mercury 𝟏𝟑𝟔𝟎𝟎 𝒌𝒈/𝒎𝟑 and the acceleration of free fall 𝟗. 𝟖 𝒎/𝒔𝟐. Sol. molecule ➃ If the volume of the part of the house filled with air is 𝟓 × 𝟏𝟎𝟒 𝒄𝒎𝟑 at 𝟏 𝒂𝒕𝒎 and the temperature is 𝟐𝟕 ℃. Calculate the number of moles and molecules of air inside the house, knowing that Avogadro’s number is 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 and Boltzmann’s constant is 𝟏. 𝟑𝟖 × 𝟏𝟎−𝟐𝟑 𝑱/𝑲 and the density of mercury 𝟏𝟑𝟔𝟎𝟎 𝒌𝒈/𝒎𝟑 and the acceleration of free fall 𝟗. 𝟖 𝒎/𝒔𝟐. Sol. molecule ➄ Calculate the number of molecules in each 𝟏 𝒄𝒎𝟑 of air at standard temperature and pressure (STP), knowing that the density of mercury is 𝟏𝟑𝟔𝟎𝟎 𝒌𝒈/𝒎𝟑 , the acceleration of free fall 𝟗. 𝟖 𝒎/𝒔𝟐 and Boltzmann’s constant is 𝟏. 𝟑𝟖 × 𝟏𝟎−𝟐𝟑 𝑱/𝑲. Sol. molecule ➅ Calculate the volume occupied by one mole of oxygen under standard conditions of temperature and pressure (STP), knowing that the density of mercury is 𝟏𝟑𝟔𝟎𝟎 𝒌𝒈/𝒎𝟑 , the acceleration of free fall 𝟗. 𝟖 𝒎/𝒔𝟐 and the universal gas constant is 𝟖. 𝟑𝟏 𝑱/𝒎𝒐𝒍. 𝑲. Sol.  PROBLEMS ON STATE EQUATION FOR IDEAL GAS ❶ A vessel with capacity 𝟐𝟎𝟎 𝑳 is filled with hydrogen gas at 𝟏𝟕 𝝄 𝑪 and a pressure of 𝟏𝟎 𝒂𝒕𝒎. Assuming that the gas is ideal, calculate the mass and number of hydrogen gas molecules, knowing that its molecular mass is 𝟐 𝒈𝒎 and the universal gas constant is 𝟖. 𝟑𝟏 𝑱/𝒎𝒐𝒍. 𝑲 , Avogadro's number 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 , the density of mercury 𝟏𝟑𝟔𝟎𝟎 𝒌𝒈/𝒎𝟑 and the acceleration of free fall 𝟗. 𝟖 𝒎/𝒔𝟐. [𝟎. 𝟏𝟔𝟖 𝒌𝒈 , 𝟓. 𝟎𝟔 × 𝟏𝟎𝟐𝟓 𝒂𝒕𝒐𝒎𝒔] ❷ The volume of the portion of the house filled with air is 𝟏𝟎𝟓 𝒄𝒎𝟑 at 𝟏 𝒂𝒕𝒎 and temperature 𝟑𝟎 𝝄 𝑪. Calculate the number of moles and molecules of air present inside the house, knowing that Avogadro’s number is 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆, Boltzmann’s constant is 𝟏. 𝟑𝟖 × 𝟏𝟎−𝟐𝟑 𝑱/𝑲, the density of mercury 𝟏𝟑𝟔𝟎𝟎 𝒌𝒈/𝒎𝟑 and the acceleration of the fall The heat 𝟗. 𝟖 𝒎/𝒔𝟐. [𝟐. 𝟒𝟐 × 𝟏𝟎𝟐𝟒 𝒂𝒕𝒐𝒎𝒔 , 𝟒 𝒎𝒐𝒍] ❸ Calculate the number of molecules in each 𝟏𝟎 𝒄𝒎𝟑 at STP, knowing that the density of mercury is 𝟏𝟑𝟔𝟎𝟎 𝒌𝒈/𝒎𝟑 , the acceleration of free fall is 𝟗. 𝟖 𝒎/𝒔𝟐 and Boltzmann’s constant is 𝟏. 𝟑𝟖 × 𝟏𝟎−𝟐𝟑 𝑱/𝑲. [𝟐. 𝟔𝟗 × 𝟏𝟎𝟐𝟎 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆] ❹ Calculate the volume occupied by 𝟐 𝒎𝒐𝒍 of oxygen gas under standard conditions of temperature and pressure (STP), knowing that the density of mercury is 𝟏𝟑𝟔𝟎𝟎 𝒌𝒈/𝒎𝟑 , the acceleration of free fall is 𝟗. 𝟖 𝒎/𝒔𝟐 , Boltzmann’s constant is 𝟏. 𝟑𝟖 × 𝟏𝟎−𝟐𝟑 𝑱/𝑲 and the universal gas constant is 𝟖. 𝟑𝟏 𝑱/𝒎𝒐𝒍. 𝑲. [𝟒𝟒. 𝟖 × 𝟏𝟎−𝟑 𝒎𝟑 ]