Molecular Nature of Gases and Gas Pressure PDF

## Molecular Nature of Gases and Gas Pressure ### Consider Soccer ball that we inflate - Inflate with pure nitrogen (one type of molecule). - Will fill the ball to an absolute pressure of 2 atm and observe that its mass increases by 12.8g. ### Given the mass Of nitrogen, we can use the molar mass...

## Molecular Nature of Gases and Gas Pressure ### Consider Soccer ball that we inflate - Inflate with pure nitrogen (one type of molecule). - Will fill the ball to an absolute pressure of 2 atm and observe that its mass increases by 12.8g. ### Given the mass Of nitrogen, we can use the molar mass to workout the numbers of moles. - n = mass / molar mass - N2 = 2(14.007) = 28.0 g - n = 12.8 g / 28.0 g/mol = 0.457 mol ### Determining the number of molecules: - 0.457 mol x 6.022 x 10^23 = 2.752054 x 10^23 molecules - Whenever it's asking for the number of molecules, always think Avogadro's number. - # molecules = moles x Avogadro's number ### The soccer ball contains approx 2.75 x 10^23 molecules - Enormous number, Soccer ball is not big, the molecules are tightly packed. ### Let's check if the impression is reasonable by determining how much space each molecule has on average, relative to its own size. - V = 5.5 L = 5.5 L x 1m^3 / 1000 L = 5. 5 x 10^-3 m^3 ### Easily estimate the average volume available to each molecule. - V / N = 5.5 x 10^-3 m^3 / 2.75 x 10^23 molecules = 2.00 x 10^-26 m^3 volume per molecule ### Volume per molecule = 2.00 x 10^-26 m^3 - Side length = √2.00 x 10^-26 m^3 = 2.7 x 10^-9 m = 27 A ### Length of nitrogen molecule = 1.1 A ## Postulates of Gas Behaviour ### Postulate: - Scientific term for a basic assumption that is used to derive some theoretical model. - All gases typically follow 2 postulates: ### 1. Gas is composed of very large numbers of molecules. ### 2. Molecules are separated by large distances relative to their own size. - **Ex.** if someone releases a substance with a strong smell into the room, such as opening a bottle of perfume, before long the smell will pervade the entire room. - The smell of the gas cannot spread unless the molecules are able to move around. - Gases spread on their own, they must be already in motion. - This gives another postulate: **Gas is composed of a very large number of molecules in ceaseless motion.** ### Summary of the 3 postulates: - Gas is composed of very large molecules. - Molecules separated by large distances relative to their own size. - Gas is composed of very large numbers of molecules in ceaseless motion. ## Molecular Motion ### Newton's First Law predict: - That the motion of an object (speed or direction) will not change unless an external force is applied to it. ### Ex. Soccer ball with nitrogen molecules. - Assume the ball is stationary. - Nitrogen molecules inside the ball initially move with the same velocity (same speed & direction). - The molecules are 2-dimensional motion. ### Initial Motion ### After Collision with the Wall - Molecules don't collide with one another when moving at the same speed and in the same direction. - Molecules collide with the curved wall of the ball, rebounding in a new direction. - Molecules rebound off the curved wall, changing their directions. - Rebounded molecules cross paths with others, leading into intermolecular collision. ### Elastic collision - Assume perfectly elastic (Momentum and kinetic energy conserved.) - Molecules exchange direction and speed remains unchanged. - Molecules appear to swap direction and speed remains unchanged. - Molecules will change direction and have different speeds post-collision. - Conservation Laws require the redistribution of Kinetic energy and momentum. ### Determining Molecular speed - **The area under the curve between 2 speeds gives the fraction of molecules predicted to have speeds in that interval.** - **The tail of the distribution are not symmetric.** - **Fixed lower bound of 0 m/s, since speed cannot be negative, while there is no clear upper limit to how fast a molecule might be travelling.** - **The shape of the distribution of molecular speed can be determined mathematically: Maxwell-Boltzmann distribution** - **Distribution describes the probability of molecules having particular speeds in gas** - **Molecules have speed around the average value.** - **Some molecules move extremely fast or slow.** - **Shape depends on the temp and mass of molecules:** - **High temp = high speed molecules.** - **Heavier molecules shift the distribution toward slower speeds.** ## Verification of the Distribution ### Theoretical predictions of the speed distribution were verified experimentally in 1955. ### Experimental Design: - Stream of aligned gas molecules was directed from a pressurized vessel. - Molecules pass through 2 rotating disks with slits. - Molecules only reach the detector if they reach both disks at the right time to pass through the slits. - Velocity of the molecules was calculated based on the disk's rotational speed. - By varying the rotational speed, the entire speed distribution was measureed. ## Maxwell-Boltzmann Speed Distributions ### Speeds of ideal gas molecules obey the Maxwell-Boltzmann speed distribution. ### Average speed - **Most probable speed:** maximum value of distribution function by setting the derivative equal to zero - **Average speed:** expected value of distribution by evaluating / *ufcuwdu*... ### Root-mean Square speed ## Effect of Mass and Temperature ### Distribution molecules depend on 2 variables: - Temperature - Molar mass ### **Ex. Comparing the distribution of O2(g) molecules at 273K and 1000K.** - The range of the speed broadens, and the distribution shifts to the right towards higher speed. ### **If we compare O2 and H2 at 273K, the lighter the molecules of gas, the broader the range of speed.** - Higher temp = higher speed. - Lighter molecules = higher speed. ## Postulates of Kinetic Molecular Theory of Gases - Gas composed of very large numbers of molecules in ceaseless motion and the molecules are separated by large distance relative to their own size. - Molecules are in ceaseless random straight-line motion; behave like hard spheres that undergo perfect elastic collision with one another. - No interaction occur during "near miss" encounters with other molecules. - If there is no force of attraction or repulsion between molecules. ### Postulates of Kinetic Molecular Theory of Gases - Gases are composed of very large numbers of molecules in ceaseless random-straight-line motion. - Molecules are separated by large distances relative to their own size. - Molecules behave like hard spheres that undergo prefect elastic collision with one another and the wall of the container. - There is no forces of attraction or repulsion between molecules. ## Gas Pressure ### P = F / A - **When the molecule collide with the ball, it rebound. X-component of its velocity change from +ve to -ve, no change in y- and z-component of velocity. Implies that the momentum change during collision of -2mVx.** - **Molecules must therefore experience an impulse. That force multiplied by time (F x t) causes the momentum to change. Pressure is due to reactive force experienced by the wall as a result of this impulse.** - Momentume Change during collision = molecule experience an impulse. Impulse is (Fxt) cause the momentume to change. Wall experience a reactive force due to this impulse. Pressure is the result of cumulative reactive force from many molecular collision with the walls. - **P = F/A**, F is the reactive force due to change in momentum molecule collide with the wall and A is the area of the wall. - **P = Fwall/A** - Fwall is reactive force due to collisions of molecules with the wall - A is the area of the wall ### Impulse associated with a single collision. - In a short time, At, any molecule with x-component of velocity, Vx will be able to reach and collide with the wall if it is within a distance VxAt from that wall. - If the wall has area A, then this defines a volume AvxAt, in which gas molecule can reach and collide with the wall. - **Because we assumed that the number of gas molecules, N, is large and the motion is random it is reasonable to expect the molecules to be evenly distributed throughout the volume: N molecules per unit volume throughout the container. # of collision with the wall in this short time, is the # of molecules per volume, times the volume containing molecules close enough to collide with the wall AvxAt.** - Motion is random, therefore only half of molecules are travelling toward the wall. - Collision during At = (1/2) x (AvxAt) - FavAt = (1/2) x (AvxAt) x (-2mvx) - Fwall At = (1/2) x (AvxAt) x (-2mvx) - Fwall = (1/2) x (Avx) x (-2mvx) - **Fwall = MNAvx²/V** - P = Fwall/A - **P = MNAvx²/AV = Nmvx²/V** ### Accounting for Distribution of Molecular speeds - We know, molecules have the same velocity and travel at different speed. - **P = MN<V²>/V** - We only considered molecular movement in one direction. Since postulated that molecular motion is random, the pressure must be the same in all direction. ### Average Square Speed: - **<V²> = <Vx²> + <Vy²> + <Vz²>.** - Since molecular motion is random, the average in each differention must be the same: <Vx²> = <Vy²> = <Vz²>. **Average Square Speed is 3 times the average square speed in any one direction: <V²> = <Vx²> + <Vy²> + <Vz²> = 3<Vx²>.** - **C² as the mean-Square molecular speed: C² = <V²> = <Vx²> + <Vy²> + <Vz²> = 3<Vx²>.** - **P = Nmc² / 3V**. N=# of molecules. m mass of the molecules. ### mtotal = Nm = RM - **P = mC² / 3V** - **PV = nMc² / 3** - **n = # of moles. M = molar mass** ## Lesson 2: The Ideal Gas Laws and Applications ### Ideal gas Law expressed as - **PV = nRT**. P-pressure, V=volume, n=amount of gas (moles), T=temp, R=gas constant ### Avogadro's Law - When temp and pressure is constant, the volume of the gas is directly proportional to the number of moles (n). - ** V α n (for fixed T & P) ** - Ex: Doubling the number of gas molecules from 30 to 60 while keeping T and P constant result in a double volume. ### Boyle's Law: - Fixed amount of gas (n) at constant temp, the pressure (p) and volume (v) are inversely proportional. - **PV = constant (for constant n & T)** - Example: reducing the volume of gas by half, double its pressure. ### Charles Law: - Constant pressure and fixed gas amount (n), the volume is directly proportional to the temperature (T) in Kelvin. - **V α T** - Example: Doubling the temp, doubles the volume of the gas. - Charles observed that when plotted on a volume vs. temp (inkelvin) graph, the lines of different gases converge at the same temp. - **Combining all the Law Leads to: PV = nRT** ### Comparing the ideal Gas Law to the Kinetic Molecular Theory model - **PV = nRT** -> Ideal Gas Law It is derived from by Avogadro, Boyle, Charles Law -> interested in large-scale behaviour of the system. - **PV = (1/3)MC²** -> Kinetic-Molecular Theory Model It is based on a set postulates or assumption about molecular matter. The average effect of large numbers of microscopic interaction. - **Kinetic Definition of temperature:** T = (2/3k) * (<MC²>) = (2/3R) * (<MC²>) = (1/3) * (MC²) ### Temperature of Ideal Gas: - Measure of its average translational kinetic energy. ### Since temp is a measure of Kinetic energy, does not matter what molecules are in the statement of the ideal gas law. - Ideal gas law could be cooled to 0K, then the molecules would have an average speed of 0 m/s, stopped moving absolute zero. ## Example problem 1 - A sample of monochlorethylene gas has a density of 2.56 g/L at 22.8°C and 1 atm. What's the molar mass? - **PV = nRT** - **n x M = m** M = mass/moles - **M = m / n = (2.56 g/L) / (0.041176 mol) = 62.17 g/mol** - **n = PV / RT = (1 atm x 1 L) / (0.08206 x 295.95K ) = 0.04117656 mol** - Assume Density can give us the mass: **m = PV = (2.56g / 1L) x 1 L = 2.56 g** - **M = MRT / PV = (2.56g x 0.08206 x 295.95K) / (1 atm x 1L ) = 62.17 g/mol** - **Note: You have to pick the right R constant according to the units given.** ## Example Problem 2 - **3.75 g sample of a mixture of KClO3 and KCl is decomposing by heat.** - **2KClO3(s) -> 2KCl(s) + 3O2(g)** - **119 ml of oxygen (O2) gas is produced, measured at 0°C and 0.98692 atm.** - **What is the percentage by mass of KClO3 in the original mixture?** - **0°C + 273.15K = 273.15K** - **PV = nRT** - **n = PV/RT = (0.98692 atm) (0.119 L) / (0.08206)(273.15K) = 5.240 x 10^-3 mol O2** - **mass of KClO3 from O2** - **m = n x molar mass** - **Ratio of KClO3:O2 - important to always do the Ratio.** - **m = (5.240 x 10^-3) (2/3) x 122.6 g/mol = 0.42839 g KClO3** - **%KClO3 = (0.42839 g) x 100 / 3.75 g = 12%** - **%KClO3 = (mass calculated x 100) / (Original mass)** ## Standard Condition - **STP = 0°C = 1 bar = IUPAC** - **SATP = 25°C = 1 bar = IUPAC** - **NTP = 20°C = 1 atm = NIST** ## Example Problem 3 - **Dry air pressure of 35.0 psi in the summer, when the temp is 30.0°C. Assume the volume is constant, what would the pressure be in winter when the temp drops -20.0°C?** - **V = Constant** - **PV = nRT, n = constant** - **Initial condition:** P1V = nRT1 - **Final Condition:** P2V = nRT2 - **PV = nRT** - **PV = nRT2** - Since V, R, n are constant we can reduce the formula. - **P1/T1 = P2/T2** - **Solving for the final P2.** - **P2 = (P1 x T2) / T1** - **P2 = (35.0 psi x 253.15K) / 303.15K = 29.2 psi** - **P1V = nRT1, P2V2=nRT2** - **R is alway constant and n is fixed** - **P1V/T1 = P2V2/T2** for constant n. ## Concept Check 4 - **6NaNO2 (aq) + 3H2SO4 (aq) → 4NO(g) + 2HNO3 + H2O(l) + 3NaSO4(aq)** - **What volume (in L) of 0.646 M aqueous NaNO2 should be used to produce 5.0 L of NO at 20°C and 0.970 atm?** - **PV = nRT** - **n = PV/RT = (0.970 atm)(5.0 L) / (0.08206)(293.15) = 0.2016138 mol NO** - **M = moles of solute / volume of solution** - **Volume = moles / M = (0.201632 mol) / (0.646 M) = 0.31181 L** - **Moles of NaNO2 from NO (known)** - **(0.201638 mol NO) * (6 mol NaNO2 / 4 mol NO) = 0.302457 mol NaNO2**