Kinetic Molecular Theory - Why the Gas Law Does Not Depend on Mass

TP The rms speed, 𝑣 “Use units, Luke!” 𝑅 1. 2. 3. 4. 5. 3𝑅𝑇/𝑀, of O2 at 25℃ is about … 8.314 8.314 / 250 m/s 500 m/s 750 m/s 1000 m/s Not sure • • • • Kinetic molecular theory Why the gas law does not depend on mass Distribution of molecular speeds Real gases • Gas law for real gases: van der Waals equation 1 2 Kinetic molecular theory Kinetic molecular theory If we assume particles of a gas collide with one another and with the walls of the container elastically, then we can show that the contribution of a single particle 𝑖 to the pressure of a gas is 𝑃 6 6 7 7 1 Kinetic molecular theory Kinetic molecular theory If we assume particles of a gas collide with one another and with the walls of the container elastically, then we can show that the contribution of a single particle 𝑖 to the pressure of a gas is If we assume particles of a gas collide with one another and with the walls of the container elastically, then we can show that the contribution of a single particle 𝑖 to the pressure of a gas is 𝑃 𝑃 Note that the units of this expression, , are correct for pressure. Note that the units of this expression, , are correct for pressure. The factor takes into account that motion takes place in three dimensions. 8 8 9 9 Kinetic molecular theory Kinetic molecular theory The total pressure is the sum of the contributions of the 𝑁 the gas, 𝑃 𝑃 𝑃 ⋯ 𝑃 𝑣 𝑣 ⋯ The total pressure is the sum of the contributions of the 𝑁 the gas, 𝑛𝑁 particles of 𝑃 𝑣 𝑃 𝑃 ⋯ 𝑃 𝑣 𝑣 ⋯ 𝑛𝑁 particles of 𝑣 We can simplify this expression by dividing the sum of squared speeds by the number of particles 𝑁 , 𝑃 10 10 ⋯ 𝑣 11 11 2 Kinetic molecular theory Kinetic molecular theory The total pressure is the sum of the contributions of the 𝑁 the gas, 𝑃 𝑃 𝑃 ⋯ 𝑃 𝑣 𝑣 ⋯ Comparing the expression for pressure from the kinetic molecular theory 𝑛𝑁 particles of ⋯ The quantity 𝑣 the individual particles. 𝑣 , 𝑃 we get the fundamental connection between temperature and the mean of the squared speeds, 𝑣 ⋯ 𝑣 to that from the ideal gas law, We can simplify this expression by dividing the sum of squared speeds by the number of particles 𝑁 , 𝑃 𝑣 𝑃 𝑣 𝑣 is the mean average of the squared speeds of 12 12 13 13 TP The rms speed, 𝑣 Kinetic molecular theory “Use units, Luke!” 𝑅 The square root of the mean of the squared speeds is known the root-meansquared speed, 1. 2. 3. 4. 5. 𝑣 14 14 3𝑅𝑇/𝑀, of O2 at 25℃ is about … 8.314 8.314 / 250 m/s 500 m/s 750 m/s 1000 m/s Not sure 15 15 3 TP The rms speed of O2 at 25℃ is about 500 m/s. Without using a calculator, use the rms speed of O2 to estimate that the rms speed of H2 at the same temperature is about … Kinetic molecular theory 𝑣 𝑅 8.314 Answer: 482 m/s 8.314 1. 2. 3. 4. 5. 6. / 500 m/s 4000 m/s 2000 m/s 1000 m/s 500 m/s 250 m/s Not sure 16 16 17 17 Why the gas law does not depend on mass Why the gas law does not depend on mass We have seen that the macroscopic behavior of gas pressure is that We have seen that the microscopic model of gas pressure is that 𝑃 𝑃 Why doesn’t 𝑃 depend on molar mass 𝑀 ? 𝑣 where 𝑣 Why doesn’t 𝑃 depend on molar mass 𝑀 ? The reason is that root-mean-square speed offsets the apparent mass dependence, 𝑃 18 18 𝑣 19 19 4 Distribution of molecular speeds Distribution of molecular speeds The distribution of speeds is a plot of the fraction of particles 𝑦 axis at each speed 𝑥 axis . Here is the distribution of speeds of O2 at 300 K, from 0 m/s to 4000 m/s, marking the position of 𝑣 on the 𝑥 axis. At 300 K, the rms speed of O2 is 𝑣 The lowest rms speed is 0 and speeds extend exponentially to high values. 500 m/s. 27 27 28 28 TP Use the rms speed of O2 at 300 K, 𝑣 500 m/s. Without a calculator, use this value to find the rms speed of O2 at 1200 K. Distribution of molecular speeds The means 𝑣 𝑣 is slightly higher than the most probable speed, 2/3 𝑣 1. 2. 3. 4. 5. 6. 0.82 𝑣 29 29 4000 m/s 2000 m/s 1000 m/s 500 m/s 250 m/s Not sure 30 30 5 𝑣rms versus temperature 𝑣rms versus temperature The rms speed of O2 at 300 K is 500 m/s and the rms speed of O2 at 1200 K is 1000 m/s. Sketch the distribution of speeds for O2 at these two temperatures on the same axes, marking the position of the two rms speeds on the 𝑥 axis. The rms speed of O2 at 300 K is 500 m/s and the rms speed of O2 at 1200 K is 1000 m/s. Sketch the distribution of speeds for O2 at these two temperatures on the same axes, marking the position of the two rms speeds on the 𝑥 axis. 31 31 32 32 TP Use the rms speed of O2 at 300 K, 𝑣 500 m/s. Without a calculator, use this value to find the rms speed of H2 at 300 K. 1. 2. 3. 4. 5. 6. 𝑣rms versus temperature Add to your sketch the distribution of speeds of H2 at 300 K. 4000 m/s 2000 m/s 1000 m/s 500 m/s 250 m/s Not sure 34 33 33 34 6 𝑣rms versus temperature Real gases Add to your sketch the distribution of speeds of H2 at 300 K. While the ideal gas law is the same for every gas, there are two effects in all real gases that are important at high gas densities due to temperatures close to condensation or very high pressures . 35 35 45 45 Real gases TP The density of H O 𝑔 is 0.0327 and the density of H O 𝑙 is 0.958 . Calculate how much space is taken up by the water molecules in 1 L of gaseous water at 100℃. The first effect is due to gas particles themselves taking up space. For example, we have seen that at 100℃, the density of H O 𝑔 is 0.0327 and the density of H O 𝑙 is 0.958 1. 2. 3. 4. 5. 6. 7. . 46 46 0.0001 L 0.0006 L 0.001 L 0.006 L 0.01 L 0.06 L Not sure 47 47 7 Real gases Real gases The density of H O 𝑔 is 0.0327 and the density of H O 𝑙 is 0.958 the volume taken up by the gaseous water as a liquid would be 0.0327 mol 𝑉 . . More generally, the volume of the container is expressed as empty space 𝑉 plus that taken up by the molecules themselves, , so 𝑉 0.000614 L This means that . 100% 𝑉 𝑏𝑛 Here the parameter van der Waals 𝑏 is the volume taken up by one mole of the molecules of the substance. 0.06% of the 1-L volume is taken up by the water molecules, 1L 𝑉 𝑉 . 48 48 49 49 Real gases 𝑉 𝑉 Real gases 𝑏𝑛 The second effect is due to attraction of the gas particles for one another. At high pressures, the volume of the molecules of a gas 𝑏𝑛 is no longer negligible relative to the volume of the container 𝑉 . This attraction lessens the impact of each particle with the container walls and so reduces the pressure. 50 50 51 51 8 Real gases Real gases The reduction in pressure is proportional to the number of particles hitting the wall and the number of neighboring particles lessening each particle’s impact with the wall, The ideal gas law relates 𝑃 𝑃 𝑃 𝑃 𝑉 Using the expressions 𝑉 𝑎 and 𝑉 , 𝑛𝑅𝑇. 𝑉 𝑏𝑛 and 𝑃 𝑃 𝑎 , we can rewrite the ideal gas law in terms of actual pressure and the actual container volume, The parameter van der Waals 𝑎 is a measure of the intermolecular attraction exerted between particles of the gas. 𝑃 𝑎 𝑉 𝑏𝑛 𝑛𝑅𝑇 This is known as the van der Waal equation. 52 52 53 53 Real gases Real gases Computations with the van der Waals equation are usually done by rearranging it for the observed pressure, Computations with the van der Waals equation are usually done by rearranging it for the observed pressure, 𝑃 𝑃 𝑎 and using tabulated values of van der Waals 𝑎 and 𝑏. This expression makes it easier to see that 𝑎 reduces pressure and 𝑏 increases pressure, relative to ideal values. 54 54 𝑎 55 55 9 Working with the van der Waals equation 𝑃 𝑎 and 𝑅 Working with the van der Waals equation 0.08314 L bar/ K mol 𝑃 For Cl2, 𝑎 6.58 bar L2/mol2 and 𝑏 0.0562 L/mol. What are the ideal and observed pressures of 3.00 mol of Cl2 confined in 4.00 L at 500. K? 𝑎 and 𝑅 0.08314 L bar/ K mol For Cl2, 𝑎 6.58 bar L2/mol2 and 𝑏 0.0562 L/mol. What are the ideal and observed pressures of 3.00 mol of Cl2 confined in 4.00 L at 500. K? 𝑃 31.3 bar confirm yourself 56 56 57 57 Working with the van der Waals equation 𝑃 𝑎 and 𝑅 Working with the van der Waals equation 0.08314 L bar/ K mol 𝑃 For Cl2, 𝑎 6.58 bar L2/mol2 and 𝑏 0.0562 L/mol. What are the ideal and observed pressures of 3.00 mol of Cl2 confined in 4.00 L at 500. K? 𝑃 𝑃 32.5 bar 3.7 bar and 𝑅 0.08314 L bar/ K mol For Cl2, 𝑎 6.58 bar L2/mol2 and 𝑏 0.0562 L/mol. What are the ideal and observed pressures of 3.00 mol of Cl2 confined in 4.00 L at 500. K? 𝑃 31.3 bar confirm yourself 𝑎 𝑎 𝑃 28.8 bar confirm yourself 31.3 bar confirm yourself 𝑎 32.5 bar 3.7 bar 28.8 bar confirm yourself Since the observed pressure is smaller than the ideal pressure, the effect of 𝑎 is more important than 𝑏 at 500. K. 58 58 59 59 10 Working with the van der Waals equation and 𝑅 Working with the van der Waals equation 0.08314 L bar/ K mol 𝑝 For Cl2, 𝑎 6.58 bar L2/mol2 and 𝑏 0.0562 L/mol. What are the ideal and observed pressures of 3.00 mol of Cl2 confined in 4.00 L at 3000. K? 𝑎 and 𝑅 0.08314 L bar/ K mol For Cl2, 𝑎 6.58 bar L2/mol2 and 𝑏 0.0562 L/mol. What are the ideal and observed pressures of 3.00 mol of Cl2 confined in 4.00 L at 3000. K? 𝑃 𝑃 187.1 bar confirm yourself 𝑎 195.3 bar 3.7 bar 191.6 bar confirm yourself Since the observed pressure is larger than the ideal pressure, the effect of 𝑏 is more important than 𝑎 at 3000. K. 60 61 61 Working with the van der Waals equation and 𝑅 0.08314 L bar/ K mol In general, the effect of 𝑏 is muted at low 𝑇 and so the effect of 𝑎 dominates there, while for high 𝑇 the effect of 𝑏 dominates. 62 11