Results from kinetic theory, 1 1. Pressure is associated with collisions of gas particles with the walls. Dividing the total average force from all the particles by the wall area gives the pressure. Increasing temperature increases pressure for two reasons. There are more collisions, and the collisions involve a larger average force. For a fixed volume and temperature, adding more particles increases pressure because there are more collisions.

Results from kinetic theory, 2 What is the average velocity of the ideal gas particles?

Results from kinetic theory, 2 What is the average velocity of the ideal gas particles? 2. The average velocity is zero, because, on average, the velocity of particles going in one direction is cancelled by the velocity of particles going in the opposite direction. When you do the analysis, you find that what really matters is the rms- speed (rms stands for root mean square). Square the speeds, take the average, and take the square root of that result.

Results from kinetic theory, 3 3. The really fundamental result of kinetic theory is that temperature is a direct measure of the average kinetic energy of the particles of ideal gas. Kinetic theory: Ideal gas law: This tells us that the average translational kinetic energy of the molecules is: Here we have a fundamental connection between temperature and the average translational kinetic energy of the atoms - they are directly proportional to one another.

Light and heavy atoms A box of ideal gas consists of light particles and heavy particles (the heavy ones have 16 times the mass of the light ones). Initially all the particles have the same speed. When equilibrium is reached, what will be true? 1. All the particles will still have the same speed 2. The average speed of the heavy particles equals the average speed of the light particles 3. The average speed of the heavy particles is larger than that of the light particles 4. The average speed of the heavy particles is smaller than that of the light particles

Light and heavy atoms Coming to equilibrium means coming to a particular temperature, which means the average kinetic energy of the particles is a particular value. The average kinetic energy of the light particles equals the average kinetic energy of the heavy particles - this can only happen if the average speed of the heavy particles is smaller than that of the light particles. Smaller by a factor of what?

Light and heavy atoms Simulation Coming to equilibrium means coming to a particular temperature, which means the average kinetic energy of the particles is a particular value. The average kinetic energy of the light particles equals the average kinetic energy of the heavy particles - this can only happen if the average speed of the heavy particles is smaller than that of the light particles. Smaller by a factor of what? 4, so that:

Three cylinders Three identical cylinders are sealed with identical pistons that are free to slide up and down the cylinder without friction. Each cylinder contains ideal gas, and the gas occupies the same volume in each case, but the temperatures differ. In each cylinder the piston is above the gas, and the top of each piston is exposed the atmosphere. In cylinders 1, 2, and 3 the temperatures are 0°C, 50°C, and 100°C, respectively. How do the pressures compare? 1. 1>2>3 2. 3>2>1 3. all equal 4. not enough information to say

Free-body diagram of a piston Sketch the free-body diagram of one of the pistons.

Free-body diagram of a piston Sketch the free-body diagram of one of the pistons. The internal pressure, in this case, is determined by the free-body diagram. All three pistons have the same pressure, so the number of moles of gas must decrease from 1 to 2 to 3.

Introducing the P-V diagram P-V (pressure versus volume) diagrams can be very useful. What are the units resulting from multiplying pressure in kPa by volume in liters? Rank the four states shown on the diagram based on their absolute temperature, from greatest to least.

Introducing the P-V diagram P-V (pressure versus volume) diagrams can be very useful. What are the units resulting from multiplying pressure in kPa by volume in liters? Rank the four states shown on the diagram based on their absolute temperature, from greatest to least. Temperature is proportional to PV, so rank by PV: 2 > 1=3 > 4.

Isotherms Isotherms are lines of constant temperature. On a P-V diagram, isotherms satisfy the equation: PV = constant

Thermodynamics Thermodynamics is the study of systems involving energy in the form of heat and work. Consider a cylinder of ideal gas, at room temperature. When the cylinder is placed in a container of hot water, heat is transferred into the cylinder. Where does that energy go? The piston is free to move up or down without friction.

Thermodynamics

The First Law of Thermodynamics Some of the added energy goes into raising the temperature of the gas (we call this raising the internal energy). The rest of it does work, raising the piston. Conserving energy: (the first law of thermodynamics) Q is heat added to a system (or removed if it is negative) is the internal energy of the system (the energy associated with the motion of the atoms and/or molecules), so is the change in the internal energy, which is proportional to the change in temperature. W is the work done by the system. The First Law is often written as

Work We defined work previously as: (true if the force is constant) F = PA, so: At constant pressure the work done by the system is the pressure multiplied by the change in volume. If there is no change in volume, no work is done. In general, the work done by the system is the area under the P-V graph. This is why P-V diagrams are so useful.

Work: the area under the curve The net work done by the gas is positive in this case, because the change in volume is positive, and equal to the area under the curve.

A P-V diagram question An ideal gas initially in state 1 progresses to a final state by one of three different processes (a, b, or c). Each of the possible final states has the same temperature. For which process is the change in internal energy larger? 1. a 2. b 3. c 4. Equal for all three 5. We can’t determine it

A P-V diagram question Because the change in temperature is the same, the change in internal energy is the same for all three processes.

Another P-V diagram question An ideal gas initially in state 1 progresses to a final state by one of three different processes (a, b, or c). Each of the possible final states has the same temperature. For which process is more heat transferred into the ideal gas? 1. a 2. b 3. c 4. Equal for all three 5. We can’t determine it

Another P-V diagram question The heat is the sum of the change in internal energy (which is the same for all three) and the work (the area under the curve), so whichever process involves more work requires more heat.

Another P-V diagram question The heat is the sum of the change in internal energy (which is the same for all three) and the work (the area under the curve), so whichever process involves more work requires more heat. Process c involves more work, and thus requires more heat.

Constant volume vs. constant pressure We have two identical cylinders of ideal gas. Piston 1 is free to move. Piston 2 is fixed so cylinder 2 has a constant volume. We put both systems into a reservoir of hot water and let them come to equilibrium. Which statement is true? 1. Both the heat Q and the change in internal energy will be the same for the two cylinders 2. The heat is the same for the two cylinders but cylinder 1 has a larger change in internal energy. 3. The heat is the same for the two cylinders but cylinder 2 has a larger change in internal energy. 4. The changes in internal energy are the same for the two cylinders but cylinder 1 has more heat. 5. The changes in internal energy are the same for the two cylinders but cylinder 2 has more heat.

Constant volume vs. constant pressure Each cylinder comes to the same temperature as the reservoir. How do the changes in internal energy compare? Which cylinder does more work?

Constant volume vs. constant pressure Each cylinder comes to the same temperature as the reservoir. How do the changes in internal energy compare? The same number of moles of the same gas experience the same temperature change, so the change in internal energy is the same. Which cylinder does more work? Cylinder 2 does no work, so cylinder 1 does more work. By the first law, cylinder 1 requires more heat to produce the same change in temperature as cylinder 2. The heat required depends on the process.

Solving thermodynamics problems A typical thermodynamics problem involves some process that moves an ideal gas system from one state to another. 1. Draw a P-V diagram to get some idea what the work is. 2. Apply the First Law of Thermodynamics (this is a statement of conservation of energy). 3. Apply the Ideal Gas Law. the internal energy is determined by the temperature the change in internal energy is determined by the change in temperature the work done depends on how the system moves from one state to another (the change in internal energy does not)

Constant volume (isochoric) process No work is done by the gas: W = 0. The P-V diagram is a vertical line, going up if heat is added, and going down if heat is removed. Applying the first law: For a monatomic ideal gas:

Constant pressure (isobaric) process In this case the region on the P-V diagram is rectangular, so its area is easy to find. For a monatomic ideal gas:

Heat capacity For solids and liquids: For gases:, where C, the heat capacity, depends on the process. For a monatomic ideal gas Constant volume: Constant pressure: In general:

Constant temperature (isothermal) process No change in internal energy: The P-V diagram follows the isotherm. Applying the first law, and using a little calculus:

Zero heat (adiabatic) process Q = 0. The P-V diagram is an interesting line, given by: For a monatomic ideal gas: Applying the first law:

Worksheet You have some monatomic ideal gas in a cylinder. The cylinder is sealed at the top by a piston that can move up or down, or can be fixed in place to keep the volume constant. Blocks can be added to, or removed from, the top of the cylinder to adjust the pressure, as necessary. Starting with the same initial conditions each time, you do three experiments. Each experiment involves adding the same amount of heat, Q. A – Add the heat at constant pressure. B – Add the heat at constant temperature. C – Add the heat at constant volume.

Worksheet A – Add heat Q to the system at constant pressure. B – Add heat Q to the system at constant temperature. C – Add heat Q to the system at constant volume. Sketch these processes on the P-V diagram. The circle with the i beside it represents the initial state of the system. One of the processes is drawn already. Identify which one, and draw the other two.

Rank by final temperature Rank the processes based on the final temperature. 1. A > B > C 2. A > C > B 3. B > A > C 4. B > C > A 5.C > A > B 6.C > B > A 7.Equal for all three Add heat Q to the system at: A - constant pressure. B - constant temperature. C - constant volume.

Rank by final temperature In process B, the temperature is constant. In process A, some of the heat added goes to increasing the temperature. In process C, all the heat added goes to increasing the temperature. C > A > B

Rank by work Rank the processes based on the work done by the gas. 1. A > B > C 2. A > C > B 3. B > A > C 4. B > C > A 5.C > A > B 6.C > B > A 7.Equal for all three Add heat Q to the system at: A - constant pressure. B - constant temperature. C - constant volume.

Rank by work In process C, no work is done. In process A, some of the heat added goes to doing work. In process B, all the heat added goes to doing work. B > A > C

Rank by final pressure Rank the processes based on the final pressure. 1. A > B > C 2. A > C > B 3. B > A > C 4. B > C > A 5.C > A > B 6.C > B > A 7.Equal for all three Add heat Q to the system at: A - constant pressure. B - constant temperature. C - constant volume.

Rank by pressure In process A, the pressure stays constant. In process B, the pressure decreases. In process C, the pressure increases. C > A > B

Example problem A container of monatomic ideal gas contains just the right number of moles so that nR = 20 J/K. The gas is in state 1 such that: P 1 = 20 kPa V 1 = 100 x m 3 (a) What is the temperature T 1 of the gas?

Example problem A container of monatomic ideal gas contains just the right number of moles so that nR = 20 J/K. The gas is in state 1 such that: P 1 = 20 kPa V 1 = 100 x m 3 (a) What is the temperature T 1 of the gas? Use the ideal gas law. PV = nRT, so: The factor of 1000 in the kPa cancels the in the volume.

Example problem A container of monatomic ideal gas contains just the right number of moles so that nR = 20 J/K. The gas is in state 1 such that: P 1 = 20 kPa V 1 = 100 x m 3 (b) If Q = 2500 J of heat is added to the gas, and the gas expands at constant pressure, the gas will reach a new equilibrium state 2. What is the final temperature T 2 ?

Example problem A container of monatomic ideal gas contains just the right number of moles so that nR = 20 J/K. The gas is in state 1 such that: P 1 = 20 kPa V 1 = 100 x m 3 (b) If Q = 2500 J of heat is added to the gas, and the gas expands at constant pressure, the gas will reach a new equilibrium state 2. What is the final temperature T 2 ? At constant pressure for a monatomic ideal gas: Therefore

Example problem A container of monatomic ideal gas contains just the right number of moles so that nR = 20 J/K. The gas is in state 1 such that: P 1 = 20 kPa V 1 = 100 x m 3 (c) Q = 2500 J of heat is added to the gas, and the gas expands at constant pressure. How much work is done by the gas?

Example problem A container of monatomic ideal gas contains just the right number of moles so that nR = 20 J/K. The gas is in state 1 such that: P 1 = 20 kPa V 1 = 100 x m 3 (c) Q = 2500 J of heat is added to the gas, and the gas expands at constant pressure. How much work is done by the gas? At constant pressure, we have: We can do this only for a constant pressure process.

Example problem A container of monatomic ideal gas contains just the right number of moles so that nR = 20 J/K. The gas is in state 1 such that: P 1 = 20 kPa V 1 = 100 x m 3 (d) Q = 2500 J of heat is added to the gas, and the gas expands at constant pressure. What is the final volume V 2 ?

Example problem A container of monatomic ideal gas contains just the right number of moles so that nR = 20 J/K. The gas is in state 1 such that: P 1 = 20 kPa V 1 = 100 x m 3 (d) Q = 2500 J of heat is added to the gas, and the gas expands at constant pressure. What is the final volume V 2 ? One approach is to bring in the ideal gas law again:

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