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>2>3 3>2>1 all equal 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.

Solving thermodynamics problems A typical thermodynamics problem involves some process that moves an ideal gas system from one state to another. Draw a P-V diagram to get some idea what the work is. Apply the First Law of Thermodynamics (this is a statement of conservation of energy). 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:

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:

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 C > A > B C > B > A Equal for all three

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 C > A > B C > B > A Equal for all three

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 C > A > B C > B > A Equal for all three

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

Worksheet A thermodynamic system undergoes a three-step process. An adiabatic expansion takes it from state 1 to state 2; heat is added at constant pressure to move the system to state 3; and an isothermal compression returns the system to state 1. The system consists of a diatomic ideal gas with CV = 5R/2. The number of moles is chosen so nR = 100 J/K. The following information is known about states 2 and 3. Pressure: P2 = P3 = 100 kPa Volume: V3 = 0.5 m3 What is the temperature of the system in state 3?

Apply the ideal gas law The system does 20000 J of work in the constant pressure process that takes it from state 2 to state 3. What is the volume and temperature of the system in state 2?

The temperature in state 2 What is the temperature of the system in state 2? 1. 200 K 2. 300 K 3. 500 K 4. 700 K 5. None of the above

Finding work The system does 20000 J of work in the constant pressure process that takes it from state 2 to state 3. What is the volume and temperature of the system in state 2? For constant pressure, we can use: Finding volume: Finding temperature: (use the ideal gas law, or …)

Complete the table For the same system, complete the table. The total work done by the system in the cycle is –19400 J. First fill in all the terms that are zero. Each row satisfies the First Law of Thermodynamics. Also remember that Process Q ΔEint W 1 to 2 2 to 3 +20000 J 3 to 1 Entire cycle -19400 J

Complete the table For the same system, complete the table. The total work done by the system in the cycle is –19400 J. Q is zero for an adiabatic process. The change in internal energy is zero for an isothermal process, and is always zero for a complete cycle. Process Q ΔEint W 1 to 2 2 to 3 +20000 J 3 to 1 Entire cycle -19400 J

Complete the table For the same system, complete the table. The total work done by the system in the cycle is –19400 J. Even the last row has to satisfy the first law: Process Q ΔEint W 1 to 2 2 to 3 +20000 J 3 to 1 Entire cycle -19400 J

Complete the table For the same system, complete the table. The total work done by the system in the cycle is –19400 J. Find the change in internal energy for the 2  3 process. Process Q ΔEint W 1 to 2 2 to 3 +50000 J +20000 J 3 to 1 Entire cycle -19400 J

Complete the table For the same system, complete the table. The total work done by the system in the cycle is –19400 J. Rows have to obey the first law. Columns have to sum to the value for the entire cycle. Process Q ΔEint W 1 to 2 -50000 J 2 to 3 +70000 J +50000 J +20000 J 3 to 1 Entire cycle -19400 J

Complete the table For the same system, complete the table. The total work done by the system in the cycle is –19400 J. Rows have to obey the first law. Columns have to sum to the value for the entire cycle. Process Q ΔEint W 1 to 2 -50000 J +50000 J 2 to 3 +70000 J +20000 J 3 to 1 -89400 J Entire cycle -19400 J

A heat engine A heat engine is a device that uses heat to do work. A gasoline-powered car engine is a good example. To be useful, the engine must go through cycles, with work being done every cycle. Two temperatures are required. The higher temperature causes the system to expand, doing work, and the lower temperature re-sets the engine so another cycle can begin. In a full cycle, three things happen: Heat QH is added at a relatively high temperature TH. Some of this energy is used to do work W. The rest is removed as heat QL at a lower temperature TL. For the cycle: QH = W + QL (all positive quantities)

Efficiency In general, efficiency is the ratio of the work done divided by the heat needed to do the work. The net work done in one cycle is the area enclosed by the cycle on the P-V diagram.

Carnot’s principle Sadi Carnot (1796 – 1832), a French engineer, discovered an interesting result that is a consequence of the Second Law of Thermodynamics. Even in an ideal situation, the efficiency of a heat engine is limited by the temperatures between which the engine operates. 100% efficiency is not possible, and most engines, even in ideal cases, achieve much less than 100% efficiency. Carnot’s principle: Ideal (Carnot) efficiency:

Heat engines running backwards Refrigerators and air conditioners are heat engines that run backward. Work is done on the system to pump some heat QL from a low temperature region TL. An amount of heat QH = QL + W is then removed from the system at a higher temperature TH. (a) Represents the cylinder in a car engine; while (b) represents a refrigerator.

A heat pump If you heat your home using electric heat, 1000 J of electrical energy can be transformed into 1000 J of heat. An alternate way of heating is to use a heat pump, which extracts heat from a lower-temperature region (outside the house) and transfers it to the higher-temperature region (inside the house). Let's say the work done in the process is 1000 J, and the temperatures are TH = 27°C = 300 K and TL = -13 °C = 260 K. What is the maximum amount of heat that can be transferred into the house? Something less than 1000 J 1000 J Something more than 1000 J

A heat pump The best we can do is determined by the Carnot relationship. Using this in the energy equation gives: For our numerical example this gives: This is why heat pumps are much better than electric heaters. Instead of 1000 J of work going to 1000 J of heat we have 1000 J of work producing 7500 J of heat.