CH 15 Zeroth and First Law of Thermodynamics
THERMODYNAMICS Thermodynamics Branch of Physics that is built upon the fundamental laws that heat and work obey. Central Heating
Objectives: After finishing this unit, you should be able to: State and apply the Zeroth and First of Thermodynamics. Demonstrate your understanding of adiabatic, isochoric, isothermal, and isobaric processes. Use a PV Diagram to solve for the work done on or by a system.
A THERMODYNAMIC SYSTEM A system is a collection of objects in which attention is being focused. A system is a closed environment in which heat transfer can take place. (For example, the gas, walls, and container. (Example: Cylinder of an automobile engine.) Work done on the gas or work done by the gas.
Surroundings The surroundings mean everything else in the environment. The system and its surroundings are separated by some kind of wall. Diathermal walls Walls that permit heat to flow through them. Adiabatic walls Wall that do not permit heat to flow between the system and its surroundings.
Four Laws of Thermodynamics. Of the 4 Laws, the Second Law was found first, the First Law was found second, then the Zeroth Law, then the Third Law was found fourth.
The Zeroth Law of Thermodynamics This is called the Zeroth Law because the First Law was already in effect and those in charge did not want to change the first law so when this law was found or stated, they numbered the new law the Zeroth Law. This Law deals with thermal equilibrium and establishes temperature as the indicator of thermal equilibrium, in the sense that there is no net flow of heat between two systems in thermal contact that have the same temperature.
The Zeroth Law of Thermodynamics Zeroth Law states: Two systems individually in thermal equilibrium with a third system are in thermal equilibrium with each other.
The Zeroth Law of Thermodynamics (a) Boxes A and B are surrounded by walls that do not conduct heat, so the heat stays in the box. Both boxes have the same temperature. (b) When A is put into thermal contact with B through a heat conducting wall, grey, no net flow of heat occurs between the boxes.
The First Law of Thermodynamics First Law of Thermodynamics states: The internal energy of a system changes from an initial value, to a final value due to heat and work.
INTERNAL ENERGY OF SYSTEM The internal energy, U, of a system is the total of all kinds of energy possessed by the particles that make up the system. Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.
Work Done by a Gas Consider the formula for work: W = Fx Because the distance is vertical, the x becomes Δy. The force, in this case, is caused by the pressure of the gas inside, using P = F/A, so F = PA. Substituting you have: W =PA(Δy). Where: Δy being the displacement of the cylinder, similar to x. (AΔy) is simply just a change in Volume. W = PΔV So we could say that work is done by the expansion in volume of a gas.
PV Graph or PV Diagram Since the above formula is similar to A = bh. The work is determined by the area UNDER a Pressure vs Volume diagram or graph. According to "Conservation of Energy", we learned that WORK can be done by a change in kinetic energy (W =ΔKE) or potential energy or both. We also learned that HEAT ENERGY (Q) can be done when there is a change in temperature. When a substance is involved in doing some form of work AND there is a temperature change in the process the INTERNAL ENERGY (U) can change.
TWO WAYS TO INCREASE THE INTERNAL ENERGY, U. + U WORK DONE BY A GAS ΔW is positive HEAT PUT INTO A SYSTEM ΔQ is positive
TWO WAYS TO DECREASE THE INTERNAL ENERGY, U. W out Q out -ΔU hot hot WORK DONE ON THE SYSTEM ΔW is negative HEAT LEAVES A SYSTEM ΔQ is negative
THERMODYNAMIC STATE The STATE of a thermodynamic system is determined by four factors: Absolute Pressure, P, in Pascals Temperature,T, in Kelvins Volume,V, in cubic meters Number of moles, n, of working gas
THERMODYNAMIC PROCESS Increase in Internal Energy, U. W out Q in Initial State: P 1 V 1 T 1 n 1 Heat input Work by gas Final State: P 2 V 2 T 2 n 2
The Reverse Process Decrease in Internal Energy, U. W in Q out Initial State: P 1 V 1 T 1 n 1 Work on gas Loss of heat Final State: P 2 V 2 T 2 n 2
SIGN CONVENTIONS FOR FIRST LAW Work BY a gas is positive Heat Q input is positive Work ON a gas is negative Heat Q out is negative +Q in -W on U +W by U U Q W -Q out Why is the formula on your Formula Sheet: U Q W
Example Problem 1. Gas in a container is at a pressure of 1.5 atm and a volume of 4 m 3. What is the work done by the gas if, at constant pressure, a. its volume expands by adding twice the initial volume, and b. it is compressed by subtracting one quarter of its initial volume?
Example Problem 2. An ideal gas is enclosed in a cylinder. There is a movable piston on top of the cylinder. The piston has a mass of 8000 g and an area of 5 cm 2 and is free to slide up and down, keeping the pressure of the gas constant. How much work is done as the temperature of 0.2 moles of the gas is raised from 20 o C to 300 o C?
Problem 3. A gas is compressed at a constant pressure of 0.8 atm from a volume of 9 L to a volume of 2 L, and in the process, 400 J of heat energy flow out of the gas. a. What is the work done by the gas? b. What is the change in internal energy of the gas?
Problem 4. A thermodynamic system undergoes a process in which its internal energy decreases by 500 J. If at the same time, 220 J of work is done on the system, find the heat transferred to or from the system.
Example 5. An ideal gas is enclosed in a cylinder. There is a movable piston on top of the cylinder. The piston has a mass of 8000 g and an area of 5 cm 2 and is free to slide up and down, keeping the pressure of the gas constant. How much work is done as the temperature of 0.2 moles of the gas is raised from 20 o C to 300 o C?
FOUR THERMODYNAMIC PROCESSES: Isobaric Process: ΔP = 0 Isochoric Process: ΔV = 0, ΔW = 0 Adiabatic Process: ΔQ = 0 Isothermal Process: ΔT = 0, ΔU = 0
ISOBARIC PROCESS: CONSTANT PRESSURE, P = 0 ΔU = ΔQ - ΔW But ΔW = P ΔV Q IN Q OUT + U Work Out - U Work In HEAT IN = W out + INCREASE IN INTERNAL ENERGY HEAT OUT = W out + DECREASE IN INTERNAL ENERGY
ISOBARIC WORK P A B V A = V B T A T B 400 J V 1 V 2 P A = P B Work = Area under PV curve. Work P V
ISOCHORIC PROCESS The substance would expand if it could, but the rigid container keeps the volume constant, so the pressure increases. More and more force is exerted on the walls of the container. No work is done, since the volume does not change. The area under the straight, vertical line of a pressure-volume graph is zero. Heat in the system increases the internal energy of the system.
ISOCHORIC PROCESS: CONSTANT VOLUME, V = 0, W = 0 0 U = Q - W so that Q = U Q IN Q OUT + U No Work Done - U HEAT IN = INCREASE IN INTERNAL ENERGY HEAT OUT = DECREASE IN INTERNAL ENERGY
ISOCHORIC EXAMPLE: No Change in volume: P 2 P 1 B A P A T A = P B T B V 1 = V 2 400 J Heat input increases P with const. V 400 J heat input increases internal energy by 400 J and zero work is done.
ADIABATIC PROCESS: Occurs without the transfer of heat, Q = 0 U = Q - W ; W = - U or U = - W W = - U U = - W U Work Out Q = 0 + U Work In Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy
ADIABATIC EXAMPLE: P A A B P B V 1 V 2 Insulated Walls: Q = 0 Expanding gas does work with zero heat loss. Work = - U
ADIABATIC EXPANSION: Q = 0 P A A B P A V A P B V B = P B V A V B T A T B 400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat exchange is ZERO. Q = 0 PV PV A A B B
ISOTHERMAL PROCESS: Constant Temperature, T = 0, U = 0 U = Q - W AND Q = W Q IN Q OUT Work Out U = 0 U = 0 Work In NET HEAT INPUT = WORK OUTPUT WORK INPUT = NET HEAT OUT
ISOTHERMAL EXAMPLE (Constant T): P A A P B B U = T = 0 V 2 V 1 P A V A = P B V B Slow compression at constant temperature: ----- No change in U.
ISOTHERMAL EXPANSION (Constant T): A P A P B B P A V A = P B V B U = T = 0 V A V B T A = T B 400 J of energy is absorbed by gas as 400 J of work is done on gas. T = U = 0
Example 6. A gas is taken through the cyclic process described in the figure. a. Find the net heat transferred to the system during one complete cycle. b. If the cycle is reversed, that is the process goes along ACBA, what is the net heat transferred per cycle?
Example 7. One mole of a gas initially at pressure of 2 atm and a volume of 0.3 L has an internal energy equal to 91 J. In its final state, the pressure is 1.5 atm, the volume is 0.8 L, and the internal energy equals 182 J. For the three paths IAF, IBF, and IF in the Figure, calculate a. the work done by the gas and b. the net heat transferred in the process.
Assignment Ch 15, Pages 466 471, #1, 5, 7, 11, 13, 23, 31, 91