The Laws of Thermodynamics Chapter 15

What is temperature? Heat? Thermal/Internal Energy? Temperature (T): A measure of the average kinetic energy of individual molecules Heat: Amount of energy transferred from one body to another at different temperature Thermal/Internal Energy (U): Total energy of all molecules in an object The sum of the translational kinetic energies of all the atoms

Average Kinetic Energy and RMS velocity The higher the temperature of a gas, the faster the molecules move! R= Gas constant =8.315 J/(mol K) M= Molar mass µ= mass of molecule kB = Boltzmann’s constant= 1.38 x 10-23 J/K T= temperature

Internal Energy of an Ideal Gas N= number of molecules n= number of moles k=Boltzmann’s Constant R=Gas Constant T= Temperature

Ideal Gas Law and Combined Gas Law P= Pressure V= Volume n= number of moles R= Gas Constant T= temperature

Types of systems A closed system: No mass enters/leaves but energy may be exchanged with the environment An open system: Mass and energy may enter/leave Isolated System: No energy in any form enters/leaves the boundaries

Thermodynamics The study of processes in which energy is transferred as heat and as work Heat is a transfer of energy due to a difference in temperature Work is a transfer of energy that is not due to a temperature difference

How does a gas do work? Let’s say there is a gas in a container with a movable piston As you heat the temperature, the gas expands and it causes the piston to move upward Since W= Fd, the gas does work on the piston Work done by a gas is equal to the product of pressure and volume!

What if the gas is compressed? If the gas is compressed by the piston, that means work is done on the gas, so W is negative! W=-PΔV (work done on the system) W=+PΔV (work done by the system) Be careful with the signs!!!!

First Law of Thermodynamics The change in internal energy of a closed system, ΔU, is equal to the heat added to the system plus the work done on the system Q = net heat added to the system (+Q) If heat leaves the system, Q is negative W = net work done on the system(-W) The work done on a system (-W) is the opposite of the work done by the system (+W)!

First Law of Thermodynamics The change in internal energy of a closed system, ΔU, is equal to the heat added to the system minus the work done by the system Q = net heat added to the system (+Q) If heat leaves the system, Q is negative W = net work done by the system(+W) The work done on a system (-W) is the opposite of the work done by the system (+W)!

Types of Processes: Isothermal Isothermal process: An idealized process that is carried out at constant temperature. Since U depends on T, there is no change in internal energy in this process Ideal Gas Law: PV=nRT=constant 1st Law: Q=W because ΔU=0

Types of Processes: Adiabatic Adiabatic Process: No heat is allowed to flow into or out of a system so Q=0, but work is done on the system. This happens if the system is extremely well insulated or if it happens so quickly that heat has no time to flow in or out 1st Law: ΔU=-W Internal Energy decreases if the gas expands (because the gas does work when it expands) so the temperature must also decrease because U is proportional to T Internal Energy increases if the gas is compressed (because work is done on the gas when it is compressed), so the temperature must also increase

Isobaric and Isochoric/Isovolumetric Processes In an isobaric process, the pressure is kept constant. In an isochoric/isovolumetric process, the volume is kept constant

PV Diagram for multiple processes (p. 447)

Using PV Diagram to Find Work The work done by a gas is equal to the area under the PV curve

Sample Problem p. 472 #9 Consider the following two step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.4 atm. Then the gas expands at constant pressure from a volume of 6.8 L to 9.3 L where the temperature reaches its original value. Calculate (a) the total work done by the gas in the process, (b) the change in internal energy of the gas in the process and (c) the total heat flow into or out of the gas.

Solve the problem. Part (a) How much work is done by the gas? From point A to point B, what kind of process is this? Volume remains constant. This is isovolumetric. W=P Δ V=0 From point B to point C, what kind of process is this? Pressure remains constant. This is isobaric W= P Δ V= Area under the curve

Solve the problem Area under the curve is PΔV P= 1.4 atm= 1.42 x 105 Pa ΔV= 9.3 L- 6.8 L= 2.5 L= 2.5 x 10-3 m3 W= P ΔV= +355 J (work done by the system) Part B…what is the change in internal energy? Initial and final temperature are the same Since ΔT=0, then ΔU= 0

Finish the Problem Part C. What is the total heat flow into or out of the gas?

Practice Problem p. 472 #10 The PV diagram in Fig 15-28 shows two possible states of a system containing two moles of a monatomic ideal gas. (P1=P2=450 Pa, V1= 2m3 , V2= 8m3) A. Draw the process which depicts an isobaric expansion from state 1 to state 2 and label this process (A) B. Find the work done by the gas and the change in internal energy of the gas in process A. C. Draw the process which depicts an isothermal expansion from state 1 to the volume V2 followed by an isochoric increase in temperature to state 2 and label this process (B). D. Find the change in internal energy of the gas for the two step process (B)

Practice Problem p. 472 #10 A. A. Draw the process which depicts an isobaric expansion from state 1 to state 2 and label this process (A) A

Practice Problem p. 472 #10 Work= P ΔV= (450 Pa)(8m3 – 6m3)= 2700 J B. Find the work done by the gas and the change in internal energy of the gas in process A. Work= P ΔV= (450 Pa)(8m3 – 6m3)= 2700 J Change in internal energy? How can we find the change in T? Use the Ideal Gas Law! PV=nRT

Practice Problem p. 472 #10 C. Draw the process which depicts an isothermal expansion from state 1 to the volume V2 followed by an isochoric increase in temperature to state 2 and label this process (B). A

Practice Problem p. 472 #10 D. Find the change in internal energy of the gas for the two step process (B) Since both paths have the same initial and final temperatures, ΔU is the same= 4050 J

Heat Engines (15-5) A heat engine converts heat into work When heat flows from high temperature to a low temperature, some of the heat can be transformed into work The input heat is used by the gas to do work Looking at a PV diagram, if the net work is positive, then the device is a heat engine

Refrigerators/Heat Pumps/ Air Conditioners (15-6) A refrigerator transfers heat from a low temperature region to a high temperature region In order to do this, work must be done on the gas to make heat flow from low temperature to high temperature Looking at a PV diagram, if the net work is negative, then the device is a refrigerator