Heat and Work. First Law of Thermodynamics 9.1. Heat is a form of energy. Calorimetry. Work. First Law of Thermodynamics.

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1 Heat and First Law of Thermodynamics 9. Heat Heat and Thermodynamic rocesses Thermodynamics is the science of heat and work Heat is a form of energy Calorimetry Mechanical equivalent of heat Mechanical work done on a system produces a rise in temperature like heat added to the system. calorie =4.84 J Heat input into a system can produce work W = F W = A = For the case of thermal epansion work is done by the system. Heat input First Law of Thermodynamics The change in the internal energy of a system U, is equal to the heat input minus the work done by the system. Heat U= W Internal Energy U W System The internal energy is the energy stored in the system. For an ideal gas the internal energy is the kinetic energy of the gas

2 Thermodynamic processes in an ideal gas diagram U U 2 State (,, ) State 2 ( 2, 2, ) State 2 The state of the ideal gas is determined by the three parameters, T. A thermodynamic process is the transition between states with input or output of heat and work with changes in internal energy. The internal energy U is a property of the state. U determined by the initial and final state and is independent of path The heat absorbed and work done in the process depend on the path. State Each point in the diagram represents a State of the system The temperature T is not plotted but is not an independent variable. T can be calculated from and using the ideal gas law. The points of constant temperature on the diagram are hyperbola = For constant T Thermodynamic processes State 2 (, 2, 2 ) ath A can be calculated if the process is reversible. ath A State 2 (, 2, 2 ) ath B State (,, ) The change in U going from to 2 is the same for the two paths. U A = U B The change in and W going from to 2 is different between the two paths. A B, W A W B State (,, ) We can calculate the work done in going from to 2 if the change is a reversible process. i.e. it goes slowly through well defined states that are in quasi-equilibrium. Reversible process T is increased slowly from 0 to 00 o C cannot be calculated for an irreversible process. ath A? State 2 (, 2, 2 ) State (,, ) The temperature is changed very slowly so that the gas sample is at a uniform temperature. We cannot calculate the work done in going from to 2 if the process is irreversible. i.e. goes rapidly through states that are not well characterized and not in quasi-equilibrium. However, we can determine U, since this is a state property. 2

3 Irreversible process done in a process The gas is not in a well defined thermodynamic state as it is heated. dw = d 2 W = dv The work is the area under the curve. uestion Find the work done in going from A to B along the line. Some thermodynamic processes epansion of ideal gas. Constant Temperature process- Isothermal Constant T T constant heat bath constant 2 B 2. Constant ressure process Isobaric. e.g heating a balloon T changes at constant A 2 3. Constant olume process. Heating a gas thermometer 4. Adiabatic process no heat echange. Rapid compression or epansion of a gas T changes at constant T changes with constant =0 No contact with heat bath Thermodynamic processes Increasing T Constant T Constant Constant Adiabatic (=0) Isothermal volume change T is constant U= 0 W= W = d = d = U is constant All the heat goes into work W = ln2 ln =W 2 W = ln Epansion 2 > work is positive work is done by the gas 2 Compression < work is negative work is done on the gas 3

4 At constant volume U = no work is done all the heat goes into internal energy We can use this to find U = nc T U= nc T Constant olume specific heat at constant volume U c = the specific heat of the gas c depends n T on the structure of the gas molecule. Units J/ mole K 2,, This is true for all processes since U is a State function and only dependent on T. Constant ressure 2 Constant ressure W = d = (2 ) = Heat input = U+ = nc T+ = nc T+ nr T = nc T Specific heat at constant pressure, c c = c + R the specific heat at constant is larger than the specific heat at constant. because work is done by the gas Adiabatic rocess Heat input =0 U= -W is done at the epense of internal energy Define Adiabatic eponent c c + R R = = = + c c c depends on the nature of the gas, =.40 for air. relationship = = cons tant T relationship T T W = = = T cons tant =0 Adiabatic compression The air (=.40) in an automobile engine is compressed quickly so that appreciable heat echange does not occur. For a compression ratio / 2 =0. Find the temperature of the gas compressed from an initial temperature of 20 o C. Adiabatic compression The air (=.40) in an automobile engine is compressed quickly so that appreciable heat echange does not occur. For a compression ratio / 2 =0. Find the work done in compressing a gas from atm if the capacity is 2.8 l. Santa Anna Conditions T= 20 o C = 85 ka = 00 ka T=? High Desert. Coast Santa Ana winds originate in the high desert region of California and is heated by adiabatic compress during the rapid descent to the coast. For the above conditions what would the temperature at the coast be? 4

5 Thermodynamic processes done in epansion One mole of air (=.40) at = 0.25 atm, = 300 K, =0.m 3 epands by a factor of 2. Find the work done if the epansion is a) isobaric,b) isothermal c) adiabatic. Which process does the most work? Defining Characteristic First Law T constant U=0 =W constant U= constant U=-W No Heat input =0 U=-W done by the gas Other relations ln( 2 / ) =constant =mc T ( 2 - ) =nc T c =c +R =constant T - =constant 2 Temperature change in epansion One mole of air (=.40) at = 0.25 atm, = 300 K, =0.m 3 epands by a factor of 2. Find the temperature change if the epansion is a) isobaric, b) isothermal c) adiabatic. Which process produces the highest temperature rise? Heat input on epansion One mole of air (=.40) at = 0.25 atm, = 300 K, =0.m 3 epands by a factor of 2. Find the heat input if the epansion is a) isobaric, b) isothermal c) adiabatic. What process results in the largest heat input? c (air) =5/2R 2 2 5