Part 3 First Law of Thermodynamics

By Prof. Dr. Said Mohamed Attia & Thermodynamics Prof. Dr. Said M. Attia Outlines

1. Thermodynamic terminology. 1. System 2. Surroundings 3. Boundary 4. Thermodynamic Processes. 5. Thermodynamic equilibrium 2. in Thermodynamics 3. Internal Energy 4. First law of thermodynamics 1. Molar Specific heat of gas at constant volume and constant pressure. 2. Isotheral process 3. Isobaric process 4. Isochoric process 5. Adiabatic process. Heat & Thermodynamics Prof. Dr. Said M. Attia Some thermodynamic Terminologies

□ System • An amount of matter in a container of a certain volume SURROUNDINGS □ Surroundings • Mass or region outside the system

SYSTEM □ Boundary • The real or imaginary surface that separates the system from the surroundings BOUNDARY Heat & Thermodynamics Prof. Dr. Said M. Attia Some thermodynamic Terminologies

Isolated System: Neither the energy nor the mass can be exchanged with the surroundings

Closed System: Mass remains constant but energy can exchanged with the surroundings

Types ofSystems Types

Open System: Both energy and matter can be exchanged with the surroundings

Heat & Thermodynamics Prof. Dr. Said M. Attia Some thermodynamic Terminologies

Macroscopic System: The word macroscopic means ‘on a large scale”. A system is said to be macroscopic when it consists of a large number of molecules, atoms or ions

Macroscopic Properties: The properties associated with a macroscopic system are called macroscopic properties. These properties are pressure, volume, temperature, density, viscosity, surface tension, refractive index, electric conductivity etc.

State of the System: When macroscopic properties of a system About the the Systems About have definite values, the system is said to be in a definite state. The state of a system (or thermodynamic state) at a specific time is defined using thermodynamic variables such as pressure, volume, temperature, and internal energy, these quantities called state variables. Heat & Thermodynamics Prof. Dr. Said M. Attia Some thermodynamic Terminologies

Properties of a system

Any characteristic of a system, e.g., temperature, pressure, volume, etc.

خواص ممتدة تعتمد على Extensive Property خواص ممركزة تعتمد على Intensive Property كمية المادة وحجمها نوع المادة وال تعتمد على كمية المادة وال حجمها It is a bulk property, meaning that it is a It depends on the amount of the material in local physical property of a system that does the system. not depend on the system size or the amount of material in the system.

E.g., Pressure, Temperature, E.g., Volume, Mass, No. of moles, Density, Viscosity, Conductivity, Energy, Heat capacity, Entropy Surface tension, Boling point, refractive index, density, etc. Heat & Thermodynamics Prof. Dr. Said M. Attia Some thermodynamic Terminologies

Thermodynamics Equilibrium A system in which the macroscopic properties do not undergo any change with time is said to be in thermodynamic equilibrium. Three types of equilibrium;

In Thermal equilibrium • There is no flow of heat from one position of the system to another. This is possible if the temperature remains the same throughout in all parts of the system. In Mechanical equilibrium • There is no mechanical work is done by one part of the system on another part of the system. This is possible if the pressure remains the same throughout in all parts of the system. In Chemical equilibrium • the composition of the various phases in the system remains the same throughout. Heat & Thermodynamics Prof. Dr. Said M. Attia Some thermodynamic Terminologies

Thermodynamics Process A may be defined as the operation by which a system changes from one state (푃1, 푉1, 푇1) to another state (푃2, 푉2, 푇2). In a process, at least one property of the system changes and give a path along which the variables of the system changes. Isothermal Process • Process take place at constant temperature. (∆푇 = 0)

Isobaric Process • Process take place at constant pressure. (∆푃 = 0)

Isochoric Process • Process take place at constant volume. (∆푉 = 0)

Adiabatic Process • Process take place without heat exchange between system and surroundings. (∆푄 = 0) Heat & Thermodynamics Prof. Dr. Said M. Attia Some thermodynamic Terminologies

Thermodynamics Process

Cyclic Process • It is the process in which the system undergoes a number of changes (process), then finally returns to its initial state again. (∆푈 = 0)

Reversible Process • It is the process that can be reversed by changes in the system without loss or waste of energy.

Irreversible Process • It is the process that can not be reversed by changes in the system without loss or waste of energy. Heat & Thermodynamics Prof. Dr. Said M. Attia Work in Thermodynamics

□ Work (W) is the scalar product of a force times the displacement 풅풙 □ 푊표푟푘 = 퐹표푟푐푒 . 푑𝑖푠푝푙푎푐푒푚푒푛푡

□ For small changes; 푨 푭

□ 푑푊 = 퐹. 푑푥 □ If the force F is applied on a piston of Area A, then the piston move a distance 푑푥, as shown in the figure. □ The work done to change the volume by Δ푉 퐹 □ 푑푊 = . 퐴d푥 = 푃. 푑푉 퐴 □ A positive sign for W defines work done on a system (compression case) □ A negative sign for W defines work done by a system against force (expansion case) Heat & Thermodynamics Prof. Dr. Said M. Attia Work in Thermodynamics

□ We can talk about the work done on or by a system when some process has occurred in which energy has been transferred to or from the system. □ The total work done on the gas as its volume change from 푉푖 to 푉푓 is given by 푉 (*) 푑푊 = −푃푑푉 ∴ 푊 = − ׬ 푓 푃푑푉 □ 푉푖 풅푽 □ Equation (*) equal to the area under the curve on the PV diagram. □ The work done on the gas from initial state to final state depends on the path between these state. Heat & Thermodynamics Prof. Dr. Said M. Attia Work in Thermodynamics

□ The value of the work done depends on the path of the thermodynamic operation. □ In the figure, the work done through the path a from point 1 to point 2 is given by

□ 푊 = 2푃표 2푉표 − 푉표 = 2푃표푉표 □ the work done through the path b from point 1 to point 2 is given by

□ 푊 = 푃표 2푉표 − 푉표 = 푃표푉표 □ The work done is not a property of the system. Heat & Thermodynamics Prof. Dr. Said M. Attia Work in Thermodynamics

Special cases for the work=PV

1- The work at constant volume • In this case, 푉 = 푐표푛푠푡푎푛푡, 푎푛푑 푑푉 = 0, and 푃 푓 푊 = − ׬ 푃푑푉 = 0 • 푖 푃푓 푓 • In this case, there is no work done,

푃푖 𝑖 푉 푉푖 = 푉푓 2- The work at zero pressure (free expansion)

• In this case, 푃 = 0, and 푃 푓 푊 = − න 푃푑푉 = 0 푖 In this case, the system expands freely, 𝑖 푓 • and there is no work done, 푃 = 0 푉 푉푖 푉푓 Heat & Thermodynamics Prof. Dr. Said M. Attia Work in Thermodynamics

Special cases for the work=PV

3- The system expands against a constant Pressure

• In this case, 푃 = 푐표푛푠푡푎푛푡, 푎푛푑 Δ푉 = 푉푓 − 푉푖, and

푓 (푊 = − ׬푖 푃푑푉 = −푃(푉푓 − 푉푖 • • The work done is negative, when the system expands, 푃 • The work done is positive, when the system contract. 𝑖 푓 푃푖 = 푃푓

푉 푉푖 푉푓 Heat & Thermodynamics Prof. Dr. Said M. Attia Work in Thermodynamics

Special cases for the work=PV

4- The work done in a cyclic process ➢In a cyclic process, the system starts and returns to the same thermodynamic state. ➢The net work involved is the enclosed area on the 푃푉 diagram. ➢ The work done is negative, If the cycle goes clockwise, the system does work on the expense of the internal energy. ➢The work done is positive, If the cycle goes counterclockwise, work is done on the system every cycle, and the internal energy increases. ➢ The work done through the 푃 푃 path 퐴퐵 and 퐶퐷 is 퐵 퐶 퐶 퐵 푃푓 푃푓 ➢ 푊퐴퐵 = 푊퐶퐷 = 0, because Δ푉 = 0. 푃 ➢ The net work done, 푖 퐴 퐷 푃푖 퐷 퐴 푉 푉 푉 푉 ➢ 푊푛푒푡 = 푊퐵퐶 − 푊퐷퐴 푖 푓 푉푓 푉푖 work done is negative work done is positive Heat & Thermodynamics Prof. Dr. Said M. Attia Work in Thermodynamics

Special cases for the work=PV

5- The system expands against a variable Pressure

• If 푇 𝑖푠 푐표푛푠푡푛푎푡, in this case, 푃 푓 • 푊 = − 푃푑푉 푓 ׬푖 푃푓 • For ideal gas 푃푉 = 푛푅푇 = 푐표푛푠푡푎푡

푛푅푇 푃푖 𝑖 • Then 푃 = , then 푉

푉푓 푛푅푇 푉푓 1 푉푓 푊 = − ׬ 푑푉 = −푛푅푇 ׬ 푑푉 = −푛푅푇 ln 푉 푉 푉 • 푉푖 푉 푉푖 푉 푉푖 푓 푉푖 푉 푃 • 푊 = −푛푅푇 ln 푓 = −푛푅푇 ln 푖 푉푖 푃푓 • The work done is negative, when the system expands, • The work done is positive, when the system contract. Heat & Thermodynamics Prof. Dr. Said M. Attia Internal energy (U)

□ Internal energy (U) is all energies associated with the kinetic and potential energies of the molecules, electronic energies, and energy of nuclei.

□ Internal Energy (U) = Kinetic Energy + Potential 푲. 푬. 풂풕 푻 Energy ퟏ

□ Kinetic energy includes □ the K.E. of random translation,

□ the K.E. of random rotation, and 푲. 푬. 풂풕 푻ퟐ □ the K.E. of random vibration motion of molecules, □ The potential energy which include all binding energy between molecules, electrons, etc. Heat & Thermodynamics Prof. Dr. Said M. Attia Enthalpy (H)

□ Enthalpy, H, is a property of a , that equal the system's internal energy plus the product of its pressure and volume, i.e., □ 퐻 = 푈 + 푃푉 (Enthalpy) □ 푑퐻 = 푑푈 + 푃푑푉 + 푉푑푃 □ That is the change in enthalpy equals the change in the internal energy and change in the work done □ At constant pressure, 푑푃 = 0, then, □ Δ퐻 = Δ푈 + 푃Δ푉 □ For liquid and solids, (Δ푉 ≈ 0), then, □ Δ퐻 = Δ푈 □ It is an extensive quantity. The unit of measurement for enthalpy in the International System of Units (SI) is the joule. Other historical conventional units still in use include the British thermal unit (BTU) and the calorie. Heat & Thermodynamics Prof. Dr. Said M. Attia Enthalpy (H)

□ In a closed system (no matter transfer), for processes at constant pressure, the heat absorbed or released equals the change in enthalpy. □ The total enthalpy, H, of a system cannot be measured directly. □ Enthalpy itself is a thermodynamic potential, so in order to measure the enthalpy of a system, we must refer to a defined reference point; therefore what we measure is the change in enthalpy, ΔH. The ΔH is and ,عملية ماصة للحرارة a positive change in endothermic Process .عملية طاردة للحرارة negative in heat-releasing or exothermic processes Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics

□ The statement of 1st law of thermodynamics: Energy can be transformed from one form to another and energy can neither be created nor destroyed For an isolated system: Or A system’s internal energy can be changed by an energy transfer to or from the system either by heat or by work. □ For an isolated system:

□ The internal energy of an isolated system is constant (Δ푈푖푠표푙푎푡푒푑 = 0) □ For a closed system: □ Heat added to system = Change in internal energy + work done on system □ The Mathematical Formula of 1st law of thermodynamics

Δ푄 = Δ푈 + Δ푊 , Δ푈 = 푈푓 − 푈푖 □ I.E., the change in quantity of heat of the system equal to the change in the internal energy + the change in the work done by the gas. □ For infinitesimal changes: 푑푄 = 푑푈 + 푑푊 = 푑푈 + 푃푑푉 Heat & Thermodynamics Prof. Dr. Said M. Attia Molar Specific heat (Molar heat capacity) of gas

at constant volume , cV and constant Pressure, cP

□ As we have mentioned before, the specific heat 푐 is the quantity of heat required to increase the temperature of 1 kg of the material by 1 oC.

□ The molar specific heat of gas (molar heat capacity), 푐푀 is the quantity of heat energy required to increase the temperature of 1 mol of the material by 1 oC. It is a heat capacity for one mole of the gas. □ The heat capacity 퐶 is the quantity of heat required to increase the temperature of the whole system (푚 kg of the material) by 1 oC. □ The change in heat Δ푄 equals

□ Δ푄 = 푚푐Δ푇 = 퐶Δ푇 = 푛푐푀Δ푇, □ where 푛 equals the number of mole, and the heat capacity is equal

퐶푀 □ 퐶 = 푚푐 = 푛푐푀 , and specific heat = 푐 = , 푊푀 = 푀표푙푎푟 푚푎푠푠 푊푀 휕푄 □ The molar specific heat is calculated using the equation 푐 = . For gases we 푀 휕푇 differentiate between the specific heat under constant volume which equal to 휕푈 휕퐻 푐 = , and the specific heat under constant pressure which equal 푐 = . 푉 휕푇 푃 휕푇 Heat & Thermodynamics Prof. Dr. Said M. Attia Molar Specific heat of gas

at constant volume , cV and constant Pressure, cP

□ Now, Suppose a 1 mol of a substance absorb an amount of heat 푑푄, How much does the temperature increase? □ From the first law of thermodynamics □ 푑푄 = 푑푈 + 푑푊

□ At constant volume (Isochoric Process)

□ 푑푉 = 0 , 푡ℎ푒푛 푑푄 = 푑푈, then at constant volume

휕푈 □ 푑푈 = 푑푄 = 푑푇 휕푇 푉

푑푄 푑푈 □ But, 푑푄 = 푐. 푑푇 푂푟 푐푉 = , then 푐푉 = 푑푇 푉 푑푇 푉

□ 푐푉is the molar specific heat at constant volume. Heat & Thermodynamics Prof. Dr. Said M. Attia Molar Specific heat of gas

at constant volume , cV and constant Pressure, cP

□ At constant Pressure (Isobaric process),

□ From the first law of thermodynamic,

□ 푑푄 = 푑푈 + 푑푃푉

□ 푑푄 = 푑 푈 + 푃푉 = 푑퐻 = 푐ℎ푎푛푔푒 𝑖푛 푒푛푡ℎ푎푙푝푦

□ 푑푄 = 푑푈 + 푃푑푉 + 푉푑푃

□ But 푃 = 푐표푛푠푡푎푛푡, then 푉푑푃 = 0, then

□ 푑푄 = 푑(푈 + 푃푉) = 푑퐻 = 푐푃. 푑푇

휕퐻 휕푄 □ 푐푃 = = 휕푇 푝 휕푇 푃

□ 푐푃 is the molar specific heat at constant pressure Heat & Thermodynamics Prof. Dr. Said M. Attia Molar Specific heat of gas

at constant volume , cV and constant Pressure, cP

□ Therefore, for 푛 mols, in isobaric process (at constant pressure) we get:

□ 푑푄 = 푑푈 + 푑푊 = 푛푐푃Δ푇 (1) □ In isochoric process, (at constant volume) we get

□ 푑푈 = 푛푐푉Δ푇 (2) □ From these two equations, we get

□ 푛푐푃Δ푇 = 푛푐푉Δ푇 + 푑푊 □ From ideal gas equation □ 푃푉 = 푛푅푇, then 푑푊 = ΔPV = nRΔ푇, then

□ 푛푐푃Δ푇 = 푛푐푉Δ푇 + 푛푅Δ푇 □ Dividing by 푛Δ푇, we get:

□ 푐푃 = 푐푉 + 푅 (3) (This is the molar specific heat)

□ For Heat capacity, where 퐶푃 = 푛푐푃, 푎푛푑 퐶푉 = 푛푐푉, then we get

□ 퐶푃 − 퐶푉 = 푛푅 (4) (This is the heat capacity) Heat & Thermodynamics Prof. Dr. Said M. Attia Molar Specific heat of gas

at constant volume , cV and constant Pressure, cP

□ Molar specific heat at constant volume for Ideal gas. 1 푑푈 □ From Eq. 2, we get 푐 = 푉 푛 푑푇 □ From Kinetic gas theory, we get 3 3 □ ∆푈 = 푃. 퐸 + 퐾. 퐸 = 0 + 푁퐾푇 = 푛푅푇 2 2 1 3푛 3 □ 푐 = 푅 Or 푐 = 푅 푉 푛 2 푉 2 □ This is the molar specific heat at constant volume for ideal gas. □ Molar specific heat at constant pressure for Ideal gas.

□ From Eq. 3, we get 푐푃 = 푐푉 + 푅 3 5 □ Then, 푐 = 푅 + 푅 = 푅 푃 2 2 □ This is the molar specific heat at constant pressure for ideal gas. Heat & Thermodynamics Prof. Dr. Said M. Attia Molar Specific heat of gas

at constant volume , cV and constant Pressure, cP

□ 휸 ratio

□ It is the ration between molar specific heat at constant pressure to the molar specific heat at constant volume. That is

퐶 □ 훾 = 푝 퐶푉 □ It is a dimensionless quantity. For ideal gas, 훾 푟푎푡𝑖표 𝑖푠

퐶 5/2푅 5 □ 훾 = 푝 = = = 1.67 퐶푉 3/2푅 3

□ For a complex gas 푐푃, 푎푛푑 푐푉 differ due to the contribution from the rotational and vibrational motion of the molecules

□ In solids and liquids 푐푃, 푎푛푑 푐푉 are approximately equal and 훾 = 1. Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases – Isochoric process

□ In an Isochoric process: □ That is, a process takes place at constant volume. In the figure, by clamping the piston at a fixed position, then 푉 = 푐표푛푠푡푎푛푡, and 푑푉 = 0. In this case the work done is zero □ 푊 = 푃푑푉 = 0 □ Then the change in the internal energy equal the quantity of heat,

□ Δ푈 = Δ푄 = 푐푉Δ푇 = 푐푉(푇2 − 푇1) □ This means if energy is added by heat to a system at constant volume, then all of the transferred energy remains in the system as an increase in its internal energy and hence increase in the temperature from 푇1 to 푇2, and the system will not do any work. □ The enthalpy 퐻 = 푈 + 푃푉, then □ 푑퐻 = 푑푈 + 푉푑푃 + 푃푑푉 , but 푑푉 = 0 □ 푑퐻 = 푑푈 + 푉푑푃

□ Δ퐻 = 푐푉 푇2 − 푇1 + 푉Δ푃 Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - Isobaric process

□ In an Isobaric process: □ That is, a process that occurs at constant pressure. □ This occurs by allowing the piston to move freely so that there is equilibrium between the net force from the gas pushing upward and the weight of the piston plus the force due to atmospheric pressure pushing down ward . In this case;

□ Δ푊 = 푃Δ푉 = 푃 (푉푓 – 푉푖) □ Then the change in the internal energy is

□ 푈2 − 푈1 = Δ푄 − 푃 (푉푓 – 푉푖), then

□ Δ푄 = 푈2 + 푃 푉푓 − 푈1 + 푃푉푖 = 퐻2 − 퐻1 = Δ퐻 = 퐶푃Δ푇

□ Where 퐶푃 𝑖푠 푡ℎ푒 ℎ푒푎푡 푐푎푝푎푐𝑖푡푦 푎푡 푐표푛푠푡푎푛푡 푝푟푒푠푠푢푟푒 = 푛푐푃

□ 푛 = 푛푢푚푏푒푟 표푓 푚표푙푠, 푐푃 is the molar specific heat at constant Pressure. □ The change in enthaliby Δ퐻 is given by

□ Δ퐻 = Δ푄 = 퐶푃Δ푇 = 푈2 − 푈1 + 푃 (푉푓 – 푉푖) = 푐푉Δ푇 + 푃푑푉 Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - Isothermal process

□ In an Isothermal process □ Isothermal process is a process that occurs at constant temperature. □ Since 푇 𝑖푠 푐표푛푠푡푎푛푡 , 푡ℎ푒푛 Δ푈 = 0 푎푛푑 Δ푄 = 푊 □ Since Δ푈 = 0, 푡ℎ푒푛 푈 = 푐표푛푠푡푎푛푡, and since 푃푉 = 푛푅푇 = 푐표푛푠푡푎푛푡. , then change in enthalpy = Δ퐻 = 0, i.e., the enthalpy is constant. B □ Note that the energy entering the gas by heat leaves by work so the temperature can remain constant. □ The ideal gas law with T constant indicates that the 푐표푛푠푡. equation of this curve is PV = constant. Then 푃 = , 푉 then

푉푓 1 푉푓 1 푉푓 푊 = 푐표푛푠푡. ׬ 푑푉 = 푃푖푉푖 ׬ 푑푉 = 푃푖푉푖 ln 푉 □ 푉푖 푉 푉푖 푉 푉푖

푉푓 푉푓 □ 푊 = 푃푖푉푖 ln = 푃푓푉푓 ln 푉푖 푉푖 □ This is the work done during isothermal process. Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - Adiabatic Process

□ In an Adiabatic Process: □ In adiabatic process; there is no exchange of thermal energy between the system and the surroundings, when the system is very well insulated, i.e., 푑푄 = 0 , and Δ푈 = −Δ푊 □ In the expansion process of the gas, the work done by the gas (is negative) on the expense of the internal energy, and the temperature decreases. □ In the compression process of the gas, the work is done on the gas and the internal energy of the gas increases , and the temperature increases.

□ The work done on the gas Δ푊 = 푃푒푥Δ푉

푑푈 푃푒푥푑푉 □ Since d푈 = 푐푉d푇, then 푑푇 = = − 푐푉 푐푉 Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - Adiabatic Process

□ then the first law of thermodynamics becomes

□ 0 = 푑푈 + 푑푊 = 푐푉푑푇 + 푃푑푉 (for one mole) □ For n mole this equation becomes 푛푅푇 □ 퐶 푑푇 + 푑푉 = 0 푉 푉 푑푇 푛푅 □ Dividing by T, we get 퐶 + 푑푉 = 0 푉 푇 푉

□ But 퐶푃 − 퐶푉 = 푛푅, then 푑푇 푑푉 □ 퐶 + (퐶 −퐶 ) = 0 푉 푇 푃 푉 푉

□ Dividing by 퐶푉, 푑푇 퐶 푑푉 푑푇 푑푉 퐶 □ then + ( 푃 − 1) = 0 Or, + (훾 − 1) = 0 , where 푃 = γ 푇 퐶푉 푉 푇 푉 퐶푉 □ By integration, 푑푇 푑푉 ׬ 0 = ׬ + ׬ 훾 − 1 □ 푇 푉 □ 푐표푛푠푡 = 퓁푛 푇 + 훾 − 1 퓁푛(푉) Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - Adiabatic Process

□ Then : □ (a) Relation between V and T □ 푐표푛푠푡 = 퓁푛 푇 푉( 훾 −1) Or 푻 푽( 휸 −ퟏ) = 풄풐풏풕풂풏풕. □ (b) Relation between P and V 푃푉 □ Since 푃푉 = 푛푅푇, then 푇 = , then the last equation becomes 푛푅 푃푉 □ 푐표푛푠푡 = 퓁푛 푉( 훾 −1) , 푛푅 □ then 푐표푛푠푡 = 퓁푛푃푉 훾 Or 푷푽휸 = 풄풐풏풔풕풂풏풕. □ (c) Relation between T and P

푛푅푇 ( 훾 −1) □ Also 푐표푛푠푡 = 퓁푛 푇 푉( 훾 −1) , then 푐표푛푠푡 = 퓁푛 푇 푃 푇훾 □ Then 푐표푛푠푡 = 퓁푛 Or 푻휸푷ퟏ−휸 = 풄풐풏풔풕풂풏풕 푃훾−1 Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - Adiabatic Process

□ The work done in an adiabatic process: 2 푊 = ׬1 푃푑푉 □ 훾 훾 훾 □ ∵ 푃푉 = 푐표푛푠푡 = 푃1푉1 = 푃2푉2 … … … = 퐴 A □ ⇒ P = 푉훾

푉 퐴 푉1− 훾 푊 = ׬ 2 푑푉 = 퐴 V2 ⇒ □ 푉1 푉훾 1−훾 푉1

푉1− 훾 푉1− 훾 □ = 퐴 2 − 1 1−훾 1−훾 훾 □ Substituting by 퐴 = 푃2푉2 , then γ 1− γ 훾 1− 훾 P V V 푃 푉 푉 푃 푉 푃 푉 □ ⇒ W = 2 2 2 − 1 1 1 = 2 2 − 1 1 1− γ 1− 훾 1−훾 1−훾 푃 푉 −푃 푉 푛푅 (푇 −푇 ) □ Then the work done by the gas is 푊 = 2 2 1 1 = 2 1 1− 훾 1− 훾 푇 −푇 □ And the work done on the gas is 푊 = 푛푅 2 1 1− γ □ The work depends on the temperature difference Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - Adiabatic Process

□ The change in the enthalpy in adiabatic process □ 푑퐻 = 푑푈 + 푃푑푉 + 푉푑푃 □ But in a diabatic process, we find □ dQ = dU + PdV = 0 □ ⇒ dH = VdP □ If 푃푉 = 푐표푛푠푡 □ 푃푑푉 + 푉푑푃 = 0 □ ⇒ 푃푑푉 = −푉푑푃 □ ⇒ d퐻 = − 푃푑푉 = − 푑푊 □ 푇ℎ푒 푐ℎ푎푛푔푒 𝑖푛 푡ℎ푒 푒푛푡ℎ푎푙푝푦 푒푞푢푎푙푠 푡ℎ푒 푤표푟푘 푑표푛푒 Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - Adiabatic Process

□ Adiabatic free expansion □ In this case, The external pressure

□ 푃푒푥 = 0 Gas Vacuum □ Δ푄 = 0 because, it is adiabatic ,

□ Δ푊 = 푃푒푥푑푉 = 0 , 푡ℎ푒푛 □ Δ푈 = Δ푄 − Δ푊 = 0 , □ Since the internal energy does not change, then the temperature dose not change, i.e., □ Δ푇 = 0 □ The process is adiabatic because it takes place in an insulated container the gas expands into a vacuum without work on or by the gas , Heat & Thermodynamics Prof. Dr. Said M. Attia Adiabatic process versus Isothermal process.

□ For PV graph, the slope of isothermal process is less than that of adiabatic process.

□ For Isothermal process, 푷 □ 푃푉 = 푛푅푇 = 푐표푛푠푡푛푎푡, then 푑푃 푃 □ 푃푑푉 + 푉푑푃 = 0, then = − 푑푉 푉 Isothermal □ For adiabatic process, Adiabatic □ 푃푉훾 = 푐표푛푠푡푛푎푡, then 푑푃 푃 □ 푉훾푑푃 + 훾푃푉훾−1푑푉 = 0, then = −훾 푽 푑푉 푉 푐 □ Since 훾 = 푃 > 1, then 푐푉 □ The slope of the isothermal curve is less than that of adiabatic process. Heat & Thermodynamics Prof. Dr. Said M. Attia Summary of the thermodynamic processes

Process Constant value The first law predicts

Isothermal 푻 = 풄풐풏풔풕풂풏풕 횫퐓 = ퟎ , 휟퐔 = ퟎ , 퐖 = 휟퐐

Isobaric 푷 = 풄풐풏풔풕풂풏풕 퐐 = 횫퐔 + 퐖 = 횫퐔 + 퐏횫퐕

횫퐕 = ퟎ 퐦퐚퐤퐞퐬 Isovolumetric 푽 = 풄풐풏풔풕풂풏풕 퐖 = ퟎ, (isochoric) 퐐 = 횫퐔

Adiabatic 휟푸 = ퟎ 횫퐔 = −퐖 Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - cyclic process

□ In a cyclic process: State 2 □ In a cyclic process, the system starts and returns to the same thermodynamic state.

□ Δ푈표푣푒푟푎푙푙 = 푈2 − 푈1 + 푈1 − 푈2 = 0 :ׯ 푑푈 = 0 (total change in internal energy), Step 1: Step 2 □ Δ푈 = −Δ푈 = ׯ 푑퐻 = 0 (total change in enthalpy), 푈 − 푈 □ 푈2 − 푈1 1 2 .(ׯ 푑푆 = 0 (total change in entropy □ □ Δ푊 = Δ푄 energy Internal □ The steeps of the cyclic process shown in the figure, (I) Isothermal expansion State 1 (II) Adiabatic expansion (III) Isothermal compression 푉1 (IV) Adiabatic compression 퐼 푉2 □ The net work involved is the enclosed area on the P-V diagram. If the cycle goes clockwise, the system does 퐼푉 work. A cyclic process is the underlying principle for an 퐼퐼 engine. 푃푟푒푠푠푢푟푒 □ If the cycle goes counterclockwise, work is done on the 푉 4 퐼퐼퐼 system every cycle. An example of such a system is a 푉3 refrigerator or air conditioner. 푉표푙푢푚푒 Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - cyclic process

□ Considering the shown figure, where 푃 퐴 1 □ The path AB: Expansion at constant 푃푓 퐵 4 temperature (푇1). 2 The path BC: Removal of heat at constant 퐷 □ 푃푖 3 퐶 volume (푉푓 ). 푉 푉 The path CD: Compression at constant 푉푖 푓 □ work done is negative temperature (푇2). □ The path DA: Addition of heat at constant 푃 volume (푉 ). 퐴 푖 푃푓 퐵

□ Because the process is cyclic, there is no change 퐷 in internal energy after each cycle. Therefore the 푃푖 퐶 net work done in each cycle equals the heat 푉 푉푖 푉푓 added to the system. We now analyze each of the work done is negative steps in the cycle. Heat & Thermodynamics Prof. Dr. Said M. Attia First Law of Thermodynamics Special cases - cyclic process

□ Step 1 - Isothermal expansion: The system does 푃 work 푊1 which equals the heat 푄1 added to the 퐴 system in the expansion, because the internal 푃 1 energy does not change. 푓 퐵 4 2 □ Step 2 - Isochoric process: The work done is 푊2 퐷 푃 = 0. Heat 푄2 is removed from the system because 푖 3 퐶 the temperature decreases from 푇 to 푇 . 1 2 푉 푉푖 푉푓 work done is negative □ Step 3 - Isothermal compression: The work 푊3 done by the system is negative, but of smaller magnitude than 푊1 because the area under the 푃푉 curve is less 푃 than that in step 1. The internal energy is does not 퐴 푃푓 퐵 change, so the heat removed is 푄3 = 푊3.

□ Step 4 - Isochoric process: The reverse of step 2. 푊4 퐷 푃 = 0, while heat 푄4 = − 푄2 is added to the system. 푖 퐶 푉 □ The net work done = The area encountered by 푉푖 푉푓 ABCD : work done is positive Heat & Thermodynamics Prof. Dr. Said M. Attia Enthalpy H, and Internal Energy U

H = Enthalpy U = Internal energy

Definition 퐻 = 푈 + 푃푉 푑푈 = 푑푄 − 푑푊

Differentiation 푑퐻 = 푑푈 + 푃푑푉 + 푉푑푃 푑푈 = 푑푄 − 푑푊

Change w.r.t 푑퐻 = 푑푄푃 at 푃 constant 푑푈 = 푑푄푉 at constant 푉 temperature 푑퐻 푑푈 푐 = 푐 = 푝 푑푇 푉 푑푇 푝 푉 푑퐻 = 푉푑푃 , 푑푈 = −푑푊, Adiabatic process 푝 푉 퐻 = න 푉푑푃 푈 = − න 푃푑푉 0 0

In Ideal gas 퐻 = න 푐푃푑푇 + 푐표푛푠푡. 푈 = න 푐푉푑푇 + 푐표푛푠푡. Heat & Thermodynamics Prof. Dr. Said M. Attia Questions and homework

□ A 28 L container contains air at 140 kPa at 20 oC. If the container is heated then the pressure becomes 345 kPa, calculate the amount of heat. 푀푊 = 29 푔/푚표푙 , 푐푉 = 0.718 푘퐽/푘푔. 퐾 . □ 2 mols of ideal gas at 300K and pressure 0.4 atm. If the pressure is increased to 1.2 atm, calculate (a) the initial and final volume of the gas, (b) the work done, (c) The amount of heat transferred to the system. □ An ideal gas has been expanded to double of its initial volume according the equation 푃 = 훼 푉2, Calculate the work done by the gas, where 훼 = 5 푎푡푚/푚6. □ Calculate the final temperature and final pressure and the work o done for 220 kg of CO2 at 1 atm, and 27 C, where the gas was pressed to 1/5 of its initial volume through an adiabatic process.