Download presentation

Presentation is loading. Please wait.

1
**PPT 108 PHYSICAL CHEMISTRY**

Semester 2 Academic session 2011/2012 NURUL AIN HARMIZA ABDULLAH

2
**CHAPTER 2 FIRST LAW OF THERMODYNAMICS**

3
**Characterized by the 4 laws of thermodynamics**

THERMODYNAMIC SYSTEM Characterized by the 4 laws of thermodynamics

5
**CHAPTER 2 First law of thermodynamics**

6
**Key Concepts Work Pressure-Volume Work (P-V Work) Heat**

First Law of Thermodynamics Enthalpy Perfect Gas Molecular Nature of Internal Energy

7
WORK Work (w) is defined as the force (F) that produces the movement of an object through a distance (d): Work is done when and object, e.g. a system's wall, moves against an opposing force. This is equivalent to an ordered motion done by the system on the surroundings or vice versa. Work = force ×distance w = F x d Work also has units of J, kJ, cal, kcal, Cal, etc.

9
**Where P is external pressure**

Pressure-volume work The example of PV work – in the cylinder of an automobile engine The combustion of the gasoline causes gases within the cylinder to expand, pushing the piston outward and ultimately moving the wheel of the car. The relationship between a volume change (ΔV) and work (w): W= -P ΔV Where P is external pressure The units of PV work are L·atm; 1 L·atm = J. If the gas expands, ΔV is positive, and the work term will have a negative sign (work energy is leaving the system). If the gas contracts, ΔV is negative, and the work term will have a positive sign (work energy is entering the system). If there is no change in volume, ΔV = 0, and there is no work done. (This occurs in reactions in which there is no change in the number of moles of gas.)

10
**Pressure-volume (P-V) work**

The work done in a volume change is called P-V work By holding the pressure constant and increasing the volume by heating the gas using Charle’s law By removing weights and decreasing the pressure and allowing the volume to adjust according to Boyle's law with no heat addition Changing the volume by 2 ways OR Initial state Final state Final state

11
**Work done by gas The units of PV work are L·atm; 1 L·atm = 101.3 J.**

if ΔV < 0, then W > 0.; increases in volume means work done BY the system on the environment. if ΔV > 0, then W < 0.; decreases in volume means work done by the environment ON the system.

12
**Work and heat are not state functions**

On a graph of pressure versus volume, the work is the area under the curve that describes how the state is changed from State 1 to State 2. A curved black line from State 1 to State 2 represents a change brought about by removing weights and decreasing the pressure and allowing the volume to adjust according to either Boyle’s law (the line is curved and the amount of work done on the gas is shown by the red shaded area below this curve) or Charle’s law (the resulting change in state proceeds from State 1 to an intermediate State "a" on the graph by heating. State "a" is at the same pressure as State 1, but at a different volume. If we then remove the weights, holding a constant volume, we proceed on to State 2. The work done in this process is shown by the yellow shaded area).

13
Using either process we change the state of the gas from State 1 to State 2. But the work for the constant pressure process is greater than the work for the curved line process. The work done by a gas not only depends on the initial and final states of the gas but also on the process used to change the state. Different processes can produce the same state, but produce different amounts of work. Notice that not only does the work done by the gas depend on the process, but also the heat transferred to the gas. In the first process, the curved line from State 1 to State2, no heat was transferred to the gas; the process was adiabatic. But in the second process, the straight line from State 1 to State "a" and then to State 2, heat was transferred to the gas during the constant pressure process. The heat transferred to a gas not only depends on the initial and final states of the gas but also on the process used to change the state.

14
**Why energy is a state function but heat and work are not?**

State Function is a thermodynamic parameter whose value does not depend on the thermodynamic process...It depends only on the initial and final states. State functions are characterized by the idea that no matter what path you take to get from point A to point B, the difference between B and A remains the same. An example of how this works can be illustrated using gravitational potential energy. Say you lift a box 5 meters vertically straight up into the air from the surface of the ground (which we shall set to a height of 0 m). You increase its energy by an amount mg h where h = 5m. Now say that you lift a box from the ground level up 1 meter above the ground, walk to the right 5 meters, lift the box 1 more meter, walk to the left 2 meters, raise the box 7 meters, walk to the left 1 more meter, and lower the box 4 meters. How much did the gravitational potential energy change? The same amount as before, mg h where h = 5m, because the box is still only 5 meters above where it started in either case and the potential energy depends only on the final height minus the initial height (it depends only on the final and initial position). In contrast, you expended a lot more energy (more work) moving the box around than you did just standing in place and lifting it straight up. So the work you did is NOT a state function. This function depends on the path. When you lifted the box straight up you did less work than your body did lugging the box around left and right while lifting it up and down. So the potential energy is a state function, but the work you did is a path function. Now in thermodynamics you usually apply these ideas to the 1st law or conservation of energy: U = Q + W In this example, Q and W are path functions...the exact amount of work you do or the heat transferred to the system depend on how you add the heat or do the work (these individual values will vary depending on whether or not you did these things in either a direct or roundabout way). However, if you add these two path functions together the overall result, U, is a state function. It only depends on U (final) – U (initial) , regardless of whether or not you did direct or roundabout Q's and W's.

15
**What are Reversible and Irreversible Processes?**

There are two main types of thermodynamic processes: the reversible and irreversible. The reversible process is the ideal process which never occurs, while the irreversible process is the natural process that is commonly found in the nature. Therefore: The reversible process is an idealization. All real processes on Earth are irreversible. P-V Work REVERSIBLE IRREVERSIBLE A reversible process is one that can be halted at any stage and reversed. In a reversible process, the system is at equilibrium at every stage of the process. An irreversible process is one where it cannot be halted at any stage and reversed and the system is not always at equilibrium at every stage of the process.

16
**What is a Reversible Process?**

The process in which the system and surroundings can be restored to the initial state from the final state without producing any changes in the thermodynamics properties of the universe is called a reversible process. In the figure below, let us suppose that the system has undergone a change from state A to state B. If the system can be restored from state B to state A, and there is no change in the universe, then the process is said to be a reversible process. The reversible process can be reversed completely and there is no trace left to show that the system had undergone thermodynamic change. For the system to undergo reversible change, it should occur infinitely slowly due to infinitesimal gradient. During reversible process, all the changes in state that occur in the system are in thermodynamic equilibrium with each other. Thus there are two important conditions for the reversible process to occur. Firstly, the process should occur in infinitesimally small time and secondly all of the initial and final state of the system should be in equilibrium with each other. If during the reversible process the heat content of the system remains constant, i.e. it is adiabatic process, then the process is also isentropic process, i.e. the entropy of the system remains constant. The phenomenon of undergoing reversible change is also called reversibility. In actual practice the reversible process never occurs, thus it is an ideal or hypothetical process.

17
**What is an Irreversible Process?**

The irreversible process is also called the natural process because all the processes occurring in nature are irreversible processes. The natural process occurs due to the finite gradient between the two states of the system. For instance, heat flow between two bodies occurs due to the temperature gradient between the two bodies; this is in fact the natural flow of heat. Similarly, water flows from high level to low level, current moves from high potential to low potential, etc. Here are some important points about the irreversible process: In the irreversible process the initial state of the system and surroundings cannot be restored from the final state. During the irreversible process the various states of the system on the path of change from initial state to final state are not in equilibrium with each other. During the irreversible process the entropy of the system increases decisively and it cannot be reduced back to its initial value. The phenomenon of a system undergoing irreversible process is called as irreversibility.

18
**or Are these examples of reversible of irreversible process?**

Gas confined in a cylinder by means of a moveable piston The cylinder of an automobile engine Are these examples of reversible of irreversible process?

19
**Calculation of PV work for reversible and irreversible processes**

A reversible process is one that can be halted at any stage and reversed. In a reversible process the system is at equilibrium at every stage of the process. An irreversible process is one where these conditions are not fulfilled. If pressure has changed continuously then the work is reversible: If the work done irreversibly by changing pressure to Pfinal when for the work we have: The work done by a reversible process represents the maximal work that the system can perform changing between the same original and final states.

20
Example for P-V WORK Inflating balloon requires the inflator to do pressure-work on the surroundings. If balloons is inflated from a volume of 0.100L to 1.85L against an external pressure of 1.00 atm, how much work is done (in joules)? Answer: ΔV = V1 -V2 = 1.85L L = 1.75L W= -P ΔV = atm x 1.75L = -1.75L.atm Convert to Joule: -1.75L.atm x J= -177 J 1L.atm The work is negative because it is being done by the system as its volume increases due to the expansion of the gas into the much bigger volume.

21
Study Example 2.2 page 44 Find the work 𝑾 𝒓𝒆𝒗 for processes (a) and (b) of Fig. 2.3 if P atm, V1 500 cm3, P atm, and V cm3. Figure 2.3 The work w done on the system in a reversible process (the heavy lines) equals minus the shaded area under the P-versus-V curve. The work depends on the process used to go from state 1 to state 2. Use this equation: Simple as the previous example. Processes (a) and (b) are expansions. Hence the system does positive work on its surroundings, and the work w done on the system is negative in these processes. Answer: 𝑊 𝑎 = J and 𝑊 𝑏 = J

23
HEAT Heat is a exchange of thermal energy between a system and its surroundings caused by temperature difference. Notice the distinction between heat and temperature. Temperature is a measure of the thermal energy of a sample matter. Heat is transfer of the thermal energy. Heat may be defined as energy in transit from a high temperature object to a lower temperature object. Anytime two substances with different temperatures come in contact with each other, there is an energy transfer. One substance loses heat energy and the other substance gains heat energy. Heat energy flows from a hotter substance to a colder substance.

24
For example, if the ice cube in the diagram is placed in the container of water, there is an energy transfer. The hotter substance loses heat energy and the colder substance gains heat energy. The water and its container lose heat energy and become cooler. The ice cube gains heat energy and becomes warmer (this causes the ice cube to melt). According to the law of energy conservation the total heat energy lost by the water and its container is equal to the total heat energy gained by the ice. Heat can be transferred reversibly or irreversibly. A reversible transfer of heat requires that the temperature difference between the two bodies be infinitesimal. When there is a finite temperature difference between the bodies, the heat flow is irreversible.

26
Heat capacity when system absorbs heat (q) its temperature changes by ΔT: Experimental measurements demonstrate that the heat absorbed by a system and its corresponding temperature change are directly proportional: qαΔT. The constant of proportionality between q and ΔT is called the heat capacity. Heat capacity (C) is the amount of heat (q) a substance must absorb to raise its temperature(ΔT) by 1 °C.-Heat capacity has units of J/°C (or J/K), and is an extensive property, depending on the sample size.

27
**Specific Heat Capacity**

The specific heat (c, or specific heat capacity, Cs) of an object, is the quantity of heat required to change the temperature of 1 gram of a substance by 1°C (or K): Specific heat has units of J/g°C, and is an intensive property, which is independent of the sample size.

28
Example Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 56oC. Answer: q = m x Cg x (Tf - Ti) m = 250g Cg = 4.18 J oC-1 g-1 (from table) Tf = 56oC Ti = 20oC q = 250 x 4.18 x ( ) q = 250 x 4.18 x 36 q = J = 38 kJ

29
**What is the difference between heat capacity and specific heat?**

Heat capacity is the amount of heat required to raise the temperature of any quantity of substance by 1 degree centigrade. Specific heat is the amount of heat required to raise the temperature of 1kg (MASS) of substance by 1 degree substance. The unit for specific heat is J/(g*K). The unit for heat capacity is J/K. Difference between specific heat capacity and heat capacity is identical to difference between concentration and amount of substance - one is intensive, other extensive property.

30
**Heat capacity at constant pressure (or isobaric heat – constant P**

An isobaric process is a constant-pressure process. In general, none of the three quantities ΔU, Q and dW is zero in an isobaric process For closed syst. in equilib., P-V work only ΔH = 𝐪 𝐩

31
**Heat capacity at constant volume (or isochoric heat capacity)-constant V**

An isochoric process is a constant-volume process. When the volume of a thermodynamic system is constant, it does zero work on the surroundings. Then W = 0, and For closed syst. in equilib., P-V work only w is zero In an isochoric process, all the energy added as heat remains in the system as an increase in internal energy. Heating a gas in a closed constant-volume container is an example of an isochoric process. (Note that there are types of work that do not involve a volume change. For example, we can do work on a fluid by stirring it. In some cases, “isochoric” is used to mean that no work is done). V f = V i : vertical line on p V diagram

32
**Is the law of conservation of Energy.**

33
**1st Law = Conservation of Energy**

The first law of thermodynamics is simply an expression of the conservation of energy principle. The principle of the conservation of energy states that energy can neither be created nor destroyed. But it can change from one type of energy to another (for example kinetic to potential) but the total amount remains fixed (The total energy of a closed system remains constant.)

34
**In thermodynamics, energy is classified into three different types:**

work (W) heat (Q) internal energy (ΔU) This allows us to write a simple form for conservation of energy (or the first law of thermodynamics) as (the change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work performed by the system on its surroundings) The standard unit for all these quantities would be the joule.

35
**Potential & Kinetic Energy**

KEY CONCEPTS: INTERNAL ENERGY HEAT WORK Internal Energy Potential & Kinetic Energy

36
ENERGY Energy is defined as the ability or capacity to do work on some form of matter (the amount of work one system is doing on another). There are several forms of energy: Potential energy is the energy that a body possesses as a consequence of its position in a gravitational field (e.g., water behind a dam). Kinetic energy is the energy that a body possesses as a consequence of its motion (e.g., wind blowing across a wind generator). It is dependent upon an object's mass and velocity (e.g., moving water versus moving air). Internal energy is the total energy (potential and kinetic) stored in molecules. It is the energy associated with the random, disordered motion of molecules; it refers to the invisible microscopic energy on the atomic and molecular scale. The First Law of Thermodynamics states that energy lost during one process must equal the energy gained during another, i.e., all energy is conserved.

37
Unit of Energy Energy is measured in Joules (J) or Calories (cal).1 J = 1 kg m2s-2 A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1°C. 1 cal = J

38
INTERNAL ENERGY The internal energy, U of a system is the sum of the kinetic and potential energies of all the particles that compose the system or the total energy of a system (the energy associated with the random, disordered motion of molecules). It involves energy on the microscopic scale. The total (internal) energy in a system includes potential and kinetic energy. Binding energies – atomic bonds (potential energy from intermolecular forces) translational kinetic energy vibrational and rotational kinetic energy

39
Example For example, a room temperature glass of water sitting on a table has no apparent energy, either potential or kinetic . But on the microscopic scale it is a seething mass of high speed molecules traveling at hundreds of meters per second. If the water were tossed across the room, this microscopic energy would not necessarily be changed when we superimpose an ordered large scale motion on the water as a whole.

40
**There are two ways to change the internal energy: with work and heat.**

Internal energy of an object can be changed by the following methods: It increases if energy is added to the system. i.e. by heating or by doing work on the system. It decreases if energy is removed from the system or work is done by the system. i.e. Thus heat and work changes the internal energy of an object.

41
**Reaction between carbon and oxygen to form carbon dioxide**

Example Reaction between carbon and oxygen to form carbon dioxide C(s) + O2(g) CO2(g) if the reactants have a higher internal energy than a products, ΔUsys is negative and energy flows out of the system into the surroundings. if the reactants have a lower internal energy than a products, ΔUsys is positive and energy flows into the system from the surroundings.

42
Internal energy is the state function, which means that its value depends only the state of the system, not the how the system arrive at the state. Example: Altitude is a state function. The change in altitude during climbing depends only on the difference between the final and initial altitudes.

43
Problems Problem 1: A gas expands against a constant pressure of 1 atm from a volume of 10 L to 20 L. During this process, the system absorbs 600 J of heat from the surroundings. Calculate the internal energy of the system. Answer: ΔU = q + w = 600 J + (-PΔV) = -413 J = 600 J + { Pa (20-10) L} = 600 J + (-1013 J) = J Problem 2 :q amount of heat is transferred to the system from the surroundings and w amount of work is done by the system. Write the expression for the internal energy. Answer: ΔU = q – w

44
Study Example 2.3 pg. 50 Calculate U when 1.00 mol of H2O goes from 25.0°C and 1.00 atm to 30.0°C and 1.00 atm. Densities of water are g/cm3 at 0°C and g/cm3 at 100°C. Refer to sildes ‘Specific Heat Capacity’, ‘P-V Work’, and the First Law equation to solve ! Answer: ΔU = q + w = 90 cal.

45
**Now, you’ve learn about heat & work, so what is the difference between HEAT & WORK?**

Heat is an energy transfer between system and surroundings due to a temperature difference. Work is an energy transfer between system and surroundings due to a macroscopic force acting through a distance.

46
**Heat and work, both, are energy**

Heat and work, both, are energy. The difference is just the amount of ordered motion during the energy transfer.

47
Work and Heat Energy transfer between system and surrounding occurs either in the form of work or heat. Work (W) → refers to mechanical work Heat (Q) → refers to energy transferred from a hot to a cold object. W = Fd A system can exchange energy with its surroundings through heat and work:

48
Summary According to the first law thermodynamic, the change in the internal energy of the system(ΔU) must be the sum of the heat transferred (q) and the work done (w): Sign of conventions for q, w, and ΔU

49
**Enthalpy vs. Energy For constant-pressure process:**

Enthalpy (H) is the heat flow in or out of a system at constant pressure (i.e., carried out in open containers at or near atmospheric pressure): where E = energy, P = pressure, and V = volume. Enthalpy depends on the amount of substance present. H = ΔU +PΔV For constant-pressure process: ΔH = 𝒒 𝒑 const. P, closed syst., P-V work only For constant-volume process: ΔU = qV const. V , closed syst., P-V work only

50
**What is the relationship between the change in energy and enthalpy?**

Energy transfer occurs as heat when little or no work gets done. This is the case in three scenarios: Reactions that do not involve gases (done at constant pressure with little or no volume change). Reactions in which the number of moles of gas does not change (when Δn = 0 then ΔV = 0). Reaction in which the volume (i.e., moles) changes but the work is negligible compared to the heat.

51
**What is Joule Thompson's experiment?**

It is an experiment in which the Joule-Thomson coefficient is measured. Basically, you are expanding a gas under adiabatic conditions to ensure constant enthalpy and you will notice that there will be a temperature change (most likely cooling). We look at the Joule Expansion to learn how to relate derivatives, such as du/dV under constant temperature and du/dT under constant volume. The Joule Expansion can be used to find these quantities.

52
Example Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the process. If one mole of HCl at 298 K and 1 atm pressure occupies 24.5 liters, find ΔU for the system when one mole of HCl dissolves in water under these conditions. Answer: In this process the volume of liquid remains practically unchanged, so ΔV = –24.5 L. The work done is w = –PΔV = –(1 atm)(–24.5 L) = 24.6 L-atm (The work is positive because it is being done on the system as its volume decreases due to the dissolution of the gas into the much smaller volume of the solution.) Using the conversion factor 1 L-atm = J mol–1 and substituting in the 1st Law Eqn we obtain: ΔU= q +PΔV = –(75300 J) + [ J/L-atm) × (24.5 L-atm)] = –72.82 kJ In other words, if the gaseous HCl could dissolve without volume change, the heat released by the process (75.3 kJ) would cause the system’s internal energy to diminish by 75.3 kJ. But the disappearance of the gaseous phase reduces the volume of the system. This is equivalent to compression of the system by the pressure of the atmosphere performing work on it and consuming part of the energy that would otherwise be liberated, reducing the net value of ΔU to –72.82 kJ.

53
**Perfect gas and the first law**

Perfect gas obeys both the following equations: For a closed system in equilibrium, the internal energy (and any other state function) can be expressed as a function of temperature and volume. However eqn. above states that for a perfect gas U is independent of volume. Therefore U of a perfect gas depends only on temperature. So: We now apply the first law to a perfect gas:

54
**Both are reversible/irreversible processes?**

Isothermal process in a perfect gas Adiabatic process in a perfect gas An isothermal process is one that occurs at constant temperature. The gas-piston container in our example could expand isothermally if it were kept immersed in a large hot-water bath while the gas expanded. Since the temperature doesn't change during an isothermal process, there is no change in internal energy. The first law then tells you that the work done by the gas is just equal to the heat that flows into the system from the bath. An adiabatic process is one that occurs without the exchange of heat with the surroundings. If the gas-piston system were insulated so that heat could not get in or out, any expansion or compression would occur adiabatically. Since Q = 0 for an adiabatic process, the first law tells you that the change in internal energy is just equal to the work done on the system. dq = 0 and q = 0

55
When a gas expands adiabatically, the work done in the expansion comes at the expense of the internal energy of the gas, causing the temperature of the gas to drop. The figure below shows P-V diagrams for these two processes. The figure compares two processes that begin with the same state and involve expansion to the same final volume. For the isothermal process, the product of P·V remains constant since T remains constant. Since the temperature must decrease for the adiabatic process, it follows that the final pressure must be less for this process. Thus the adiabat lies below the isotherm. Let's look at one more example that incorporates many of the AP points of emphasis.

56
Study Example 2.5 pg. 60 A cylinder fitted with a frictionless piston contains 3.00 mol of He gas at P 1.00 atm and is in a large constant-temperature bath at 400 K. The pressure is reversibly increased to 5.00 atm. Find w, q, and U for this process. Refer to the slide of Isothermal process in a perfect gas to solve ! Answer: W = 1.61x J q = x J U = 0

57
**THE MOLECULAR NATURE OF INTERNAL ENERGY**

The molecular description of internal energy is outside the scope of thermodynamics, but a qualitative understanding of molecular energies is helpful. Consider first a gas. The molecules are moving through space. A molecule has a translational kinetic energy mv2, where m and v are the mass and speed of the molecule. A translation is a motion in which every point of the body moves the same distance in the same direction. If each gas molecule has more than one atom, then the molecules undergo rotational and vibrational motions in addition to translation. A rotation is a motion in which the spatial orientation of the body changes, but the distances between all points in the body remain fixed and the center of mass of the body does not move (so that there is no translational motion). Besides translational and rotational energies, the atoms in a molecule have vibrational energy. In a molecular vibration, the atoms oscillate about their equilibrium positions in the molecule. A molecule has various characteristic ways of vibrating, each way being called a vibrational normal mode.

58
**Figure 2.14 shows translational, rotational, and vibrational motions in CO2.**

59
**The limitations of the first law of Thermodynamics.**

No restriction on the direction of the flow of heat: the first law establishes definite relationship between the heat absorbed and the work performed by a system. The first law does not indicate whether heat can flow from a cold end to a hot end or not. For example: we cannot extract heat from the ice by cooling it to a low temperature. Some external work has to be done. Does not specify the feasibility of the reaction: first law does not specify that process is feasible or not for example: when a rod is heated at one end then equilibrium has to be obtained which is possible only by some expenditure of energy. Practically it is not possible to convert the heat energy into an equivalent amount of work. To overcome this limitations, another law is needed which is known as second law of thermodynamics. The second law of thermodynamics helps us to predict whether the reaction is feasible or not and also tell the direction of the flow of heat. It also tells that energy cannot be completely converted into equivalent work.

60
….End of Chapter 2….

Similar presentations

OK

Chemical Thermodynamics Lecture 1. Chemical Thermodynamics Prepared by PhD Halina Falfushynska.

Chemical Thermodynamics Lecture 1. Chemical Thermodynamics Prepared by PhD Halina Falfushynska.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google