B.2 Thermodynamics - notes from kognity PDF

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Internal energy and work done Internal energy and work done are physical quantities which we have met previously in the course. In this section, we will explore how to calculate the internal energy of an ideal gas, and the work done during the expansion or compression of a gas. Internal energy of an ideal gas In a previous section, we introduced the assumptions which underpin the kinetic theory of ideal gases. One of these is that the gas molecules exert no forces on each other except when in collision. This means that the molecules of an ideal gas possess only kinetic energy - there is no potential energy associated with intermolecular forces. We also learnt in another section that internal energy is the sum of the random kinetic energy of the molecules and their total potential energy. Hence, for an ideal gas the internal energy is just the total kinetic energy of all its molecules. We know that the average kinetic energy (E¯K) a gas molecule is proportional to the absolute temperature of the gas (T) only and is given by: E¯K=32kBT where kB is the Boltzmann constant. So if the gas contains N molecules, the internal energy (U) will be: U=N×32kBT. Using R=kB×NA where Ris the ideal gas constant and NA is Avogadro’s number: U=32NRNAT. But the number of moles of gas (n) is given by n=NNA so: U=32nRT. This equation can be found in the Physics Data Booklet.

Be Aware In this section, we will restrict ourselves to monatomic gases. Example 1 Calculate the internal energy in 1 mole of a monatomic gas at 273 K. » Hide solution U=32nRT=32×1×8.31×273=3400J Work done on/by a gas Consider a fixed mass of gas in a container with a frictionless piston which is free to move.

A fixed mass of gas in a container with a frictionless piston expands slightly, doing work. Figure 1. A fixed mass of gas in a container with a frictionless piston expands slightly, doing work. Suppose the gas expands slightly, pushing the piston outwards by a small distance Δx. The change in the gas pressure is negligible in this case and the force on the piston is pA where A is the cross-sectional area. Thus we can deduce that the work done (W) by the gas during the expansion is given by: W=force×distance moved indirection theforce W=pA×Δx and as the increase in the volume of the gas (ΔV) is AΔx so: W=pΔV. This equation is in the Physics Data Booklet.

Be Aware If the gas is compressed, work is done on it and W, the work done by the gas, is given a negative sign. Example 2 A gas occupying 0.010m3is compressed to half its volume, with the gas pressure remaining constant at 1×105 Pa. What is the work done by the gas? » Hide solution W=1×105×(−0.005)=−500 J. What if the pressure changes during the thermodynamic process? Then we have to calculate the work done from the graph of pressure versus volume. This will be discussed in more detail in later section when we show how thermodynamic processes can be represented in pV diagrams.

The first law of thermodynamics The first law of thermodynamics is a statement of the law of conservation of energy as applied to thermodynamic systems. Throughout this section, we will be discussing "systems" and "surroundings".

Definition A thermodynamic system is a quantity of matter (e.g. a fixed mass of gas) or a group of objects (e.g. a galaxy) under investigation.

Definition The region external to the system with which it interacts is called the surroundings. Thermal energy and work may be exchanged between the system and its surroundings. If part of the surroundings is designed to provide heat or absorb heat, we refer to this as a reservoir. We can identify three distinct types of system: an isolated system - one which cannot exchange thermal energy, work or matter with its surroundings; an open system - one which can exchange thermal energy, work and matter with its surroundings; a closed system - one which can exchange thermal energy or work with its surroundings, but not matter. For example, if water is boiled in an open saucepan, the steam which leaves the pan exchanges thermal energy and matter with the surroundings. This would be an example of an open system. However if a lid is placed on the pan, the lid and the surrounding air will get hot, so thermal energy is exchanged with the surroundings, but as no steam escapes, there is no matter transfer. This is an example of a closed system. If hot water is placed in an insulated thermos flask, no thermal energy or matter is exchanged with the surroundings and this forms an isolated system. If a quantity of energy Q is supplied to a gas, it can gain internal energy ΔU and/or do work on its surroundings W. So, from the law of conservation of energy: Q=ΔU+W. This equation can be found in the Physics Data Booklet. The direction of energy transfer is represented by the sign of the relevant quantity, as detailed in Table 1. Table 1. Sign conventions used in applying the first law of thermodynamics.

Example 1 In a thermodynamic process, 75 J of work is done on an ideal gas when it is compressed. If its internal energy decreases by 14 J, how much thermal energy is removed from a gas. » Hide solution Using the sign conventions in Table 1: WΔU=−75J=−14J Hence substituting in the first law of thermodynamics: Q=−75−14=−89J meaning that 89 J of thermal energy has been removed from the gas. Example 2

An ideal gas of volume 0.020 m3 expands to a volume 0.025 m3 at constant pressure of 1.1x105 Pa by applying a thermal energy of 9.0x103 J. What is the change in the internal energy of the gas? » Hide solution First calculate the work done by the gas as it expands: W=pΔV=1.1×105×(0.025−0.020)=550J. Applying the first law of thermodynamics: 9.0×103ΔU=ΔU+550=8450J. Rounding to two significant figures to match the precision of the data given in the question, ΔU=8500J. This has a positive sign, meaning that the internal energy of the gas has increased. Amongst the various changes which a gas can experience, there are four specific processes which warrant special consideration. Isothermal processes:

Definition During an isothermal process, temperature remains constant. This means that internal energy remains constant. If we apply the first law of thermodynamics to an isothermal process, as ΔU=0 then Q=W. The thermal energy supplied to the gas is equal to the work done by the gas as it expands. The thermal energy extracted from the gas is equal to the work done on the gas as it contracts.

Be aware For a process to be isothermal, the system must be in contact with a large reservoir and it must take place slowly to ensure that the gas always remains in thermal equilibrium with its surroundings. Isovolumetric processes:

Definition During an isovolumetric process, volume remains constant. This means that no work is done on or by the gas. If we apply the first law of thermodynamics to an isovolumetric process, as W=0 then Q=ΔU. Supplying thermal energy to the gas causes the internal energy to increase and the temperature will rise. Extracting thermal energy from the gas causes the internal energy to decrease and the temperature will fall. Isobaric processes:

Definition During an isobaric process, pressure remains constant. For an isobaric process, it is necessary to consider the changes in the all the variables in the first law, Q=ΔU+W. Isobaric changes are associated with the free expansion or contraction of a gas as thermal energy is transferred to or from it. For example, consider fuel in a container with a piston which is free to move. When the fuel is burnt, the pressure inside the container would normally increase but if the piston moves to allow the gases to expand, pressure can be kept constant. Such a process forms part of the cycle in a petrol or diesel engine. Adiabatic processes:

Definition

During an adiabatic process, no thermal energy is transferred between a gas and its surroundings. If we apply the first law of thermodynamics to an adiabatic process, as Q=0 then ΔU=−W. For an adiabatic expansion, W is positive so ΔUmust be negative, and the temperature of the gas decreases. For an adiabatic compression, W is negative so ΔU must be positive, and the temperature of the gas increases.

Be aware An adiabatic process must take place quickly and the gas must be in a well-insulated container. For a fixed quantity of an ideal, monatomic gas, when its state changes adiabatically from pressure p1 and volume V1 to pressure p2 and volume V2 then: p1V5/31=p2V5/32. In other words, for an adiabatic change of an ideal, monatomic gas: pV53=constant. where p is the pressure and V is the volume of the gas. This equation can be found in the Physics Data Booklet. From the ideal gas law, we know that p1V1=nRT1 and p2V2=nRT2 where T1 and T2 are the temperatures of the gas in the two different states. Hence p1=nRT1V1andp2=nRT2V2 so: (nRT1V1)×V5/31=(nRT2V2)×V5/32 or: nRT1V2/31=nRT2V2/32 and: T1V2/31=T2V2/32. If the volume of a gas increases as it expands adiabatically (V2>V1) then the gas temperature must decrease. If the volume of a gas decreases as it is compressed adiabatically then its temperature must increase.

Be aware An adiabatic expansion results in cooling. An adiabatic compression results in heating. Example 3 An ideal monatomic gas doubles in volume during an adiabatic expansion. By what factor does its pressure change? » Hide solution p1V5/31=p2(2V1)5/3 So: p2=p125/3=0.31p1

and the pressure changes by a factor of 0.31. Example 4 By what factor does the volume of an ideal monatomic gas need to change in order to increase the temperature by a factor of three during an adiabatic compression? » Hide solution Substituting: T1V2/31=(3T1)×V2/32 then rearranging: (V2V1)2/3=13 and solving: V2V1=(13)3/2 gives V2=0.19V1 so the volume must decrease by a factor of 0.19.

pV diagrams The state of a fixed mass of gas is specified in terms of its pressure (p), volume (V) and temperature (T). To represent changes in the state of the gas graphically, we would have to draw three-dimensional graphs of p, Vand T, and these are difficult to interpret. Instead, we draw graphs to show changes in pressure and volume, called pV diagrams, although we must remember that in general, temperature will also change. To understand how temperature changes are represented on a pV diagram, let us first consider isothermal (fixed temperature) processes. Three isothermal processes, occurring at different temperatures, can be seen in Figure 1.

A \(pV\) diagram representing changes in pressure and volume at constant temperature (isothermal processes). Each curve tracks the \(pV\) state of a gas at a different temperature. The blue curve represents a higher temperature than the red curve, which is in turn at a higher temperature than the green curve (\(T_1>T_2>T_3\)). Figure 1. A pV diagram representing changes in pressure and volume at constant temperature (isothermal processes). Each curve tracks the p−V state of a gas at a different temperature. The blue curve represents a higher temperature than the red curve, which is in turn at a higher temperature than the green curve (T1>T2>T3).

Boyle's Law (section 3.2.1) explains that the pressure of a fixed mass of gas is inversely proportional to its volume, at fixed temperature, and the curves, called isotherms, show this relationship (p∝1V). As temperature increases, the isotherms move further from the origin. It is important to visualise these isotherms on a pV diagram even when they are not displayed. If a process causes a transition from one isotherm to another, then the temperature of the gas will be changing. Figure 2 shows how four different processes would be represented on a pV diagram. In all cases, the gas begins with the pressure, volume and temperature of point X.

A \(pV\) diagram representing four different processes: \(X\rightarrow A\) is isobaric (constant pressure), \(X\rightarrow B\) is isovolumetric (constant volume), \(X\rightarrow C\) is isothermal (constant temperature) and \(X\rightarrow D\) is adiabatic (no energy transfer with the surroundings). Figure 2. A pV diagram representing four different processes: X→A is isobaric (constant pressure), X→B is isovolumetric (constant volume), X→C is isothermal (constant temperature) and X→D is adiabatic (no energy transfer with the surroundings). In Table 2, the change of state which the gas undergoes for each of the four processes is detailed, in terms of changes to its pressure, volume or temperature.

The gas at point C is at the same temperature as the gas at point X. Remembering that isotherms closer to the origin represent lower temperatures (see Figure 1) , the gas at point D must be at a lower temperature than the gas at point C.

Important On a pV diagram, a curve representing an adiabatic process is steeper than a curve representing an isothermal process.

Examiner Tip It is important to stress that although during an adiabatic process there is no thermal energy transfer with the surroundings, the temperature of the gas will change because work has been done. As discussed in a previous section, the work done during a thermodynamic process is W=pΔV. If the pressure is changing throughout the process, we need to divide the process into infinitesimal volume changes (where the pressure is approximately constant), calculate the work done for each of these and then add the values up. This is equivalent to calculating the area under the curve on the pV diagram. If

you recall, we did something similar in section 2.3.1 when calculating the work done as the area under a force-time graph. From Figure 2, it can be seen that the work done for the process X→B is identically equal to zero.

Be Aware The work done by or on a system during a thermodynamic process is equal to the area under the corresponding curve on a pV diagram.

Examiner Tip If the area under the curve is not comprised of simple geometrical shapes (e.g. rectangles or triangles), an accurate calculation of the work done requires counting squares. If the exam question asks for an estimate, however, then you can approximate the area using geometrical shapes. Watch out for the units!

Entropy and the second law of thermodynamics The traditional game of "pick-up sticks" (https://en.wikipedia.org/wiki/Mikado_(game)) starts with a player taking a bundle of sticks and tossing them into a pile on the ground. Each player then has to remove a stick from the pile in turn, without making the other sticks move. The winner is the player who collects the most sticks.

The game of "pick-up sticks" starts by tossing a pile of sticks onto the ground. Players have to pick up sticks one at a time without making the other sticks move. The winner is the player who collects the most sticks. Figure 1. The game of "pick-up sticks" starts by tossing a pile of sticks onto the ground. Players have to pick up sticks one at a time without making the other sticks move. The winner is the player who collects the most sticks. When the player tosses the sticks to the ground, there is no way of knowing how they will fall. It is a random process. It is most likely that the sticks will end up in a pile, with sticks on top of sticks, at various angles. However, there is a very small, but finite, probability that the sticks will land in perfectly straight lines, with an equal separation between sticks. Thankfully the disordered system of a randomised pile is more likely than the ordered system of sticks in a line, or this would make for a very boring game! This concept of order and disorder can be applied to many different physical systems. A gas, for example, is a more disordered state of matter than a liquid. In a gas, the molecules move freely and randomly, but in a liquid, the bonds between the molecules restrict their movement to a more coherent liquid flow. In turn, a solid is a more ordered state of matter than a liquid, because in a solid, the molecules are essentially fixed in their lattice positions, barring their small vibrational motion. If we raise the temperature of a substance, then the average kinetic energy of its molecules increases. The result is to increase the disorder of the system. Similarly, a fixed quantity of gas occupying a large volume is more disordered than the same quantity of gas in a small volume. In the large volume, we know less about the position of the individual gas molecules.

Definition Entropy (S) is a measure of the amount of disorder in a system. We can define the change of entropy (ΔS) which occurs when heat (ΔQ) is transferred to or from a system at constant temperature (T, measured in Kelvin) as:

ΔS=ΔQT. This equation can be found in the Physics Data Booklet. The unit of entropy is J K-1. When heat is added to a system, ΔQ is positive and ΔS is positive. Removing heat from a system gives a negative value of ΔQ and ΔS. For example, in melting ice, heat is added to the system and the change in entropy must, therefore, be positive. This is illustrated in Figure 2.

In ice, the water molecules are bound together in a regular crystal lattice. This represents a highly ordered state. By constrast, the water molecules in the liquid state are relatively free to move, giving a more disordered state. Entropy increases when melting ice. Figure 2. In ice, the water molecules are bound together in a regular crystal lattice. This represents a highly ordered state. By constrast, the water molecules in the liquid state are relatively free to move, giving a more disordered state. Entropy increases when melting ice. Be aware The formula ΔS=ΔQT can only be used for changes which occur at constant temperature. For other changes, entropy must be calculated using integral calculus. Example 1 What is the change in entropy when 55 J of energy is used to boil water at 100ºC ? » Hide solution The boiling of the water takes place at constant temperature so: ΔS=55373=0.15JK−1 The second law of thermodynamics can be understood in terms of entropy changes which take place in an isolated system. However, it is also inherently a statement of the irreversibility of thermodynamic processes. Let us begin with an example. Suppose that one day, whilst doing the washing up, you drop a plate. It falls to the floor and smashes into several pieces. This is our analogue of a thermodynamic process. The reverse process would involve the pieces of the plate spontaneously reassembling themselves into the unbroken plate. Clearly, no matter how long you wait, this will not happen. The process of breaking a plate is irreversible (unless you intervene to glue the pieces back together!) Let us consider an example related to heat transfer. On a warm day, you leave a container of ice cream on the table in the kitchen. When you return half an hour later, the ice cream has melted. Thermal energy has been transferred from the hot surroundings to the cold ice cream. However if you leave the melted ice cream on the table, it will not spontaneously freeze and in doing so transfer thermal energy to the surroundings. This process does not violate the principle of conservation of energy, but it just does not occur. The process of the ice cream melting under these conditions is irreversible. Of course, the ice cream can be frozen once more if it is placed in a freezer. The frozen ice cream is in a more ordered state than the melted one. However the process of transforming the ice cream from liquid to solid requires the freezer to do work in removing thermal energy. In doing so, the motor of the freezer heats the surrounding air. In thermodynamics, a reversible process is one whose direction is reversed by an infinitesimal change to a property of the system without increasing its entropy. The system must be continually in equilibirum with its surroundings. As it would take an infinite amount of time for a reversible process to be completed, no real thermodynamical processes are reversible. In practice, a process can be deemed to be

reversible if it responds to the applied change faster than the time in which the change takes place. All natural processes are irreversible. This is an ...