what happens to the internal energy of a gas when it expands (with no heat flow)

17. Heat and the Get-go Police of Thermodynamics

17.1. Heat

Ii bodies brought in thermal contact volition change their temperature until they are at the same temperature. In the process of reaching thermal equilibrium , rut is transferred from one body to the other.

Suppose we take a system of interest at temperature T Due south surrounded by an environment with temperature T E . If T S > T E oestrus flows from the organization to the environment. If T S < T E oestrus flows from the environment into the system. Heat, presented by the symbol Q and unit Joule, is called to be positive when heat flows into the organisation, and negative if heat flows out of the system (see Figure 17.i). Heat menstruation is a results of a temperature difference between two bodies, and the menstruation of heat is zero if T Due south = T E .

Figure 17.1. Heat Catamenia.

Heat is not the simply style in which energy can be transferred betwixt a organisation and its environment. Energy can also be transferred between a organisation and its environment past ways of piece of work (Westward). The unit of measurement of piece of work is the Joule.

Some other commonly used unit is the calorie . The calorie is defined as the amount of heat that would raise the temperature of 1 k of water from 14.5 � C to fifteen.5 � C. The Joule and the calorie are related as follows: 1 cal = four.1860 J

17.2. Heat Chapters

When heat is added to an object, its temperature increases. The modify in the temperature is proportional to the amount of oestrus added

The abiding C is called the rut capacity of the object. The heat capacity of an object depends on its mass and the type of material of which it is made. The heat capacity of an object is proportional to its mass, and the rut capacity per unit of measurement mass , c, is unremarkably used. In that case

where one thousand is the mass of the object. The tooth heat chapters is the heat capacity per mole of fabric. For most materials the molar heat chapters is 25 J/mol Thousand.

In order to determine the heat capacity of a substance we not only demand to know how much estrus is added, but also the weather under which the heat transfer took place. For gases, adding heat under abiding pressure level and nether constant temperature will atomic number 82 to very different values of the specific heat capacity.

17.3. Heat of Transformation

When rut is added to a solid or a liquid, the temperature of the sample does non necessarily rise. During a phase change (melting, boiling) heat is added to the sample without an increase in temperature. The amount of rut transferred per mass unit during a phase alter is called the heat of transformation (symbol L) for the process. The amount of heat needed/released is given by

where m is the mass of the sample.

Trouble 28P

What mass of steam of 100 � C must be mixed with 150 thousand of ice at 0 � C, in a thermally insulated container, to produce liquid h2o at l � C ?

We first with calculating the heat required to transform 150 k of ice at 0 � C to 150 g of liquid at 0 � C. The estrus of transformation of water is 333 kJ/kg (come across Tabular array 20-2, page 555). The transformation of ice into water therefore requires a total heat given by

The heat required to modify the temperature of 150 g of water from 0 � C to 50 � C is given by

The total rut that needs to be added to the system is therefore equal to 81.5 kJ. This rut must be supplied by the steam. Heat volition exist released when the steam is transformed into liquid, The heat of transformation for this process is 2260 kJ/kg. Suppose the mass of the steam is thou. The total heat released in the conversion of steam into water is given by

The estrus released when the steam cools down from 100 � C to 50 � C is given by

The full rut released by the cooling of the steam is therefore equal to 2470 m kJ. The full heat required is 81.5 kJ, and we therefore conclude that the mass of the steam must be equal to 33 thou.

17.4. Work

Suppose a system starts from an initial state described by a pressure p i , a volume V i , and a temperature T i . The final state of the system is described by a pressure p f , a volume V f , and a temperature T f . The transformation from the initial state to the concluding state can be achieved in a diverseness of ways (see for case Figure 17.2). In Figure 17.2a both pressure and volume change simultaneously. In Figure 17.2b the pressure of the system is start lowered while keeping the volume abiding (this can for example be achieved by cooling the sample) and afterwards, the volume is increased while keeping the pressure constant (this tin be achieved by heating the gas while increasing the volume).

If the pressure of a gas increases it can motility a piston (this happens in an engine). In this case, work is done by the organisation equally the expanding gas lifts the piston. On the other hand, if nosotros increment the weight of the piston, work will be done on the system as the piston falls. The forcefulness exerted by the gas on the piston is equal to p A, where A is the area of the piston and p is the gas pressure. If the piston is displaced by a distance ds, the amount of piece of work done can be calculated as follows:

Effigy 17.2. 2 possible ways to get from the initial state to the final state.

The full work done during a finite deportation of the piston is now easy to summate

If Westward is positive, work was done by the system (for instance, the expanding gas lifts the piston). A negative value of Due west tells you that piece of work was done on the system (the piston is pressed down in society to compress the gas).

The amount of piece of work done is equal to the area under the curve in the pV diagrams shown in Figure 17.2. Clearly, the corporeality of work done depends on the path chosen. The work W for the path shown in Figure 17.2a is significantly more than the work Due west for the path shown in Figure 17.2b. Any modify in the system in which the volume does not change will not produce/cost any work. The work done for the paths shown in Figure 17.two can be calculated hands

No piece of work is done for the path shown in Figure 17.2b between (p i , V i ) and (p f , Five i ) since there is no change in volume. The work done to motility from (p f , V i ) to (p f , 5 f ) is calculated easily

Clearly, W 2b is always less then Westward 2a , and we can make the amount of work done every bit small or as large as nosotros want. For case no work would exist done if the transition follows the following path:

(p i , V i ) � (0, V i ) � (0, Five f ) � (p f , Five i )

A system can be taken from a given initial state to a given final state by an infinite number of processes. In general, the work W and also the heat Q will have different values for each of these processes. Nosotros say that oestrus and piece of work are path-dependent quantities.

From the previous discussion neither Q nor Due west represents a change in some intrinsic properties of the system. Experimentally, all the same, it is observed that the quantity Q - Westward is the same for all processes. It depends only on the initial and final states and it does not matter at what path is followed to get from one to the other. The quantity Q - W is called the change in the internal free energy U of the organization:

�U = U f - U i = Q - W

This equation is called the kickoff constabulary of thermodynamics . For small changes the first law of thermodynamics can be rewritten as

dU = dQ - dW

17.4.1. Adiabatic Processes

If a system is well insulated, no transfer of heat volition occur between it and its surround. This means that Q = 0, and the get-go law of thermodynamics shows that

�U = - Due west

If work is done by the system (positive West) its internal energy decreases. Conversely, if work is done on the arrangement (negative W) its internal energy will increment. For gases, the internal energy is related to the temperature: a higher internal energy means a higher temperature. Adiabatic expansion of a gas volition lower its temperature; adiabatic compression of a gas will increase its temperature .

17.4.2. Constant Volume Processes

If the volume of a arrangement is held constant, the system tin practise no work (W = 0 J). The first law of thermodynamics then shows that

�U = Q

If estrus is added to the organization its internal free energy will increment; if estrus is removed from the system its internal energy will subtract .

17.four.3. Cyclical Processes;

Processes which, subsequently sure interchanges of oestrus and work, are restored to their initial state are called cyclical processes. In this example, no intrinsic properties of the organisation are inverse, and therefore �U = 0. The commencement law of thermodynamics now immediately yields

Q = Westward

17.four.4. Complimentary Expansion

Costless expansion is an adiabatic procedure in which no work is washed on or by the organization. This means that Q = W = 0 J, and the commencement law of thermodynamics at present requires that

�U = 0 J

17.five. Transfer of Heat

The transfer of heat between a organisation and its surround tin can occur in a variety of ways. Three different mechanisms of heat transfer will now exist discussed: conduction , convection , and radiation .

17.5.1. Conduction

Consider the slab of cloth shown in Figure 17.3. The left end of the beam is maintained at a temperature T H ; the correct end of the beam is maintained at a temperature T C . As a result of the temperature difference heat will flow through the slab, from its hot cease to its common cold end. Experimentally it is shown that the rate of oestrus transfer (Q/t) is proportional to the cross-sectional area of the slab, proportional to the temperature difference, and inversely proportional to the length of the slab

Here, k is the thermal conductivity , which is a constant that depends on the blazon of material. Large values of thousand define practiced heat conductors. The thermal resistance R is related to the thermal electrical conductivity k in the following manner

Figure 17.iii. Conduction.

Thus, the lower the thermal conductivity of the material, the college the thermal resistance R. From the definition of R information technology immediately follows that

Consider a blended slab is fabricated up out of ii different materials, with length L 1 and L two , and with thermal conductivity k 1 and k 2 , is placed between two heat baths (see Figure 17.4). Suppose that the temperature of the interface between the two slabs is equal to T x . The amount of heat flowing from T H to T x is given by

The amount of heat flowing from T x to T C is given past

Of course, the heat flowing through slab ane must equal the heat flowing through slab 2. Thus

Figure 17.4. Heat transport through a composite slab.

This equation can be used to obtain the temperature at the interface between slab ane and slab 2:

The heat flowing through the slab can now be calculated easily

A composite slab therefore has a thermal resistance equal to the sum of the thermal resistance of each of the individual slabs.

17.v.2. Convection

Heat transfer past convection occurs when a fluid, such as air or water, is in contact with an object whose temperature is college than the temperature of its environs. The temperature of the fluid increases and (in almost cases) the fluid expands. Being less dense than the surrounding cooler fluid, information technology rises because of buoyant forces. The surrounding cooler fluid falls to take the identify of the rising warmer fluid and a convective circulation is fix.

17.5.3. Radiation

Every object emits electromagnetic radiation . The energy spectrum of the radiations emitted depends on the temperature of the object; the boilerplate energy increases with increasing temperature.

Problem 57P

A container of h2o has been outdoors in cold atmospheric condition until a five.0 cm-thick slab of ice has formed on its surface. The air to a higher place the ice is at -10 � C. Summate the rate of germination of ice (in centimeters per hour) on the bottom surface of the ice slab. Take the thermal conductivity of ice to be 0.0040 cal/s . cm . � C and the density to be 0.92 1000/cm3.

The water at the boundary betwixt the water and the ice will be at a temperature of 0 � C. The heat transported through five cm of ice is equal to

This rut is released when h2o is transformed into ice. The oestrus of transformation of this process is 79.v cal/yard. Suppose a mass one thousand of water is transformed each 2nd into ice. This produces a total estrus equal to

H = 79.v k cal/s

This must be equal to the oestrus flow through the ice:

79.5 m = 0.0080 A

A mass grand of water ice (covering an surface area A) will accept a thickness d, where d is given by

Combining the last two expression we obtain for the rate of water ice germination:

Send comments, questions and/or suggestions via email to wolfs@pas.rochester.edu and/or visit the home folio of Frank Wolfs.

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Source: http://teacher.pas.rochester.edu/phy121/lecturenotes/Chapter17/Chapter17.html