Thermal Physics Topic 10.2 Processes

The First Law of Thermodynamics Is a statement of the Law of Conservation of Energy in which the equivalence of work and thermal energy flow is taken into account. It can be stated as the heat added to a closed system equals the change in the internal energy of the system plus the work done by the system.

That is,  Q = U +  W = U + pV or U = Q - W

Explanation Where `+ Q' is the thermal energy added to the system and `+ W' is the work done by the system. If thermal energy leaves the system, then Q is negative. If work is done on the system, then W is negative. For an isolated system, then W = Q = 0 and U = 0.

Example If 22 J of work is done on a system and 3.4 x 102 J of heat is added, what is the change in internal energy of the system?

Solution Using the formula, Q = U + W We have that 340 J = U + (-22) J So that U = 340 J + 22 J = 362 J That is, the change in internal energy of the system is 3.6 x 102 J.

Isobaric A graph of pressure as a function of volume change when the pressure is kept constant is shown below

p -V diagram p V p V Isobaric process V1 V2 Area = work done = p (V1 - V2)

Work Done Such a process is said to be isobaric. Note that the work done by the gas is equal to the area under the curve An isobaric transformation requires a volume change at constant pressure For this to occur, the temperature needs to change to keep the pressure constant

Example 6.0 dm3 of an ideal gas is at a pressure of 202.6 kPa. It is heated so that it expands at constant pressure until its volume is 12 dm3. Find the work done by the gas.

Solution Using the formula W = pV we have that W = 202.6 kPa x(12 - 6.0) dm3 = 202.6 x 103 Pa x (12 - 6.0) x 10-3m3 = 1.216 x 103 J That is, the work done by the gas in the expansion is 1.2 x 103 J.

Isochoric / Isovolumetric A graph of pressure as a function of volume change when the volume is kept constant is shown below p V Isochoric / Isovolumetric

Work Done Such a process is said to be isochoric. When the volume is kept fixed, the curve of the transformation is said to be an isochore. Note that the work done by the gas is equal to zero as V = 0. There is zero area under the curve on a p-V diagram.

p - V diagram p V Isochoric / Isovolumetric

However The temperature and pressure can both change and so such a transformation will be accompanied by a thermal energy change.

Example A thermal system containing a gas is taken around the cycle as in the next slide (Cyclic processes such as this one have important applications in heat engines that convert internal energy into useful mechanical energy).

p V B C 6 2 A D 4 10 Starting at point A on the diagram, describe the cycle, and, calculate the work done by the system in the completion of one cycle.

Solution From A to B, the volume is kept constant (isovolumetric) as the pressure increases. This can be achieved by heating the gas. Since V = 0, then no work is done by the gas, W = 0.

From B to C, the gas expands (volume increases) (isobaric expansion) while the pressure is kept constant. The amount of work done by the gas is given by the area under the 6 kPa isobar. Now, we have that W = p V So that W = 6 kPa. (10 - 4) m3 = 36,000J

From C to D, the gas is cooled to keep the volume constant as the pressure is decreased (isovolumetric) Again V = 0 and no work is done by the gas, W = 0

From D to A, the gas is compressed (volume decreases) (isobaric compression) while the pressure is kept constant. The amount of work done on the gas is given by the area under the 2 kPa isobar. Again, using the fact that W = p V we have that W = 2 kPa. (4 -10) m3 = - 12,000 J

That is, the net work done by the gas is therefore 36,000 J – 12,000 J =24,000 J. Because the cycle is traced in a clockwise direction, the net work done on the surroundings is positive. If the cycle was traced in a counter-clockwise direction, the work done would be negative

Isothermal Processes A thermodynamic process in which the pressure and the volume are varied while the temperature is kept constant is called an isothermal process

In other words, when an ideal gas expands or is compressed at constant temperature, then the gas is said to undergo an isothermal expansion or compression.

The figure below shows three isotherms for an ideal gas at The figure below shows three isotherms for an ideal gas at different temperatures where Tl < T2 < T3 p V T1 T2 T3 Isothermal process

Isothermal Processes The curve of an isothermal process represents a Boyle's Law relation pV =constant = nRT the moles of gas n, the molar gas constant R, the absolute temperature T are constant

How? In order to keep the temperature constant during an isothermal process the gas is assumed to be held in a thin container with a high thermal conductivity that is in contact with a heat reservoir - an ideal body of large mass whose temperature remains constant when heat is exchanged with it. e.g.. a constant-temperature water bath.

And the expansion or compression should be done slowly so that no eddies are produced to create hot spots that would disrupt the energy equilibrium of the gas.

An Ideal Gas Consider an ideal gas enclosed in a thin conducting vessel that is in contact with a heat reservoir, and is fitted with a light, frictionless, movable piston If an amount of heat Q is added to the system which is at point A of an isotherm, then the system will move to another point on the graph, B.

The heat taken in will cause the gas to expand isothermally and will be equivalent to the mechanical work done by the gas Because the temperature is constant, there is no change in internal energy of the gas

That is, T = 0 and U = 0 Q = W The work done by gas is equal to the Area under the curve between A and B

Isothermally If the gas expands isothermally from A to B and then returns from B to A following exactly the same path during compression Then the isothermal change is said to be reversible.

Adiabatic Processes An adiabatic expansion or contraction is one in which no heat Q is allowed to flow into or out of the system For the entire adiabatic process, Q = 0

How To ensure that no heat enters or leaves the system during an adiabatic process it is important to: make sure that the system is extremely well insulated carry out the process rapidly so that the heat does not have the time to leave the system

p V T1 T2 Adiabatic compression Adiabatic Expansion

Example The compression stroke of an automobile engine is essentially an adiabatic compression of the air-fuel mixture The compression occurs too rapidly for appreciable heat transfer to take place

What happens In an adiabatic compression the work done on the gas will lead to an increase in the internal energy resulting in an increase in temperature. U=Q - W but Q=O therefore U= - W

In an adiabatic expansion the work done by the gas will lead to a decrease in the internal energy Resulting in a decrease in temperature.

Work done p V p V p V Area = work done on the gas as it is compressed Area between curves = Net work done by the gas Area = work done by the gas expanding

Example For the compression stroke of an experimental diesel engine, the air is rapidly decreased in volume by a factor of 15, the compression ratio. The work done on the air-fuel mixture for this compression is measured to be 550 J (a) What type of thermodynamic process is likely to have occurred? (b) What is the change in internal energy of the air-fuel mixture? (c) Is the temperature likely to increase or decrease?

Solution (a) Because the compression occurs rapidly appreciable heat transfer does not take place, and the process can be considered to be adiabatic, Q = 0

 U = Q - W = 0 - (-550) J Therefore, the change in internal energy is 550J

(c) The temperature rise will be very large resulting in the spontaneous ignition of the air-fuel mix

Thermodynamic Cycle and Engines A thermodynamic engine is a device that transforms thermal energy to mechanical energy Cars, Steam trains, jets and rockets have engines that transform fuel energy (chemical energy) into kinetic energy of their motion

Efficiency In all engines the conversion is accompanied by the emission of exhaust gases that waste some of the thermal energy These engines are not very efficient as only part of the thermal energy is converted to mechanical energy

The Engine By Vija and Bombo Presenters The Engine By Vija and Bombo Internal Combustion By Adil and Suhayb Schematic of a Heat Engine By Ham and Heisei Refrigerators By Aleem and Bhavik Heat Pumps By Alex and Said Schematic for a Refrigerator by Karim and Chilla Carnot´s Theorem By Martin and Mitul

The Engine Has two crucial features: By Vija and Bombo Has two crucial features: It must work in cycles to be useful the cyclic engine must have more than one heat reservoir

The Thermodynamic Cycle Is the process in which the system is returned to the same state from which it started That is, the initial and final states are the same in the cyclic process

Combustion Engines The steam engine and the steam turbine are examples of external combustion engines The fuel is burnt outside the engine and the thermal energy is transferred to the piston or a turbine chamber by means of steam

The Steam Engine Water is heated in a boiler heat reservoir (high temperature) to produce steam As the piston returns to its original position, the intake valve is closed and the exhaust valve opens

The Steam Engine cont... The steam passes into an open intake valve where it expands causing a piston to move outwards This allows the exhaust gas to be forced out into a condenser heat reservoir (low temperature)

The Steam Engine

Steam Turbines Most electricity is produced using steam engines The piston is replaced with a rotating turbine that contains blades The rotating turbine converts mechanical energy to electrical energy via a generator The steam is cooled in cooling towers

Internal Combustion By Adil and Suhayb

The Intake Stroke With the exhaust valve closed, a mixture of petrol vapour and air from the carburetor is sucked into the combustion chamber through the inlet valve as the piston moves down

The Compression Stroke Both valves are closed and the piston moves up to squeeze the mixture of petrol vapour and air to about 1/8th its original volume

The Power Stroke Both valves are closed the mixture is ignited by a spark from the spark plug The mixture burns rapidly and the hot gases then expand against the piston

The Exhaust Stroke The exhaust valve opens as the piston moves upwards Exhaust gases are expelled Then the cycle begins again

Schematic of a Heat Engine B By Ham and Heisei Hot reservoir at TH Cold reservoir at TL Engine Fluids enter carrying energy QH QH Engine does useful work, W W Fluids leave carrying energy QL = QH - W QL

Efficiency The thermal efficiency of a heat engine, , is defined as the ratio of work it does, W to the heat input QH  = W / QH  = (QH - QL) / QH It is clear that the efficiency will be greater if QL can be made small

Question An engine absorbs 230J of thermal energy from a high temperature reservoir, does work and exhausts 140J to a cold temperature reservoir. What is the efficiency? How much work is done? Answer 39% and 90J

Refrigerators By Aleem and Bhavik A refrigerator is a device operating in a cycle that is designed to extract heat from its interior so as to achieve or maintain a lower temperature inside The heat is exhausted to the surroundings normally at a higher temperature

How the Refrigerator Works A motor driving a compressor pump provides the means by which a net amount of work can be done by the system for a cycle Even though the cabinets are well insulated, heat from the surroundings leaks back inside

Heat Pumps By Alex and Said Any device that can pump heat from a low temperature reservoir to a high temperature reservoir is called a heat pump Examples of heat pumps include refrigerator and reverse cycle air-conditioning devices used for space heating and cooling

How.. A volatile liquid called Freon is circulated in a closed system of pipes by the compressor pump By the process of evaporative cooling, the vaporised Freon is used to remove the heat The compressor maintains a high pressure difference across a throttling valve

How.. Evaporation of the Freon occurs in several loops called the evaporator pipes that are usually the coldest part of the fridge As the liquid evaporates on the low pressure, low temperature side, heat is added to the system

How.. In order to turn from a liquid to a gas the Freon requires thermal energy equal to the latent heat of vaporization This energy is obtained from the fridge

How.. On the high pressure, high temperature side of the throttling valve, thermal energy is removed from the system The vaporized Freon in the compressor pipes is compressed by the compressor pump It gives up its latent heat of vaporization to the air surrounding the compressor pipes

How.. The heat fins act as a heat sink to radiate the thermal energy to the surroundings at a faster rate The fins are painted black and have a relatively large surface area for their size

A Typical Small Refrigerator

Schematic for a Refrigerator Karim and Chilla Hot reservoir at TH Cold reservoir at TL QH W QL

Refrigerator The previous figure shows the energy transfers that occurs in a refrigerator cycle By doing work on the system, heat QL is added from the low temperature TL reservoir A greater amount of energy QH is exhausted to the high temperature TH

Reverse Cycle Heat Pump room evaporator condenser TL TH WINTER room condenser TH TL evaporator SUMMER

Winter The evaporator heat exchanger is outside the room It exhausts its heat to the inside air

Summer The evaporator heat exchanger is inside the room It exhausts its heat to the outside air

In Both Cases Thermal energy is pumped from a low-temperature reservoir to a high-temperature reservoir The efficiency of refrigerators and heat pumps is defined in a different manner to heat engines

Coefficient of Performance, K For a refrigerator and a heat pump used for cooling K = QL / W Ratio of heat absorbed from inside the refrigerator and the work supplied to the refrigerator in one cycle

Coefficient of Performance, K For a heat pump used for heating K = QH / W Ratio of the total positive heat exhausted to the inside and the work supplied to the heat pump in one cycle

Carnot´s Theorem By Martin and Mitul No engine working between two heat reservoirs can be more efficient than a reversible engine between those reservoirs He argued that if thermal energy does flow from a cold body to a hot body then work must be done Therefore no engine can be more efficient than an ideal reversible one

And... All such engines have the same efficiency Therefore only a simple equation is needed to calculate the efficiency Consider an ideal perfectly insulated, frictionless engine that can work backwards as well as forwards The p-V diagram would have the form...

The Carnot Engine

Explain The net work is the area enclosed by ABCDA Thermal energy is absorbed by the system at the high temperature TH and is expelled at the low temperature reservoir TL

and Work is done by the system as it expands along the top isotherm from A to B And also along the adiabat from B to C Work is then done on the system to compress it along the bottom isotherm from C to D And also along the left adiabat from D to A

Carnot’s Efficiency Carnot was able to obtain an expression for the efficiency in terms of temperature  = 1 - TL / TH Therefore the efficiency of the Carnot cycle depends only on the absolute temperatures of the high and low temperature reservoirs The greater the temperature difference the greater the efficiency will be

But engine efficiency is  = 1 - QL / QH Therefore TC/TH = QC/QH Or QC/TC = QH/TH What would happen if the temperature difference was not maintained?

Real Life In practice, friction and other mechanical losses and thermal losses reduce the overall efficiency Nearly 2/3 of the heat generated by modern power stations is released into the environment

Topic 10.3 Second Law of Thermodynamics and Entropy THANK YOU NEXT..... Thermal Physics Topic 10.3 Second Law of Thermodynamics and Entropy 87