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Heat and Thermodynamics | Physics
Long answer Questions:
1. Explain the meaning of heat and work with suitable examples.
Meaning of heat and work:
● When we rub our hand against each other the temperature of the hands increases. We have done some work on our hands by rubbing. The temperature of the hands increases due to this work.
● If we place our hands on the chin, the temperature of the chin increases.
● This is because the hands are at higher temperature than the chin.
● In the above example, the temperature of hands is increased due to work and temperature of the chin is increased due to heat transfer from the hands to the chin.
● By doing work on the system, the temperature in the system will increase and sometimes may not.
● Like heat, work is also not a quantity and through the work, energy is transferred to the system.
● So we cannot use the word ‘the object contains more work' or ‘less work'.
● Either the system can transfer energy to the surrounding by doing work on surrounding or the surrounding may transfer energy to the system by doing work on the system.
● For the transfer of energy from one body to another body through the process of work, they need not be at different temperatures.
2. Discuss the ideal gas laws.
Boyle's law, Charles' law and ideal gas law:
● For a given gas at low pressure (density) kept in a container of volume V, experiments revealed the following information.
● When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume.
P ∝ 1/v known as Boyle's law.
● When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature. V ∝ T. Known as Charles's law.
● By combining these two equation we have PV = CT. Here C is a positive constant.
C is proportional to the number of particles in the gas container.
● If we take two container of same type of gas with same volume V, same pressure P and same temperature T, then the gas in each container obeys the above equation. PV = CT.
● If the two containers of gas is considered as a single system, then the pressure and temperature of this combined system will be same but volume will be twice and number of particles will also be double.
● For this combined system, V becomes 2V, so C should also double to match with the ideal gas equation P(2V)/T = 2C.
● It implies that C must depend on the number of particles in the gas and also should have the dimension of [PV /T]= JK−1.
● We can write the constant C as k times the number of particles N. Here k is the Boltzmann constrit (1.381 × 10−23 JK−1) and it is found to be a universal constant. So the ideal gas law can be stated as follows PV = NkT.
3. Explain in detail the thermal expansion.
● Thermal expansion is the tendency of matter to change in shape, area and volume due to a change in temperature.
● All three states of matter (solid, liquid and gas) expand when heated. When a solid is heated, its atoms vibrate with higher amplitude about their fixed points. The relative change in the size of solids is small.
● Railway tracks are given small gaps so that in the summer, the tracks expand and do not buckle.
● Railroad tracks and bridges have expansion joints to allow them to expand and contract freely with temperature changes.
● Liquids, have less intermolecular forces than solids and hence they expand more than solids. This is the principle behind the mercury thermometers.
● In the case of gas molecules, the intermolecular forces are almost negligible and hence they expand much more than solids. For example in hot air ballons when gas particles get heated, they expand and take up more space.
● The increase in dimension of a body due to the increase in its temperature is called thermal expansion
● The expansion in length is called liner expansion. Similarly the expansion in area is termed as area expansion and the expansion in volume is termed as volume expansion.
In solids, for a small change in temperature ∆T, the fractional change in length [∆L/Lo] is directly proportional to ∆T. (∆L/ Lo) = αL ∆T
Therefore, αL = ∆L / Lo∆T ; Where, αL = coefficient of linear expansion .
∆L = Change in length ; Lo = Original length ; ∆T = Change in temperature,
For a small change in temperature ∆T the fractional change in area [∆A / Ao] of a substance is directly proportional to ∆T and it can be written as ∆A/ Ao = αA∆T
Therefore, αA [∆A / Ao∆T];
Where αA = coefficient of area expansion;
∆A = Change in area;
Ao = Original area;
∆T = Change in temperature
For a small change in temperature ∆T the fractional change in volume
[∆A / Vo] of a substance is directly proportional to ∆T.
∆V / Vo = αV ∆T, Therefore, αV [∆V/Vo∆T].
Where, αV = coefficient of volume expansion ; ∆V = Change in volume ;
Vo = Original volume ; ∆T = Change in temperature. Unit of coefficient of linear, area and volumetric expansion of solids is °C−1 or K−1.
4. Describe the anomalous expansion of water. How is it helpful in our lives?
Anomalous expansion of water:
● Liquids expand on heating and contract on cooling at moderate temperatures.
● But water exhibits an anomalous behavior.
● It contracts on heating between 0°C and 4°C.
● The volume of the given amount of water decreases as it is cooled from room temperature, until it reach 4°C.
● Below 4°C the volume increases and so the density decreases.
● The water has a maximum density at 4°C.
● This behavior of water is called anomalous expansion of water.
● In cold contries during the winter season, the surface of the lakes will be at lower temperature than the bottom.
● Since the solid water (ice) has lower density than its liquid form, below 4°C, the frozen water will be on the top surface above the liquid water (ice floats)
● This is due to the anomalous expansion of water.
● As the water in takes and ponds freeze only at the top the species living in the lakes will be safe at the bottom.
5. Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.
● Calorimetry means the measurement of the amount of heat released or absorbed by thermodynamic system during the heating process.
● When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the cold body.
● No heat is allowed to escape to the surroundings.
● It can be mathematically expressed as Qgain = − Qlost ; Qgain + Qlost = 0
● Heat gained or lost is measured with a calorimeter.
Usually the calorimeter is an insulated container of water.
● A sample is heated at high temperature (T1) and immersed into water at room temperature (T2) in the calorimeter. After some time both sample and water reach a final equilibrium temperature Tf. Since the calorimeter is insulated, heat given by the hot sample is equal to heat gained by the water.
Qgain = − Qlost
● The heat lost is denoted by negative sign and heat gained is denoted as positive.
● From the definition of specific heat capacity
Qgain = m2s2(Tf −T2)
Qlost = m1s1 (Tf – T1)
● Here s1 and s2 specific heat capacity of hot sample and water respectively.
m2s2(Tf −T2) = m1s1 (Tf – T1)
m2s2Tf − m2s2T2 = - m1s1Tf + m1s1T1
m2s2Tf + m1s1Tf = m2s2T2 + m1s1T1
The final temperature Tf = [ m1s1T1 + m2s2T2 ] / [ m1s1 + m2s2 ]
6. Discuss various modes of heat transfer.
Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. Thermal conductivity depends on the nature of the material.
Convection: Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.
Radiation: Radiation is a form of energy transfer from one body to another by electromagnetic waves. Radiation which requires no medium to transfer energy from one object to another.
Example: 1 Solar energy from the sun, 2. Radiation from room heater.
7. Explain in detail Newton’s law of cooling.
Newton's law of cooling:
● Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
dQ/dt ∝ − (T−TS) ------------- (1)
● The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where, T = Temperature of the object Ts = Temperature of the surrounding.
● The rate of cooling is high initially and drecreases with falling temperature.
● Let us consider an object of mass m and specific heat capacity s at temperature T. Let Ts be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,
dQ = msdT ----------------- (2)
Dividing both sides of equation (2) by
dQ / dt = msdT / dt ------------- (3)
● From Newton's law of cooling dQ/dt ∝ − (T−TS)
dQ / dt = − a (T−TS) ------------- 4
● Where is some positive constant. From equation (3) and (4)
− a (T − Ts) = ms (dt/ dt)
dt / (T−TST) = − (a/ ms) dt ------------- 5
Integrating equation (5) on both side
ʃ∞0 [dt / (T - Ts)] = −ʃt0 (a/ms) dt
In (T − Ts) = − (a/ms)t + b1
● Where b1 is the constant of integration, taking exponential on both sides, we get,
T = Ts+ b2e–(a / ms) t .
Here b2 = eb1 = Constant.
8. Explain Wien’s law and why our eyes are sensitive only to visible rays?
● When's law states that, the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body.
λm ∝ 1/T
λm = b / T
● Where, b is known as Wien's constant. Its value is 2.898 × 10-3 mk
● The Sun is approximately taken as a black body.
● Since any object above 0 K will emit radiation. Sun also emits radiation.
● Its surface temperature is about ; 5700 K. By substituting this value in the equation (1).
λm = b / T = 2.858×10-3 / 5700 = 508 nm
● It is the wavelength at which maximium intensity is 508nm.
● The visible part of the spectrum lies between 400nm to 700nm.
● The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible not in infrared or X-ray ranges in the spectrum.
9. Discuss the
a. thermal equilibrium
b. mechanical equilibrium
c. Chemical equilibrium
d. thermodynamic equilibrium.
a. Thermal equlibrium: Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.
b. Mechanical equilibrium:
● Consider a gas container with piston.
● When some mass is placed on the piston, it move downward due to downward gravitational force and after certain, humps and jumps, the piston will come to rest at a new position.
● When the downward gravitational force given by the piston is balanced by the upward force exerted by the gas, the system is said to be in mechanical equilibrium.
● A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermo dynamic system or on the surrounding by thermodynamic system.
c. Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.
d. Thermodynamic equilibrium: If two systems are set to be in thermodynamic equilibrium, then the system are at thermal, mechanical and chemical equilibrium with each other. In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.
10. Explain Joule’s Experiment of the mechanical equivalent of heat.
● Joule showed that mechanical energy can be converted into internal energy and vice versa.
● Two masses were attached with a rope and a paddle wheel.
● When these masses fall through a distance h hue to gravity, both the masses lose potential energy equal to 2mgh.
● When the masses fall, the paddle wheel turns.
Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel.
● This causes a rise in temperature of the water. This implies that gravitational potential energy is converted to internal energy of water.
● The temperature of water increases due to the work done by the masses.
● The mechanical work has the same effect as giving heat.
● Joule found that to raise 1 g of an object by 1°C, 4.186 J of energy is required. In earlier days the heat was measured in calorie. 1 cal = 4.186 J. This is called Joule's mechanical equivalant of heat.
11. Derive the expression for the work done in a volume change in a thermodynamic system.
Work done in volume changes:
● Consider a gas contained in the cylinder fitted with a movable piston. Suppose the gas is expanded quasi- statically by pushing the piston by a small distance dx.
● Since the expansion occurs quasi-statically the pressure, temperature and internal energy will have unique values at every instant.
● The small work done by the gas on the piston.
dW = Fdx ……….. (1)
● The force exerted by the gas on the piston F = PA. Here A is area of the piston and P is pressure exerted by the gas on the piston.
● Equation (1) can be rewritten as dW = PA dx ……. (2)
● But Adx = dV = change in volume during this expansion process. So the samll work done by the gas during the expansion is given by dW = PdV ………. (3)
● dV is positive since the volume is increased. So dW is positive.
The work done by the gas by increasing the volume from Vi to Vf is given by
W = VfʃVi PdV ………… (4)
● If the work is done on the system, then Vi > Vf. Then, W is negative. It implies that while the system is doing work, the pressure need not be constant.
To evaluate the integration, the pressure is first expressed function of volume and temperature using the equation of state.
12. Derive Mayer’s relation for an ideal gas.
● Consider μ mole of an ideal gas in a container with volume V, pressure P and temperature T.
● When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy.
● Let the change in internal energy be dU.
● If Cv is the molar specific heat capacity at constant volume.
dU = μCv dT ……… (1)
● Suppose the gas is heated at constant pressure so that the temperature increases by dT. If 'Q’ is the head supplied in this process and 'dv' the change in volume of the gas.
If' 'Q' = μCP dT ………. (2)
● If W is the work done by the gas in this process, then W = PdV …………. (3)
● But from the first law of thermodynamics.
Q = dU + W ….. (4)
● Substituting equations (1), (2) and (3) in (4), we get.
μCP dT = μCv dT + PdV ……. (5)
● For mole of ideal gas, the equation of state is given by PV = μRT
=> PdV + VdP = μRdT ……. (6)
Since the pressure is constant, dP = 0
CPdT = CvdT + RdT
CP = Cv + R (or) CP − Cv = R
This relation is called Meyer's relation.
13. Explain in detail the isothermal process.
● It is a process in which the temperature remains constant but the pressure and volume of a thermodynamic system will change. The ideal gas equation is
PV = μRT, Here, T is constant for this process.
● So the equation of state for isothermal process is given by PV = constant ………..(1)
● This implies that if the gas goes from one equilibrium state (P1, V1) to another equilibrium state (P2, V2) the following relation holds for this process
P1V1= P2V2 ……..(2)
● Since PV = constant, P is inversely proportional to P ∝ 1/V.
● This implies that PV graph is a hyperbola. The pressure-volume graph for constant temperature is also called isotherm.
● For an ideal gas the internal energy is a function of temperature only. For an isothermal process since temperature is constant, the internal energy is also constant. This implies that dU or ∆U = 0.
● For an isothermal process, the first law or thermodynamics can be written as follows,
Q = W ………. 3
● The heat supplied to a gas used to do only external work.
● The isothermal compression takes place when the piston of the cylinder is pushed. This, will increase the internal energy which will flow out of the system through thermal contact.
14. Derive the work done in an isothermal process
● Consider an ideal gas which is allowed to expand quasi-statically at constant temperature from initial state (P1, V1) to the final state (Pf, Vf). The work done by the gas.
W = VfʃVf pdV ------------(1)
● As the process occurs quasi-statically, a every stage the gas is at equilibrium with the surroundings. Since it is in equilibrium at every stage the ideal gas law is valid.
● Writing pressure in terms of volume and temperature.
P = μRT / V ----------- (2)
● Substituting equation (2) in (1) we get,
● μRT is constant throughout the isothermal process.
● By performing the integration in equation (3)
We get W = μRT in (Vf / Vi) ----------------- (4)
● Since we have an isothermal expansion, Vf /Vi, > 1.
So In (Vf / Vi) > 0
● The work done by the gas during an isothermal expansion is positive.
● The above result in equation (4) is true for isothermal compression also. But in an isothermal compression Vf / Vi < 1. So In (Vf / Vi ) < 0. As a result the work done on the gas in an isothermal compression is negative.
● In the PV diagram the work done during the isothermal expansion is equal to the area under the graph.
● For an isothermal compression, the area under the PV graph is equal to the work done on the gas which turns out to be the area with a negative sign.
15. Explain in detail an adiabatic process.
● This is a process in which no heat flows into or out of the system (Q=0). But the gas can expand by spending its internal energy or gas can be compressed through some external work.
● So the pressure, volume and temperature of the system may change in an adiabatic process.
● The equation of state for an adiabatic process is given by PVγ = Constant ……… (1).
● Here γ is called adiabatic exponent (γ = Cp/Cv) which depends on the nature of the gas.
● If the gas goes from an equilibrium state (P1, V1) to another equilibrium state (Pf, Vf) adiabatically then it satisfies the relation.
PiViγ = PfVf γ ………… (2)
● The PV diagram for an adiabatic process is also called adiabat.
● The PV diagram for isothermal and adiabatic process look similar. But actually the adiabatic curves is steeper than isothermal curve.
● To rewrite the equation (1) in terms of T and V. From ideal gas equation, the pressure P = μRT / V. Substituting this equation in the equation (1), we have [ μRT /V ] Vγ = Constant or [T/V] Vγ = Constant / (μR)
● So it can be written as TVγ−1 = Constant ……. (3)
● If the gas goes from an initial equilibrium state (Ti Vi) to final equilibrium state (Tf Vf) adiabatically then it satisfies the relation Ti Vi γ−1 = Tf Vf γ−1 …….. (4)
● The equation of state for adiabatic process can also be written in terms of T and P as TγP1-γ = Constant …… (5)
16. Derive the work done in an adiabatic process
● Consider μ moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base.
● A frictionless and insulating piston of cross sectional area A is fitted in the cylinder. Let W be the work done when the system goes from the initial state (Pi, Vi, Ti) to the final state (Pf Vf, Tf) adiabatically, W = VfʃViPdV ……….(1)
● The adiabatic process occurs quasi-statically, at every state the ideal gas law is valid.
● Under this conditions the adiabatic equation of state is PVγ = constant (or) P = constant / Vγ can be substituted in the equation (1), we get Wadia = VfʃVi constant / Vγ = dV
= Constant VfʃVi Vγ dV
But, = P1V1 = PfVf = constant
Wadia =1 / (1 – γ) [ (PfVf γ / Vfγ - 1 ) – ( PiViγ / Viγ – 1) ]
Wadia = (1 / 1–γ) [PfVf – P1V1] ……………… 2
From ideal gas law, PfVf = RTf and P1V1 = μRT1
Substituting in equation (2), we get, Wadia = μR/(γ–1) [Ti–Tf] ……………..(3)
● In adiabatic expansion, work is done by the gas. i.e., Wadia is positive. As Ti >Tf, the gas cools during adiabatic expansion.
● In adiabatic compression, work is done on the gas. i.e., Wadia is negative.
As Ti >Tf, the temperature of the gas increases during adiabatic compression.
17. Explain the isobaric process and derive the work done in this process
● This is a thermodynamic process that occurs at constant pressure. Even though pressure is constant in this process, temperature, volume and internal energy are not constant. From the ideal gas equation, we have
V = [ μR / P ] T ……… 1
Here μR / P = Constant
● In an isobaric process the temperature is directly proportional to volume.
V ∝ T (Isobaric process) ……….. (2)
● This implies that for a isobaric proces, the V-T graph is a straight line passing through the origin.
● If a gas goes from a state (Vi , Ti) to (Vf, Tf) at constant pressure, then the system satisfies the following equation
Tf / Vf = Ti / Vi
● The work done in an isobaric process: Work done by the gas W = vfʃvi PdV
● In an isobaric process, the pressure is constant W = P vfʃvi dV , W = P [Vf − Vi ] = P ΔV ……….. (3)
● Where ΔV denotes change in the volume. If ΔV is negative, W is also negative. This implies that the work is done by the gas.
● The equation (3) can also be rewritten using the ideal gas equation.
● From ideal gas equation PV = μRT and V = μRT / P
● Substituting this in equation (3) we get, W = μRTf (1 − Ti/Tf)
● In the PV diagram, area under the isobaric curve is equal to the work done in isobaric process.
● The first law of thermodynamics for isobaric process is given by ΔU = Q − PΔV.
18. Explain in detail the isochoric process.
This is a thermodynamic process in which the volume of the system is kept constant.
But pressure, temperature and internal energy continue to be variables.
The pressure - volume graph for an isochoric process is a vertical line parallel to pressure axis.
● The equation of state for an isochoric process is given by P = (μR/V)T.
Whrere, (μR/V)= Constant
● The pressure is directly proportional to temperature.
● This implies that the P-T graph for an isochoric process is a straight line passing through origin.
● If a gas goes from state (Pi, Ti) to (Pf, Tf) at constant volume, then the system satisfies the following equation = Pi / Ti = Pf / Tf
● For an isochoric process, ΔV = 0 and W = 0. Then the first law becomes.
ΔU = Q
● Implying that the heat supplied is used to increase only the internal energy.
Aa a result the temperature increases and pressure also increases.
● Suppose a system loser heat to the surroundings through conducting walls by keeping the volume constant, then its internal energy decreases.
● As a result the temperature decreases; the pressure also decreases.
19. What are the limitations of the first law of thermodynamics?
Limitations of first law of thermodynamics:
● The first law of thermodynamics explains well the inter convertibility of heat and work.
But it does not indicate the direction of change.
● When a hot object is in contact with a cold object, heat always flows from the hot object to cold object but not in the reverse direction.
● According to first law, it is possible for the energy to flow from hot object to cold object or from cold object to hot object.
● The direction of heat flow is always from higher temperature to lower temperature.
● When brakes are applied, a car stops due to friction and the work done against friction is converted into heat.
● But this heat is not reconverted to the kinetic energy of the car.
● So the first law is not sufficient to explain many of natural phenomena.
20. Explain the heat engine and obtain its efficiency.
● Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.
A heat engine has three parts: (a) Hot reservoir, (b) Working substance, (c) Cold reservoir.
Hot reservoir (or) source: It supplies heat to the engine. It is always maintained at a high temperature TH
● It is substance like gas or water, which converts the heat supplied into work.
● A simple example of a heat engine is a steam engine. In olden days steam engines were used to drive trains.
● The working substance in these is water which absorbs heat from the burning of coal.
● The heat converts the water into steam.
● This steam does work by rotating the wheels of the train, thus making the train move.
Cold reservoir (or) Sink: The heat engine ejects some amount of heat (QL) in to cold reservoir after it doing work. It is always maintained at a low temperature TL.
● For example: In the automobiles engine, the cold reservoir is the surroundings at room temperature. The automobile ejects heat to these surroundings through a silencer.
● The heat engine works in a cyclic process. After a cyclic process it returns to the same state.
● Since the heat engine returns to the same state after it ejects heat, the change in the internal energy of the heat engine is zero.
● The efficiency of the heat engine is defined as the ratio of the work done (output) to the heat absorbed (input) in one cyclic process.
● Let the working substance absorb heat QH units from the source and reject QL units to the sink after doing work W units.
Input heat = Work done + ejected heat
QH = W + QL
W = QH − QL
Then the efficiency of heat engine η = outPut / Input = W/QH = (QH− QL) / QH
η = Output / Input = W / QH = 1 – (QL/QH)
● QH, QL and W all are taken as positive, a sign convention followed in this expression.
Since QL < QH, the efficiency (η) always less than 1.
● This implies that heat absorbed is not completely converted into work.
● The second law of thermodynamics placed fundamental restrictions on converting heat completely into work.
21. Explain in detail Carnot heat engine.
● A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine.
● The Carnot engine has four parts which are given below.
Source: It is the source of heat maintained at constant high temperature TH. Any amount of heat can be extracted from it, without changing its temperature.
Sink: It is a cold body maintained at a constant low temperature TL. It can absorb any amount of heat.
Insulating stand: It is made of perfectly non-conducting material. Heat is not conducted through this stand.
Working substance: It is an ideal gas enclosed in a cylinder with perfectly non-conducting walls and perfectly conducting bottom. A non-conducting and frictionless piston is fitted in it.
● The working substance is subjected to four successive reversible process called Carnot's cycle.
● Let the initial pressure, volume of the working substance be P1 ,V1.
Step A to B: Quali-static isothermal expansion from (P1,V1, TH) to (P2, V2, TH).
● The cylinder is placed on the source. The heat (QH) flows from source to the working substance (ideal gas) through the bottom of the cylinder. Since the process is isothermal, the internal energy of the working substance will not change.
● The input heat increases the volume of the gas. The piston is allowed to move, out very slowly (quasi-statically).
● W1 is the work done by the gas in expanding from volume V1 to volume V2 with a decrease of pressure from P1 to P2.
● This is represented by the P-V diagram along the path AB.
● Then the work done by the gas (working substance) is given by
QH = WA→B = v2ʃv1 PdV
● Since the process occurs quasi-statically, the gas is in equlibrium with the source till it reaches the final state. The work down in the isothermal expansion is given by the equation.
WA→B = μRTH In ( V2/V1 ) = Area under the curve AB.
Step B to C: Quasi-static adiabatic expansion from (P2, V2, TH) to (P3, V3, TH)
● The cylinder is placed on the insulating stand and the piston is allowed to move out.
● As the gas expands adiabatically from volume V2 to volume V3 the pressure falls from P2 to P3.
● The temperature falls to TL. This adiabatic expansion is represented by curve BC in the P-V diagram.
● This adiabatic process also occurs quasi-statically and implying that this process is reversible and the ideal gas is in equilibrium throughout the process.
● The work done by the gas in an adiabatic expansion is given by,
WB→C = v3ʃv2 PdV = μR/(γ−1) [ TH – TL] = Area under the curve BC.
Step C - D : Quasi-static isothermal compression from (P3, V3, T1,) to (P4, V4, TL).
● The cylinder is placed on the sink and the gas is isothermally compressed until the pressure and volume become P4 and V4 respectively.
This is represented by the curve CD in the PV diagram.
● Let WC→D be the work done on the gas.
● According to first law of thermodynamics.
WC→D = v4ʃ v3 PdV = μRTL In ( V4/V3 ) = − μRTL In ( V3 / V4) = − Area under curve CD
V3 is greater than V4. So the work done is negative, implying work is done on the gas.
Step D-A: Quasi-static adiabatic compression from (P4, V4, T1) to (P1, V1, TH).
● The cylinder is placed on the insulating stand again and the gas is compressed adiabatically till it attains the initial pressure P1, volume V1 and temperature TH. This is shown by the curved DA in the P-V diagram.
WB→C v1ʃv4 PdV = -μR / (γ −1) [ TL – TH] = Area under the curve DA
● In the adiabatic compression also work is done on the gas so it is negative. Let 'W' be the net work done by the working substance in one cycle.
∴WA→B + WB→C + WC→D + WD→A since WD→A = WB→C
= WA→B + WC→D
● The net work done by the Carnot engine in one cycle W = |W|A→B − |W|C→D
● The network done by the working substance in one cycle is equal to the area (enclosed by ABCD) of the P-V diagram.
● After, one cycle the working substance returns to the initial temperature TH. This implies that the change in internal energy of the working substance after one cycle is zero.
22. Derive the expression for Carnot engine efficiency.
Efficiency of a Carnot engine:
● Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.
η = Work done / Heat extracted = W/ QH ……..(1)
● From the first law of thermodynamics, W = QH − QL
η = [ QH - QL ] / QH = 1 [QL / QH] ………..(2)
● Applying isothermal condition, we get,
QH = μRTH In (V2 / V1); QL = μRTL In (V3 / V4 ) ………..(3)
● The amount of (QL) ejected into the sink, (QL/QH ) = [TL In (V3 / V4 ) ] / [ TH In (V2 / V1) ] ………..(4)
● By applying adiabatic conditions, we get, TH V2γ−1 = TL V3γ−1
● By dividing the above the two equations, we get, (V2 / V1) γ−1 = (V3 / V4 ) γ−1
● Which implies that, V2/V1 = V3/V4 ………. (5)
● Substituting equation (5) in (4), we get, QL/QH = TL/TH
● The efficiency η = 1 − TL/TH
23. Explain the second law of thermodynamics in terms of entropy.
● The quantity QH / TH is equal to QL / QH the quantity is called entropy.
● It is a very important thermodynamic property of system.
● It is also state variable QH / TH is the entropy received by the Carnot engine from hot reservoir and QL/ TL is entropy given out by the Carnot engine to the cold reservoir.
● For reveersible engines (Carnot Engine) both entropies should be same.
● So that the change in entropy of the Carnot engine in one cycle is zero. But for all practical engines like diesel and petrol engines which are not reversible engines, they satisfy the relation QL/TL > QH/TH
● "For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change".
● Entropy determines the direction in which natural process should occur.
● Because entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object.
● Entropy will decrease leading to violation of second law thermodynamics.
● Entropy is also called 'measure of disorder'. All natural process occur such that the disorder should always increases.
● Consider a bottle with a gas inside. When the gas molecules are inside the bottle it has less disorder. Once it spreads into the entire room it leads to more disorder.
● In order words when the gas is inside the bottle the entropy is less and once the gas spreads into entire room, the entropy increases. From the second law of thermodynamics, entropy always increases.
● If the air molecules go back in to the bottle, the entropy should decrease, which is not allowed by the second law of thermodynamics.
● The same explanation applies to a drop of ink diffusing into water. Once the drop of ink spreads, its entropy is increased.
● The diffused ink can never become a drop again. So the natural processes occur in such a way that entropy should increase for all irreversible process.
24. Explain in detail the working of a refrigerator.
Refrigerator: A refrigerator is a Carnot's engine working in the reverse order.
● The working substance (gas) absorbs a quantity of heat Q1 from the cold body (sink) at a lower temperature TL.
● A certain amount of work W is done on the working substance by the compressor and a quantity of heat QH is rejected to the hot body (source) ie. the atmosphere at TH.
● From the first law of thermodynamics.
● QL + W = QH AS a result, the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.
Co-effeicient of performance (COP) (β):
● Measure of the efficiency of a regrigerator.
● The ratio of heat extracted from the cold body (sink) to the enternal work done by the compressor W.
COP = β = QL/W
β = QL / [QH − QL]
β = 1 / [ (QH − QL) − 1]
● QH / QL = TH / TL
β = TL / [TH − TL]
● Generater the COP, the better is the condition of the refrigerator.
● A typical refrigerator has COP around 5 to 6
● Lesser the difference in the temperature of the cooling chamber and the atmosphere, COP is higher.
● Without external work heat cannot flow from cold object to hot object.
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