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2. THERMODYNAMICS The energy that is transferred from one system to another by force moving its point of application in its own direction It is the study of interelations between heat and other forms is called work. of energy dx Thermodynamic System : A collection of large number of molecules of matter (solid, liquid or gas) which are so
arranged that these posses certain values of pressure, System
volume and temperature forms a thermodynamic system. Fs = P s A • The parameters pressure, volume, temperature, internal energy etc which determine the state or condition of system are called thermodynamic state variables. In thermodynamics we deal with the thermodynamic systems Work done by the system F dx as a whole and study the interaction of heat & energy during the change of one thermodynamic state to another. Ps Adx
2.1 Thermal Equilibrium Ps dV The term ‘equilibrium’ in thermodynamics implies the state Where P Pressure of system on the piston. This work done when all the macroscopic variables characterising the s by system is positive if the system expands & it is negative system (P, V, T, mass etc) do not change with time. if the system contracts. • Two systems when in contact with each other come to • Work and Heat are path functions whereas internal energy thermal equilibrium when their temperatures become same. is a state function. • Based on this is zeroth law of thermodynamics. According • Heat & work are two different terms through they might to zeroth law, when the thermodynamics systems A and B look same. are separately in thermal equilibrium with a third thermodynamic system C, then the systems A and B are in 2.3 Important Thermodynamics Terms thermal equilibrium with each other also. State Variables : P, V, T, moles 2.2 Heat, Work and Internal Energy They can be extensive or intestive.
Internal Energy is the energy possessed by any system Equation of State : The equation which connects the due to its molecular K.E. and molecular P.E. Here K.E & P.E pressure (P), the volume (V) and absolute temperature (T) of a gas is called the equation of state. are with respect to centre of mass frame. This internal energy depends entirely on state and hence it is a state PV = constant (Boyle’s Law) variable. For 1 real gases internal energy is only by virtue V of its molecular motion. = constant (Charle’s Law) T nf RT Units for ideal gases where PV = NRT 3 Thermodynamic Process : A thermodynamic process is n = number of moles said to take place when some changes occur in the state of f = Degree of fredom a thermodynamic system, i.e., the thermodynamic parameters of the system change with some important time. R = Universal Gas Constant Types of these thermodynamic process are Isothermal, T = Temperature in Kelvin Adiabatic, Isobaric and Isocboric Internal Energy can be change either by giving heat energy Quasi Static Process : A thermodynamic process which is or by performing some work. infinitely slow is called as quasi-static process. Heat Energy is the energy transformed to or from the system • In quasi static process, system undergoes change so slowly, because of the difference in temperatures by conduction, that at every instant, system is in equilibrium, both thermal convection or radiation. and mechanical, with the surroundings. HEAT & THERMODYNAMICS
• Quasi-static process is an idealised processed. We generally 2.5.1 Isothermal Process assume all the processes to be quasistatic unless stated. Description : A thermodynamic process in which Indicator P-V, Diagram : A graph between pressure and temperature remains constant volume of a gas under thermodynamic operation is called P-V. diagram. Condition : The walls of the container must be perfectly conducting to allow free exchange of heat between gas
P and its surroundings. The process of compression or expansion should be slow a so as the provide time for exchange of heat. b These both conditions are perfectly ideal. c d v Equation of State : T = Constant or Pv = Constant Indicator Diagram : a Isobaric
b Isothermal P P c Adiabatic d Isochoric • Area under P – V diagram gives us work done by a gas.
st v 2.4 1 Law of Thermodynamics T Let Q = Heat supplied to the system by the surroundings W = Work done by the system on the surroundings P Slope of PV at any point. U = Change in internal energy of the system. V First law of thermodynamics states that energy can neither U = 0 (Temperature remains constant) be created nor be destroyed. It can be only transformed v2 from the form to another. W = Pg dV Mathematically : Q = U + Q v2 Sign Conventions : v2 nRT dV • When heat is supplied to the system, then Q is positive = VV [Using PV = nRT) and when heat is withdrawn from the system, Q is v2 negative. v • When a gas expands, work done by the gas is positive & nRTl n 2 = v when a gas contracts then w is negative 1 • U is positive, when temperature rises and U is negative, First Law of Thermodynamics when temperature falls. Q = U + W Remember here we always take work done by the system.
In chemistry, work done on the system is considered. Hence v2 st nRTl n there is some different look of 1 law of Thermodynamics in Q = v1 chemistry. Q + W = U Remarks : All the heat supplied is used entirely to do work against external sorroundings. It heat is supplied then the where Q, U have same meanings but W stands for work gas expands & if heat is withdrawn then the gas contracts. done on the system Practical Examples : 2.5 Application of the First of Law of Thermodynamics Melting of ice at 0°C Here we see how 1st Law of Thermodynamics is applied to Boiling of H O at 100°C various thermodynamic processes. 2 HEAT & THERMODYNAMICS
2.5.2 Adiabatic Process cons tan t 1 1 Description : When there is not heat exchange with = r 1 r 1 1 r V2 V1 surroundings Conditions : The walls of the container must be perfectly Also we know non-conducting in order to prevent any exchange of heat r r between the gas and its surroundings. PV1 1 PV 2 2 constant The process of compression or expansion should be rapid 1 PVPVr r so there is no time for the exchange of heat. 2 2 1 1 r 1 r 1 1 r V2 V1 These conditions are again ideal condition and are hard to obtain PVPV nR T T Equation of State : W = 2 2 1 1 1 2 1 r r 1 Pvr = constant First Law of Thermodynamics or TVr –1 = constat Q = UT + W r or PT1 r cons tan t Substituting the values Indicator Diagram We get Q = 0 Q = 0 is as expected P Remarks : It gas expands adiabatically then its temp decreases & vice vers a isotherm Practical Example • Propagation of sound waves in the form of compression & v rarefaction • Sudden bursting of a cycle tube.
rp 2.5.3 Isochoric Process Slope of adiabatic curve V Description : Volume remains constant • As shown in graph adiabatic curve is steeper than Condition : A gas being heated or cooled inside a rigid isothermal curve. container. nf RDT nR T2 T 1 PVPV2 2 1 1 P U Equation of State : V = constant or = constant 2 r 1 r 1 T
Work Done by Gas : If a gas adiabatically expands from V1 P P P to V2 V2 dV W = r V1 V v T T V2 dV cons tan t = Vr V1 nf R T U = r 2 PV cons tan t cons tan t p n R T2 T 1 r = V r 1
r 1 V2 V PVPV nR T = cons tan t 2 2 1 1 1 r U = V1 r 1 r 1 HEAT & THERMODYNAMICS
Work First Law of Thermodynamics W = O as gas does not expands Q = U + W First Law of thermodynamics nf R T Q = nR T Q = U + W 2 nf R T Q = ...(7) fR 2 Q = RT ...(10) 2 Remarks : Since we have studied earlier, that when heat is Remarks : Similar to C , we can define molar heat capacity supplied to any process. its temp increases according to V relation at constant pressure Q = nCT Q C = ...(11) P n T Q P C = n T From equation 10 & 11 Now this C depends upon external conditions for gases. f R R We get CP = ...(12) Q 2 Here it is refered as ...(8) n T v From equation 9 & 12
i.e. Molar heat capacity at constant volume f R Replacing by C we get Comparing equation 7 and 8 2 V C = C = R ...(13) fR P V We getC = ...(9) v 2 which is also called as Mayer’s Relation. • Similar to molar specific heat at constant pressure and molar 2.5.4 Isobaric Process specific heat at constant volume, we can define molar Description : When pressure remains constant specific heat for any process. Condition : When in one container, the piston is free to For example : move and is not connected by any agent. C adiabatic = 0 Equation of State : P = constant C isothermal =
V Basically gas does not possess a unique specific heat. = constant T Mainly we have CP & CV Indicator Diagram : Specific Heat at Constant Volume : It is defined as the amount of heat required to raise the temperature of 1g of a P P gas through 1C°, when its volume is kept constant. It is
denoted as CV. Specific Heat at Constant Pressure : It is defined as the amount of heat required to raise the temperature of 1g of a v T gas through 1C° keeping its pressure constant. It is denoted
as CP. nf R T • Please Note C , C means Molar heat Capacity & C , C U = same as always P V P V 2 means specific heat capacity
•CV = MCV & CP = MCP where M stands for molar mass of any W = PdV P V sample.
(as pressure is constant) R CC • PV = PV2 – PV = nR T M HEAT & THERMODYNAMICS
2.5.5 Melting Process 2.7 Heat Engines
In any case first law is always applicable It is a device that converts heat energy into mechanical energy. Q = mL as learned earlier. f Key Elements : W = 0 • A source of heat at higher temperature (In the change of state from solid to liquid we ignore any expansion or contraction as it is very small) • A working substance According to first law of thermodynamics • A sink of heat at lower temperature. U = Q – W Working : U = mL • The working substance goes through a cycle consisting of f several processes. Remark : The heat given during melting is used in • In some processes it absorbs a total amount of heat Q from increasing the internal energy of any substance 1 the source at temperature T1. 2.5.6 Boiling Process • In some processes it rejects a total amount of heat Q2 to the
Here sink at some lower temperature T2. • The work done by the system is a cycle is transferred to the Q = mLV environment via some arrangement. W = P[V2 – V1] Schematic Diagram (Pressure is constant during boiling and it is equal to
atmospheric pressure) W U = Q – W U = mLf – P(V2 – V1) Source or Q1 Working Q2 Sink or 2.5.7 Cyclic Process Hot Reservoir Substamnce Cold Reservoir A cyclic process is one is which the system returns to its T T initial stage after undergoing a serves of change 1 2 Example Indicatgor Diagram First Law of Thermodynamics P Energy is always conserved A Q1 = W + Q2 ...(14) Thermal Efficiency of a heat engine is defined of the ratio B of net work done per cycle by the engine to the total amount V of heat absorbed per cycle by the working substance from the source. It is denoted by . U = mLV O W = Area under the loop. W = Q ...(15) Q = W as per First Law of thermodynamics 1 Here W is positive if the cycle is clockwise & it is negative Using equation 14 and 15 we get if the cyclic is anti clockwise. Q2 = 1 ...(16) 2.6 Limitations of the First Law of Thermodynamics Q1 • The first law does not indicate the direction in which the Ideally engines shuld have efficiency = 1 change can occur. Remarks : The mechanism of conversion of heat into work • The first law gives no idea about the extent of change varies for different heat engines. • The first law of thermodynamics gives no information about • The system heated by an external furnace, as in a steam the source gives no information about the source of heat. engine. Such engines are called as external combustion i.e. whether it is a hot or a cold body. engine. HEAT & THERMODYNAMICS
• The system is which heat is produced by burning the fuel Though all the statements are the same in their contents, inside the main body of the engine. Such an engine is called the following two are significant. as Internal Combustion Engine. Kelvin Pnek Statement : No process is possible whose 2.8 Refrigerator and Heat Pumps sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work. A refrigertor or heat pump is a device used for cooling Calcius Statement : No process is possible whose sole things. result is the transfer of heat from a colder object to a hotter Key Elements : object.
• A cold reservoir at temperature T2. Significance : 100% officiency in heat engines or infinite • A working substance. CoP in refrigerators is not possible.
• A hot reservoir at temperature T1 2.10 Reversible and Irreversible Process Working Reversible Process : A thermodynamic process taking a • The working substance goes through a cycle consisting of system from initial state i to final state f is reversible, if the several process. process can be turned back such that both, the system and • A sudden expansion of the gas from high to low pressure the surroundings return to their original states, with no which cool it and converts it into a vapour-liquid mixture. other change anywhere else in the universe. • Absorption by the cold fluid of heat from the region to be Conditions for reversibility : cooled converting it into vapour. 1. Process should proceed at an extremely slow rate, i.e., • Heating up of the vapour due to external work done on the process is quasistatic so that system is in equilibrium with working substance. surroundings at every stage. • Release of heat by the vapour to the sorroundings bringing 2. The system should be free from dissipative forces like it to the initial state and completing the cycle. friction, inelasticity; viscosity etc. Sychematic Diagram. Examples : No process exactly reversible, though a slow expansion of an ideal gas is approximately reversible. W Irreversible Process : A process which does not satisfy any of the conditions for reversible is called an irreversible
Hot Reservoir Q1 Q2 Cold Reservoir process. T T 1 2 Causes : • Spontaneous process First Law of Thermodynamics • Presence of friction, viscosity and such dissi-ptive forces Q + W = Q ...(17) 2 1 Significance of Reversibility : Coefficient of Performance of refrigerator () is defined as • Main concern of thermodynamics is the efficiency with the ratio of quantity of heat removed per cycle from contents which the heat is converted into Mechanical Energy. of the refrigerator (Q2) to the energy spent per cycle (W) to remove this heat • Second Law of Thermodynamics rules out the possibility of a perfect heat engine with 100% efficiency. Q 2 ...(18) • It turns out that heat engine based on idealised reversible W processes achieves the highest possible efficiency. Using equation 17 and 18 we get 2.11 Carnot Engine Q 2 Sadi Carnot devised on ideal cycle of operation for a heat QQ1 2 engine called as carnot cycle. Ideally heat pumps should have = Engine used for realising this ideal cycle is called as carnot heat engine. 2.9 Second Law of Thermodynamics Constructions : The essential parts of an ideal heat engine There are number of ways in which this law can be stated. or Carnot heat engine are shown in figure. HEAT & THERMODYNAMICS
Consider one gram mole of an ideal gas enclosed in the
cylinder. Let V1, P1, T1 be the initial volume, pressure and temperature of the gas. The initial state of the gas is represented by the point Aon P–V. diagram, We shall assume Cylinder Ideal that all the four processes are quasi-static and non Gas dissipative, the two conditions for their reversibility. Steps Source at Insulating Sink at
Temp. T1 Pad Temp. T2 1. Isothermal Expansion : The cylinder is placed on the source and gas is allowed to (i) Source of heat : The source is maintained at a fixed higher expand by slow outward motion of piston. Since base is
temperature T1, from which the working substance draws perfectly conducting therefore the process is isothermal. heat. The source is supposed to possess infinite thermal Now capacity and as such any another of heat can be drawn from it without changing its temperature. U1 = O (ii) Sink of heat : The sink is maintained at a fixed lower V2 temperature T , to which any amount of heat can be rejected q = W = RT1l n. = Area ABmKA 2 1 1 V by the working substance. It has also infinite thermal 1 capacity and as such its temperature remains constant at q Heat absorbed by gas T , even when any amount of heat is rejected to it. 1 2 w work done by gas (iii) Working substance : A perfect gas acts as the working 1 substance. It is contained in a cylinder with non-conducting 2. Adiabatic Expansion : sides but having a perfectly conducting-base. This cylinder The cylinder is now removed from source and is placed on is fitted with perfectly non-conducting and frictionless the perfectly insulating pad. The gas is allowed to expand piston. further from B(P2, V2) to C (P3, V3). Since the gas is thermally Apart from these essential parts, there is a perfectly insulated from all sides, therefore the processes is adiabatic insulating stand or pad on which the cylinder can be placed. q2 = 0 It would isolate the working substance completely from the surroundings. Hence, the gas can undergo adiabatic RTT U 2 1 changes. 2 r 1 The Carnot cycle consists of the following four stages : RTT 1. Isothermal exdpansion W 3 1 = Area BCNMB 2 r 1 2. Adiabatic expansion 3. Isothermal Compression : 3. Isothermal compression The cylinder is now removed from the insulating pad and is 4. Adiabatic compression. placed on the sink at a temperature T2. The piston is moved The cycle is carried out with the help of the Carnot engine slowly so that the gas is compressed until is pressure is P4 as detailed below : and volume is V4. U = 0 P 3
V4 A(V1 , P 1 ) W2 RT 2l n = – Area CDLNC V3
B(V2 , P 2 )
Pressure (P) Q1 T1
D(V4 , P 4 ) V 4 C(V3 , P 3 ) q3 RT 2l n V3 T2 Q2 X O KLMN q3 = Heat absorbed in this process Volume (V) w3 = Work done by Gas HEAT & THERMODYNAMICS
4. Adiabatic Compression : QT 1 1 The cylinder is again placed on the insulating pad, such QT2 2 that the process remains adiabatic. Here the gas is further
T2 compressed to its initial P1 and V1. Carnot 1 T1
RTT1 2 U4 Division r 1 • Carnot engine - depends only upon source temperature nd sink RTT temperature. W 2 1 = – area DAKLD 4 r 1 • Carnot engine =1 only when T2 = 0 K or T1 = which is impossible to attain. q4 = 0 • If T = T = Heat cannot be converted to mechanical w = work done by the gas 2 1 4 energy unless there is same difference between the Analysis : temperature of source and sink. Total work done by the engine per cycle. 2.12 Carnot Theorem = W + W + W + W 1 2 3 4 Statement : = W1 + W3 (a) Working between two given temperatures, T 1 of hot
VV reservoir (the source) and T2 of cold reservoir (the sink, no 2 4 W = RT1l n RT2 l n engine can have efficiency more than that of the Carnot VV1 3 engine.
Q1 = Total heat absorbed = q1 (b) The efficiency of the Carnot engine is independent of the nature of the working substance. V2 = RT1l n ...(19) Engine used for realising this ideal cycle is called as carnot V1 heat engine.
Q2 = Total heat released = –q3 Proof :
[q3 = Heat absorbed & not heat released] Step - 1 : Imagine a reversible engine R and an irreversible
engine-I working between the same source (hot reservoir T1) V3 = RT2l n ...(20) and sink (cold reservoir T2). V4 Step - 2 : Couple two engines such that I acts like heat We can see that for heat engine engine and R acts like refrigerator.
W = Q1 – Q2 Step - 3 : Let engine I absorb Q1 heat from the source deliver work W1 and release the balance Q – W1 to the sink in one = Area under ABCDA 1 cycle. Efficiency of Carnot Engine
1 W Q W 1 2 1 QQ1 1 Q1 Q–W1 1
Now steps 2 is adiabatic 2 step 4 is also adiabatic T1 T2 Q1 Q–W1
r 1 r 1 R TVTV1 2 2 3
r 1 r 1 Step - 4 : Arrange R, such that it returns same heat Q, to and TVTV1 1 2 4 the source, taking Q2 from the sink and requiring work
V V W = Q1 – Q2 to be done on it. 2 3 ...(21) VV1 4 Step - 5 : Supppose R < I (i.e.) If R were to act as an engine it would give less work output than that of I (i.e.) From equation 19, 20 and 21 we get 1 1 W < W for a given Q1 and Q1 – W > Q1 – W HEAT & THERMODYNAMICS
Step - 6 : In totality, the I-R system extracts heat (r1 – W) – • There is no force of attraction or repulsion amongst the 1 1 (Q1 – W ) = W – W & delivers same amount of work in one molecules of an ideal gas. cycle, without any change in source or anywhere else. This 3.2.2 Real Gas is against second Law of Thermodynamics. (Kelvin - Planck statement of second law of thermodynamics) All gases are referred to as real Gases. All real gas near the ideal gas behaviour at low pressures Hence the assertion q1 > qR is wrong. and temperatures high enough, where they cannot be • Similar argument can be put up for the second statement of liquified. carnot theorem, (ie) Carnot efficiency is independent of working substance. 3.2.3 Ideal Gas Laws
We use ideal gas to else are calculating but the relation. Avogadro Hypothesis : Equal volumes of all gas under identical conditions of pressure and temperature would QT 1 1 contain equal number of molecule. QT will always hold true for any working substance 2 2 Perfect Gas Equation : used in a carnot engine. PV = nRT
3. KINETIC THEORY OF GASES SRT and P M In this topic, we discuss the behaviour of gases and how are the various state variable like P, V, T, moles, U are inter- where n = Number of moles.
related with each other r = Universal Gas constant = NxkB where N = Avagadro No. 3.1 Molecular Nature of Matter D
kB = Boltzman constant Same as Atomic Theory given by Delton. According to R = 8031 J/mol K. him, atoms are the smallest constituents of elements. All atoms of one element are identical, but atoms of different R = 1.98 Cal/mol K. element are different. Boyle’s Law : When temperature of a given mass of gas is In solids : Atoms tightly packed, interatomic spacing about kept constant, its pressure varies inversely as the volume of gas. Å . Interatomic force of attraction are strong. (i.e.) PV = constant In liquids : Atoms are not as rigidly fixed as in solids. Interatomic spacing is about the same 2Å. Interatomic for a Charles Law : When pressure of a given mass of kept attraction are relative weaker. constant, volume of the gas varies directly as the temperature of the gas. In Gases : Atoms very free. Inter atomic spacing is about tens of Angstroms. Interatomic forces are much weaker in (i.e.) V T gases than both in solids and liquids. Dalton’s Law of Partial Pressures : The total pressure of a mixture of ideal gases is the sum of partial pressures exerted In this chapter, we mainly focus on gases. by the individual gases in the mixture. 3.2 Molecular Nature of Matter P–V = (n1 + n2 + n3 + ...)RT
3.2.1 Ideal Gas RT P = (n + n + n + ...) 1 2 3 V That gas which strictly obeys the gas laws, (such as Boyle’s Law, Charles, Gay Lussac’s Law etc.) P = P1 + P2 + ...... Characteristics n RT where P = 1 Pressure of gas • The size of the molecule of an ideal gas is zero. 1 V HEAT & THERMODYNAMICS
Neviation of Real Gas from Ideal Gas : volume occupied by the molecules is negligible in comparison to the volume of the gas. Ideal Gas 4. The molecules do not exert any fore of attraction or 1 repulsion on each other, except during collision. T 1 5. The collisions of the molecules with themselves and with
T2 T1 > T 2 > T 3 > the walls of the vessel are perfectly elastic. As such, that momentum and the kinetic energy of the molecules are
T3 conserved during collisions, though their individual velocities change. 0 200 400 600 800 6. There is not concentration of the molecules at any point P (atm) inside the container i.e. molecular density is uniform throughout the gas. 1.6 7. A molecule moves along a straight line between two 1.4 successive collisions and the average straight distance P 1.2 covered between two successive collisions is called the
1.0 T1 mean free path of the molecules. T > T > T 0.8 1 2 3 8. The collisions are almost instantaneous, i.e., the time of
T1 collision of two molecules is negligible as compared to time 0.6 interval between two successive collisions. 0.4 T T2 T 2 3 3.4 Pressure of an Ideal Gas and its Expression 0.2 T Pressure exerted by the gas is due to continuous 0 3 20 60 100 140 160 220 bombardment of gas molecules against the walls of the V container. Expression : 1.2 Consider a gas enclosed in a cube of side 1. Take the axes P1 > P 2 > P 3 1.0 to be parallel to the sides of the cube, as shown in figure. A molecule with velocity ( , , ) hits the planar wall parallel T P x y z 0.8 1 to yz-plane of area A (= l2). Since the collision is elastic, the 0.6 molecule rebounds with the same velocity; its y and z
P2 components of velocity do not change in the collision but 0.4 P 3 the x-component reverses sig. That is, the velocity after 0.2 collision is (– x,vz, vy). The change in momentum of the 0 molecule is : –m x – (m x) – 2m x. By the principle of 0 100 200 300 400 500 conservation of momentum, the momentum imparted to the V wall in the collision = 2m x. 3.3 Kinetic Theory Postulates To calculate the force (and pressure) on the wall, we need to calculate momentum imparted to the wall per unit time, if 1. A gas consists of a very large number of molecules (of the 23 it is within the distance x t from the wall. that is, all other of Avogadro’s number. 10 ), which are perfect elastic molecules within the volume, A x t only can hit the wall in spheres. For a given gas they are identical in all respects, time T is 1/2A x t n where n is the number of molecules but for different gases, they are different. per unit volume. The total momentum transferred to the 2. The molecules of a gas are in a state of incessant random wall by these molecules in time t is : motion. They move in all directions with different speeds., Q = (2m ) (1/2 n Av t) ( of the order of 500 m/s) and obey Newton’s laws of motion. x x The force on the wall is the rate of momentum transfer Q/t 3. The size of the gas molecules is very small as compred to and pressure is force per unit area : the distance between them. If typical size of molecule is 2 P = Q/(A t) = nm 2 Å, average distance between the molecules is 20 Å. Hence x HEAT & THERMODYNAMICS
Actually, all molecules in a gas do not have the same 1 2 PVM ...(23) velocity; there is a distribution in velocities. The above 3 equation therefore, stands for pressure due to the group of with equation 23 and Ideal gas equation molecules with speed x in the x-direction and n stands for the number density of that group of molecules. The total 1 nRT M 2 pressure is obtained by summing over the contribution due 3 to all groups: 1 2 2 nRT Nm P nm x 3
2 2 n 3RT 1 2 where x is the average of x . Now the gas is isotropic, m N 2 2 i.e. there is no preferred direction of velocity of the molecules Also N = nN in the vessel. Therefore by symmetry, A 3 R 1 2 2 2 2 x y z m 2 NA 2
1 2 2 2 1 2 3 x y z Average KE of translation per molecule of the gas kB T 3 3 2 3.5 Kinetic Interpretation of Temperature where is the speed and 2 denotes the mean of the squared speed. Thus From above equations, we can easily see that KE of one molecule is only dependent upon its Temperature.
1 2 P nm KE of molecule will cease if, the temperature of the gas 3 molecules become absolute zero. Absolute zero of a temperature may be defined as that 12 1 M 1 2 P mn S ...(22) temperature at which the root mean square velocity of the 2 2 V 2 gas molecule reduces to zero. M = Total mass of gas moleculus All the ideal gas laws can be derived from Kinetic Theory V = Total volume fo gas molecules of gases.
3.4.1 Relation between Pressure and KE of Gas Molecules 3.6 Derivation of Gas Laws from Kinetic Theory
From equation 22 3.6.1 Boyle’s Law
1 2 2 PS We know that PVNK 3 3
2 where K is the average kinetic energy of translation per PE 3 gas molecule. At constant temperature. K is constant and Pressure exerted by an ideal gas is numerically equal to two for a given mass of the gas. N is constant. third of mean kinetic Thus, PV = constant for given mass of gas at constant temperature, which is the Boyle’s Law. 3.4.2 Average KE per molecule of the Gas 3.6.2 Charle’s Law From equation 22 We know 2 We know that PVNK 3 1 M P 2 3 For a given mass of gas, N is constant. HEAT & THERMODYNAMICS
r 3 rms Since K k T, K T and as such PV T. 2 B 3P 3P 1 But as , r or r If P is constant, V T, which is the Charles’ Law. rms 3.6.3 Constant Volume Law Therefore, if r1 and r2 are the rates of diffusion of two gases 2 of densities and respectively, We know that PVNK 1 2 3 r For a given mass of gas, N is constant. Since 1 2 r2 1 3 K kB T, K T The rates of diffusion of two gases are thus inversely 2 proprotional to the square roots of their densities which is Thus, PV T Graham’s Law of diffusion. If V is constant, P T, which the constant volume law. 3.6.7 Dalton’s Law of Partial Pressures 3.6.4 Ideal Gas Equation The kinetic theory of gases attributes the gas pressure to 2 3 the bombardment of the walls of the containing vessel by As PVNK and K kB T 3 2 molecules. In a mixture of ideal gases, we might therefore expect the total pressure (P) to be the sum of the partial 2 3 PV N kB T or PV = Nk T pressures (p , p , ...) due to each gas, i.e., 3 2 B 1 2 which is the ideal gas equation. 2NN1 2 2 P p1 p 2 ... K1 K2 ... 3 V1 3 V2 3.6.5 Avogadro’s Law Consider two gases 1 and 2. We can write 2 NN1 2 or P K1 K2 ... 3 V1 V2 2 2 PVNK,PVNK 1 2 1 1 3 1 2 2 3 2 In equilibrium, the average kinetic energy of the molecules If their pressures, volumes and temperatures are the same, of different gases will be equal, i.e., then 3 K1 K 2 ...K kB T P1 = P2, V1 = V2, KK1 2 . 2
Clearly, N1 = N2 Thus : Thus, Equal volumes of all ideal gases existing under the same 2 3 conditions of temperature and pressure contain equal P n1 n 2 ... kTB n1 n 2 ...kT B ...(24) 3 2 number of molecules which is Avogadro’s Law or hypothesis. NN 1 2 This law is named after the Italian physicist and chemist, where n1 , n2 ... VV1 2 Amedeo Avogadro (1776–1856). Equation (24) represents Dalton’s Law of partial pressures PV Alliter : As PV NkB T, N which states that : kB T The resultant pressure exerted by a mixture of gases or If P, V and T are constants, N is also constant. vapours which do not interact in any way is equal to the 3.6.6 Graham’s Law of Diffusion sum of their individual (i.e., partial) pressures. Figures shows a model explaning kinetic theory of gases. It The rate (r) of diffusion of a gas through a porous pot or has been constructed in accordance with theory on one into another gas is determined by the rms speed of its molecules, i.e., hand and real experimental observations on the other hand. HEAT & THERMODYNAMICS
N = Number of degrees of freedom of the system 1 mv2 K 2 N = 3A – R I PV = nRTnRTPV 3 III kB T 3.8.1 Monoatomic Gases 2 Based on Based on combining The molecules of a monoatomic gas (like neon, argon, two mathematical experimental helium etc) consists only of one atom. observations Based on models of a gas in of gases Newtonian boxes I and II 1 Nm 2 A = 1 II P v mechanocal 3 V treatment of gas as R = 0 collection of particles N = 3 Here 3 degrees of freedom are for translational motion 2 PV NK 3.8.2 Diatomic Gases 3 = NkB T A = 2 Based on substituting Assuming the distance between the two molecules is fixed and rearranging the expression for K back into equation in box II then R = 1 The above piece of logic is tempting but false. This is due N = 3 × 2 – 1 = 5 to the reason that though the equation in box IV is useful, Here 5 degrees of freedom implies combination of 3 it does not tell us anything new, since it results from translational energies and 2 rotational energies. combining equations in boxes II and III. 3.7 Internal Energy As studied in thermodynamics, Internal Energy of any substance is the combination of Potential Energies & Kinetic Energies of all molecules inside a given gas. • In real gas Internal Energy = P.E of molecules + K.E of Molecules Y Y • In real gas Internal Energy = K.E of Molecules A1 X X A1 A2 Here PE of molecules is zero as assumed in Kinetic theory A2 postulates; There is no interaction between the molecules (i) (ii) hence its interactional energy is zero. Z Z (a) (b) Y 3.8 Degree of Freedom A The number of degrees of freedom of a dynamical system is 1 defined as the total number of coo-ordinates or independent X A2 A3 quantities required to describe completely the position & configuration of the system. Z (c) Example : If vibrational motion is also considered then [only at very • A particle moving in straight line, say along X-axis need high temperatures only x coordinate to define itself. It has only degree of N = 7 freedom. where 3 for translational • A particle in a plane, needs 2 co-ordinates, hence has 2 degree of freedom. 2 for rotational In general if 2 for vibrational A = number of particles in the system 3.8.3 Triatomic Gas R = number of independent relations among the particles HEAT & THERMODYNAMICS
Linear dy where vibrational velocity A = 3 dt R = 2 ky2 N = 3 × 3 – 2 = 7 and = Energy due to configuration 2 Non-Linear According to Law of Equipartition
1 Energy per degree of freedom k T 2 B
1 1 k T K T K T is energy for 2B 2 BB
A = 3 complete one vibrational mode. R = 2 3.10 Specific Heat Capacity N = 3 × 3 – 3 = 6 With the knowledge of law of equipartition, we can predict • Here again vibrational energy is ignored. the heat capacity of various gases.
3.8.4 Polyatomic Gas 3.10.1 Monoatomic Gas
A polyatomic gas has 3 translational, 3 rotational degrees Degree of freedom = 3. of freedom. Apart from them if there v vibrational modes Average Energy of a molecule at temperature T then there will be additional 2v vibrational degrees of freedom. 1 E 3 kB T Total degree of freedom 2 n = 3 + 3 + 2V = 6 + 2V Energy for one mole E × NA
3.9 Law of Equipartition of Energy 3 U kBA N T Statement : According to this law, for any dynamical system 2 in thrmal equilibrium, the total energy is distributed equally 3 amongst all the degrees of freedom, and the energy U RT associated with each molecule per degree of freedom is 2 1 In thermodynamics, we studied k T , where k is Boltzman constant and T is temperature 2 B B QU C of the system. V [ W = O for constant v] TTV k T Application : U f B where = Total degree of freedom. 2 3R CV This law is very helpful in determining the total internal 2 energy of any system be it monoatomic, diatomic or any C 5 polyatomic. Once the internal energy is know we can very 5R r P CP & . 2 CV 3 easy predict Cv & CP for such systems. Remark : In case vibrational motion is also there in any 3.10.2 Diatomic Gases system, say for diatomic molcule, then there should be When no vibration energy due to vibrational as well given by Degree of freedom = 5 2 1 dy 1 2 EV m ky 5 2 dt 2 Average energy for one mole = RT 2 HEAT & THERMODYNAMICS
U 5 Therefore the atoms do not possess any translational or CR V T 2 rotational degree of freedom. On the other hand, the molecules do possess vibrational 7R C motion along 3 mutually perpendicular directions. P 2 Hence for 1 mole of a solid, threre are NA number of atoms. C 7 r P The energy associated with every molecule CV 5 1 When vibration is present. 3 2 kB T 3KB T 2 There is only one mode of vibration between 2 molecules. U = 3 Rt for one mole Degree of freedom = 7 QU 7 C 3R U RT TT 2 • The above equation is called as Dulong & Petit’s Law. 7 • At low temperatures the vibrational made may not be that CRV 2 active hence, heat capacity is low at low temperatures for 9 solids. and CR P 2 C 3R 9 and r 7 3.10.3 Polyatomic Gases T Degree of freedom 300 K = 3 for translational 3.10.5 Specific heat capacity of Water + 3 for rotational Water is treated like solid. + 2v for vibrational Water has three atoms, 2 of hydrogen and one of oxygen = 6 + 2v Total degree of freedom for every atom if v = Number of vibrational modes = 3 × 2 = 6 RT Total degree of freedom for every molecule of water U 6 2v K 2 = 3 × 2 = 18 C = (3 + V)R V 1 C = (4 + V) R 18 R T P QU 2 C TTT 4 V and r 3 V C = 9R 3.10.4 Specific heat capacity of solids 3.11 Maxwell Law of Distribution of Molecular • In solids, there is very less difference between heat capacity Assumptions of Maxwell Distribution at constant pressure or at constant volume. Therefore we • Molecules of all velocities between 0 to are present. do not differentiate between C & C for solids. P V • Velocity of one molecule, continuously changes, though QU fraction of molecules in one range of velocities is constant. C TT Result
{As solids hardly expand or expansion is negligible} 2 3 / 2 mv M Now in solid the atoms are arranged in an array structure N 4 N V2 e 2kB T V 2 k T and they are not free to move independently like in gases. B HEAT & THERMODYNAMICS
Expression : dN V Where NV Mean Free Path dV
Where d N V = Total number of molecules with speeds between V & V + dV V t N = Total number of molecules.
N V d
d
V Vmp Vav Vrms
Based on this we define three types of speed f molecules of gas Suppose the molecules of a gas are spheres of diameter d. Focus on a single molecule with the average speed
3.12 Mean Free Path Remark : Mean free path depends inversely on the number density and size of the molecule. The path tranversed by a molecule between two successive collisions with other molecule is called the free path 3.13 Brownian Motion The irregular movement of suspended particles like tiny Total distance travelled by a molecule l dust particles or pollen grains in a liquid is called Brownian No. of collisions it makes with other molecules. Motion. FEB . 20 . 21 . 22
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