physics 11th Heat Transfer

HEAT TRANSFER

1. Introduction

• We already know that heat is the energy transferred from from one systm to another or from one part of the system to its another part , arising due to temperature difference.
• Heat can be transferred from one place to other by through three different modes conduction,convection & radiation.

2. Thermal Conduction

• Conduction of heat takes place in a body when diffrent part of body are at diffrent temperature.
• To notice conduction of heat put one end of metal rod on flame and another end on your hand. After some time you will feel hotness in your hand also. 
• Here heat transfer takes place from hot end on flame to cold end in your hand through conduction.
• How heat transfers through conduction is given in steps below
  1)Molecules at hot end of the rod begin to vibrate as there is an increase in the energy of vibration as temperature of the end of rod on flame increases.
  2)These vibrating molecules then collides with the nearest neighbour sharing their energy with them and increasing their energy.
  3)These neighbouring molecules further pass their energy to molecules on colder end of the rod i.e farther from the end put on flame.
  4)This way energy of thermal motion is passed along from one molecule to the next keeping their original position fixed.
• Metals are good conducters of electricty as well as heat.

3.Thermal Conductivity

• Ability of Material to conduct heat is measured by thermal conductivity of that material.
• Consider a slab of uniform crossection A and length L also one face of slab is kept at Temperature T₁ and another at T₂and remaining surface area is covered with a non conducting material to avoid transfer of heat.
  
  
• After sufficient time slab reaches steady state temperarure at every point will remain unchanged.
• In steady state, rate of flow of the heat through any crosssection of slab is
  a) directly propertional to area A
  b) directly propertional to temperature diffrence (T₂-T₁)
  c) inversely propertional to length
• Thus if H is the quantity of heat flowing through slab per unit time then
  
  where k is a constant whose numerical value depends on the material and is called thermal conductivity of the material.
• S.I. unit of thermal conductivity is Js^-1m^-1K^-1.
• For small amount of heat dQ flowing between two faces of slab in small time interval dt,
  
• Materials for which K is large are good conductors of heat, while small value of K for a material implies material is poor conductor of heat. 

4. Convection

• Convection is transfer of heat by actual motion of matter
• If material is forced to move by a blower or pump the process is called forced convection.
• If the material flows due to difference in density for example that caused by thermal expansion then the process is called natural of free convection.
• Meahanism of heat transfer in human body is forced convection. Here heart serves as the pump and blood as the circulating fluid.
       

5. Radiation

• Radiation process does not need any material medium for heat transfer.
• Term Radiation refers to the continous emission of energy from surface of all bodies and this energy is called radient energy.
• Radiant energy is in the form of Electro Magnetic waves. 
• Radiant energy emitted by a sunface depends on the temperature and nature of the surface.
• All bodies whether they are solid, liquid or gas emit radiant energy.
• EM radiations emitted by a body by virtue of increased temperature of a body are called thermal radiation.
• Thermal radiation falling on a body can partly be absorbed and partly be reflected by the body and this absorption and reflection of radiation depends on the color of body.
• Thermal radiation travels through vacuum on straight line and with the velocity of light.
• Thermal radiations can be reflected and refracted.

6. Black Body Radiation

• A body that absorbs all the radiation falling on it is called a black body.
• Radiation emitted by black body is called Black Body radiation.
• A black body is also called an ideal radiator.
• For practical purpose black body can be considered as an enclosure painted black from inside and a small hole is made in the wall.
  
     
                      
• Once radiation enters the enclosure it has very little chance to come out of the hole and it gets absorbad after multiple refrections inside th enclosure.
• Concept of a perfact black body is an ideal one.

7. Stefan Boltzmann law

• The rate u_rad at which an object emits energy via EM radiation depends on objects surface area A
  and temperature T in kelvin of that area and is given by
       u_rad = σεAT⁴               (2)
  Where
       σ= 5.6703×10^-8 W/m²K⁴
  is stefan boltzmann constant and ε is emissivity of object's sunface with value between 0 and 1.
• Black - Body radiator has emissivity of 1.0 which is an ideal limit and does not occuer in nature.
• The rate u_abs at which an object absorbs energy via thremal radiation from its environment with temperature T_env (in kelvin) is
       u_abs = σεA(T_env)⁴               (3)
  Where ε is same as in equation 2
• Since an object radiate energy to the enviornment and absorbe energy from environment its net energy enchange due to thermal radiation is
       u=u_abs-u_rad
        = σεA{(T_env)⁴-T⁴}     (4)
• u is positive if net energy is being absorbed via radiation and negative if it is being lost via radiation.

8. Nature of thermal Radiation

• Radiation emitted by a black body is a mixture of waves of different wavelengths and only a small range of wanelength has significant countribution in the total radiation.
• A body is heated at different temperature and Enegy of radiation is plotted agains wavelength is plotted for different temperature we get following curves.
  
   
• These curves show
  (i) Energy is not uniformly distributed in the radiation spectreum of black body.
  (ii) At a given temperature the intensity of radiations increases with increase in wavelength, becoms maximum at particular wavelength and further increase in wavelngth leads to decrease in intensity of heat radiation.
  (iii) Increase in temperature causes increase in energy emission for all wavelengths.
  (iv) Increase in temperature causes decrease in λ_m, where λ_m is wavelenght corresponding to highest intersity. This wavelength λ_m is inversily properational to the absolute temperature of the emitter.
       λ_mT = b                    (5)
  Where b is a constant and this equation is known as Wein's displacement law.
       b=0.2896×10^-2 mk for black body and is known as Wien's constant.

9. Kirchoff's law

• Good absorbers of radiation are also good radiaters this statement is quantitatively explained by Kirchoff's law.
  (i) Emissive Power -
  Emissive power denotes the energy radiated per unit area per unit solid angle normal to the area.
  
       E = Δu/ [(ΔA) (Δω) (Δt)]
  where, Δu is the energy radiated by area ΔA of surface in solid angle Δω in time Δt.
  
  (ii) Absorptive Power -
  Absorptive power of a body is defined as the fraction of the incident radiation that is absorbedby the body
       a(absorptive power) = energy absorbed / energy incident
  
  (iii) Kirchoff's Law
  "It status that at any given temperature the ratio of emissive power to the absorptive power is constant for all bodies and this constant is equal to the emissive power of perfect B.B. at thesame temperature.
       E_/abody=EB.B.
• From kirchoff's law we can say that a body having high emissive power should have high absorptive power and those having low emiesive power should have law absorptive power so as to keep the ratio E/a same.

10. Newton's Law of Cooling

• Consider a hot body at temperature T₁ is placed in surrounding at temperature T₂.
• For small temperature difference between the body and surrounding rate of cooling is directly proportional to the temperature difference and surface area exposed i.e.,
       dT/dt = - bA (T₁ - T₂)
• This is known a Newton's law of cooling.
  b depends on nature of surface involved and the surrounding conditions. Negative sign is to indicate that T₁>T₂ , dT/dt is negative and temperature decreases with time
• According to this law, the rate of cooling is directly prospertional to the excess of temperature.

Solved examples

Question : 
The surface of a body has a emissivity of .55 and area of 1.5 m²
Find out the following 
a. What rate of heat is radiated from the body if the temperature is 50°C
b. At what rate is radiation absorbed by the radiater when the surrounding temperature is 22°C
c What is the net rate of radiation from the body
Given σ=5.67 *10^-8

Solutions 
a) Rate of radiation radiated=eσAT_b⁴=(.55)(5.67 *10^-8)(1.5)(323)⁴=509W
b) Rate of radiation absorbed=eσAT_s⁴=(.55)(5.67 *10^-8)(1.5)(295)⁴=354W
c)Net =Rate of radiation radiated-Rate of radiation absorbed=155 W

Physics 11th Thermodynamics

THERMODYNAMICS

1. Introduction

• Thermodynamics is that branch of physics which is concerned with transformation of heat into mechanical work.
• It deals with the concepts of heat, temperature and interconversion of heat into other forms of energy i.e., electrical, mechnical, chemical magnetic etc.
• Thermodynamics does not take any account of atomic or molecular constitution of matter and it deals with the bulk systems.
• State of any thermodynamic system can be described in terms of certain know macroscopic variables known as thermodynamic variables. 
• Thermodynamic variables determine the thermodynamic behaviour of a system . Quantities like pressure(P), volume(V), and temperature(T) are thermodynamic variables. Some other thermodynamic variables are entropy, internal energy etc. described in terms of P, V and T 
• A thermodynamic system is said to be in thermal equilibrium if all parts of it are at same temperature.
• Thus two systems are said to be in thermal equilibrium if they are at same temperature.

2. Concept of Heat

• Heat may be defined as energy in transit.
• Word heat is used only if there is a transfer of energy from one thermodynamic system to the another.
• When two systems at different temperatures are kept in contect with each other then after some time temperatures of both the syatems become equal and this phenomenon can be described by saying that energy has flown from one system to another.
• This flow of energy from one system to another on account of temperature difference is called heat transfer.
• Flow of heat is a non-mechanical mode of energy transfer.
• Heat flow depends not only on initial and find states but also on path it's.

3. P-V Indicator Digram

• Only two thermodynamic variables are sufficient to describe a system because third vaiable can be calculated from equation of state of the system.
• P-V Indicator Digram is just a graph between pressure and volume of a system undergoing an operation.
• When a system undergoes an expansion from state A (P₁ V₁) to a state B (P₂V₂) its indicator digram is shown as follows.
   
• In case of compression system at state A(P₁ V₁) goes to a state B(P₂V₂) its indicator digram is as follows.
  
   
• Intermediate states of system are represented by points on the curve.
• The pressure volume curve for a fixed temperature is called isotherm.

4. Work in volume changes

• Consider a cylinder filled with gas and equiped with a movable piston as shown in fig below
  
  
  fig - Force exerted by a system during small expansion.
  Suppose,
       A - Cross Sectional area of cylinder
       P - Pressure exerted by piston at the piston face.
       PA - Force exerted by the system.
• If piston moves out by a distance dx then work done by this force is dW given by
            dW = PAdx
             = PdV                    (1)       
  since V = Adx and dV is change in volume of the system.
• In a finite volume change from V₁ to V₂
       W=∫PdV                         (2)
  where limits of integration goes from V₁ to V₂
  Graphically this relationship is shown below
  
   
• Thus eqn (2) can be interpreted graphically as area under the curve between limits V₁and V₂.
• If pressure remains constant while the volume changes, then work is
       W = P(V₂-V₁)          (3)
• Work done not only depends on initial and final states but also on the intermediate states i.e., on the path.

Learning:Work done in a process is given by area under the process on the PV diagram 

5. Internal Energy and first law of thermodynamics

• Internal energy can be described as the sum of kinetic and potantial energies of individual movecules in the material.
• But in thermodynamics one should keep in mind that U is simply a macrosopic variable of the system.
• U is thermodynamic state variable and its value depends only on the given state of the system and not on path taken to arrive the state.
• Transfer of heat and performance of work are two mean of adding or subtracting energy from a system.
• On transfer of energy, system is said to have undergone a change in internal energy.
• Thus the sum of heat put into the system plus work done on the system equals increase in internal energy of the system for any process.
  if, U₁ is internal energy of state 1 and U₂ is internal energy of state 2 than change in internal energy would be
        ΔU=U₂ - U₁
• If W is the work done by the system on its surroundings then -W would be the work done on the system by the surroundings .
• If Q is the heat put into the system then,
       Q+(-W)=ΔU
  usually written as
       Q=ΔU+W                     (4)
• Equation (4) is then know as first law of thermodynamics and it can be applied when
        Q, W and U are expressed in same units.
  Some Imp stuff      (1) Q is positive when heat is given to the system and Q is negative when heat is taken from the system
       (2) W is positive when system expands and does work on surroundings
• Hence we may say that when a certain amount of heat Q is given to the system then some part of it is used in increasing internal energy ΔU of the system while remaining part leaves the system in form of work done by the system on its surroundings.
• From equation 4 we see that first law of thermodynamics is a statement of conservation of energy stated as
  ' The energy put into the system equals the sum of the work done by the system and the change in internal energy of the system'
• If the system undergoes any process in which ΔU=0 i.e., charge in internal energy is zero then from (4)
       Q = W
  that is heat supplied to the system is used up enterely in doing work on the surroundings.

6. Specific heat capacity of an ideal gas

• We have defined specific heat capacity and molar specific heat capacity earlier in the previous chapter.
• There are two specific heats of ideal gases.
  (i) Specific heat capacity at constant volume
  (ii) Specific heat capacity at constant pressure
  C_p and C_v are molar specific heat capacities of ideal gas at constant pressure and volume respectively for C_p and C_v of ideal gas there is a simple relation.
       Cp-Cv = R                         (7)
  where R- universal gascontant
• This relation can be proved as follows.
  from first law of thermaodynamics for 1 mole of gas we have
       ΔQ =Δ U+PΔV                    (8)
• If heat is absorbed at constant volume then ΔV = 0 and
        C_V=(ΔQ/ΔT)_V=(ΔU/ΔT)_V     (9)
  If Q in absorbed at constant pressure than
       C_P=(ΔQ/ΔT)_P=(ΔU/ΔT)_P+P(ΔV/ΔT)_P
  now ideal gas equation for 1 mole of gas is
       PV = RT
        = P(ΔV/ΔT) = R                    (10)
  from (9) and (10)
       C_P - C_V=(ΔU/ΔT)_P-(ΔU/ΔT)_V+P(ΔV/ΔT)_P
• Since internal energy U of ideal gas depands only on temperature so subscripts P and V have no meaning.
  =>          C_P - C_V = R
  which is the desired relation

7. Thermodynamic Processes

(a) Quasi static Processes

:

• In Quasi static process deviation of system from it's thermodynamic equilibrium is infinitesimally small.
• All the states through which system passed during a quasi static process may be regarded as equilibrium states.
• Consider a system in which gas is contained in a cylinder fitted with a movable piston then if the piston is pushed in a infinitely slow rate, the system will be in quiscent all the time and the process can be considered as quasi-static process.
• Vanishingly slowness of the process is an essential feature of quasi-static process.
• During quasi-static process system at every moment is infinitesimally near the state of thermodynamic equilibrium.
• Quasi static process is an idealized concept and its conditions can never be rigoursly satisfied in practice.

(b) Isothermal Process

:

• In isothermal process temperature of the system remains constant throughout the process.
• For an iso-thermal process equation connecting P, V and T gives.
       PV = constant
  i.e., pressure of given mass of gas varies inversly with its volume this is nothing but the Boyle's law.
• In iso thermal process there is no change in temperature, since internal energy for an ideal gas depends only on temperature hence in iso thermal process there is no change in internal energy.
  Thus,
       ΔU=0
  therefore,     ΔQ =ΔW
• Thus during iso thermal process
       Heat added (or substacted) from the system = wok done by (or on) the system

(c) Adiabatic Process

:

• Process in which no heat enters or leaves a system is called an adiabatic process
• For every adiabatic process Q=0
• Prevention of heat flow can be acomplished by surrounding system with a thick layer of heat insulating material like cork, asbestos etc.
• Flow of heat requires finite time so if a process is perfomed very quickly then process will be pratically adiabatic.
• On applying first law to adiabatic process we get
       ΔU=U₂ - U₁= - ΔW               (adiabatic process)
• In adiabatic process change in internal energy of a system is equal in magnitade to the work by the system.
• If work is done on the system contracts i.e. ΔW is nagative then.
       ΔU = ΔW
  and internal energy of system increases by an amount equal to the work done on it and temperature of system increases.
• If work is done by the system i.e., ΔW is negative
       ΔU = -Δ W
  here internal energy of systems decreases resulting a drop in temperature.

(d) Isochoric process v:

• In an isochoric process volume of the system remain uncharged throughout i.e. ΔV = O.
• When volume does not change no work is done ; ΔW = 0 and therefore from first law
       U₂ - U₁ = ΔU =ΔQ
• All the heat given to the system has been used to increase the intenal energy of the system.

(e) Isobaric Process

:

• A process taking place at constant pressure is called isobaric process.
• From equation (3) we see that work done in isobaric process is
       W = P(V₂ - V₁) nR (T₂-T₁)
  where pressure is kept constant.
• Here in this process the amount of heat given to the system is partly used in increasing temperature and partly used in doing work.

8. Work done in Isothermal process

• In an isothermal process temperature remains constant.
• Consider pressure and volume of ideal gas changes from (P₁, V₁) to (P₂, V₂) then, from first law of thermodynamics
       ΔW = PΔV
  Now taking ΔV aproaching zero i.e. ΔV�� and suming ΔW over entire process we get total work done by gas so we have
       W = ∫PdV
  where limits of integration goes from V₁ to V₂
  as PV = nRT we have P = nRT / V
       W = ∫(nRT/V)dV
  where limits of integration goes from V₁ to V₂
  on integrating we get,
       W=nRT ln(V₂/V₁)               (3)
  Where n is number of moles in sample of gas taken.

9. Work done in an Adiabatic process

• For an adiabatic process of ideal gas equation we have
       PV^γ = K (Constant)               (14)
  Where γ is the ratio of specific heat (ordinary or molar) at constant pressure and at constant voluume
       γ = C_p/C_v
• Suppose in an adiabatic process pressure and volume of a sample of gas changs from (P₁, V₁) to (P₂, V₂) then we have
       P₁(V₁)^γ=P₂(V₂)^γ=K
  Thus, P = K/V^γ
• Work done by gas in this process is
       W = ∫PdV
  where limits of integration goes from V₁ to V₂
  Putting for P=K/V^γ, and integrating we get,
       W = (P₁V₁-P₂V₂)/(γ-1)          (16)
• In and adiabatic process if W>0 i.e., work is done by the gas then T₂< T₁
• If work is done on the gas (W<0) then T₂ > T₁ i.e., temperature of gas rises.

10. Heat Engine and efficiency

• Any device which convents heat continously into mechenical work is called a heat engine.
• For any heat engine there are three essential requirements.
  (i) SOURCE : A hot body at fixed temperature T₁ from which heat engine can draw heat
  (ii) Sink : A cold body, at a fixed lower temprature T₂, to which any amount of heat can be rijectd.
  (iii) WOEKING SUBTANCE : The material, which on being supplied with heat will do mechanical work.
• In heat engine, working substances, could be gas in cylinder with a moving piston.
• In heat engine working substance takes heat from the sorce, convents a part of it into mechanical work, gives out rest to the sink and returns to the initial state. This series of operations constitutes a cycle.
• This cycle is represented in fig below
  
   
• Work from heat engine can be continously obtained by performing same cycle again and again.
• Consider,
       Q₁ - heat absorbed by working substance from sorce
       Q₂ - heat rejected to the since
       W - net amount of work done by working substance
       Q₁-Q₂ - net amount of heat absorbed by working substance.
       ΔU = 0 since in the cycle Working Substance returns to its initial condition.
  So on application of first law of thermodynamics
       Q₁-Q₂ = W
• Thermal efficiency of heat engine
       η= work output in energy units / Heat input in same energy units
        = W / Q₁ = (Q₁-Q₂ )/ Q₁
  Or, η = 1-(Q₂/Q₁)                    (17)
  from this equation it is clear that
       Q = 1 for Q₂=0
  and there would be 100% conversion of heat absorbed into work but such ideal engines are not possible in practice.

11. Principle of a Refrigerator

• Refrigerators works in reverse direction of heat engines.
• In refrigerators working substance extracts heat Q₂ from sink at lower temperature T₂
• Some external work is performed by the compressor of refrigerator and then heat Q₁ is rejected to the
  source, to the radiator of the refrigerator.
  
  
  Coefficent of performance :
       β= Amount of heat absorbed from the cold reservoir / work done in running the mechinery
  Q₂ - heat absorbed from cold reservoir.
  Q₁ - heat rejected to hot reservoir during one complete cycle
  W = (Q₁-Q₂ ) is the work done in running the machinery
  thus,
       β= Q₂/W =Q₂/(Q₁-Q₂)          (18)
• Like heat engines refrigerators can not work without some external work done on the system. Hence coefficent of performance can not be infinite.

12. Second law of thermodynamics

• First law of thermodynamics states the equivalance of heat and energy.
• It does not state anything about the limitation in the conversion of heat into work or about the condition necessary for such conversion.
• Second law of thermodynamics is generalization of certain experience and observation and is concerned with tine direction in which energy flow takes place.
• This law can be stated in number of ways. Although differently said, they are essentially equvalent.
  (i)Kelvin Plank Statment :
  "It is impossible to construct a device which, operating in a cycle, has a sole effect of extracting heat from a reservoir and performing an equivalent amount of work".
  (ii)Clasius Statement :
  "It is impossible for a self acting machine, unaided by enternal agency, to transfer heat from a colder body to a hotter body".
• It can ne proved that these two statements of second law are completely equivalent and voilation of Kelvin Plank statement leads to voilation of Clasius statement and vice-versa.

13. Reversibility and irreversibility

• Reversible process is the one which can be retraced in opposite order by changing external conditions slightly.
• Those processes which can not be retraced in opposite order by reversing the controling factors are known as irreversible process.
• It is a consequence of second law that all the natural processes are irreversible process.
• Conditions for reversibility of a process are
       (i) Process is performed quasi-statically
       (ii) it is not accompained by any dissipative effects.
• It is impossible to satisfy these two conditions perfectly, thus requessible process is purely an ideal abstraction.

14. Carnot's Heat Engine

• According to second law of thermodynamics, no heat engine can have 100% efficiency
• Carnot’s heat engine is an idealized heat engine that has maximum possible efficiency consistent with the second law.
• Cycle through which working substance passed in Carnot’s engine is known as Carnot’s Cycle.
• Carnot's engine works between two temperatures
       T₁ - temperature of hot reservoir
       T₂ - temperature of cold reservoir
• In a Complete Carnot's Cycle system is taken from temperature T₁ to T₂ and then back from
  temerature T₂ to T₁.
• We have taken ideal gas as the working substance of cornot engine.
• Fig below is an indicator digram for Cornot Cycle of an ideal gas
  
  
  (i) In step b→c iso thermal esepansion of gas taken place and thermodynamic variables of gas changes from (P₁, V₁,T₁) to (P₂,V₂,T₁)
• If Q₁ is the amount of heat absorbed by working substance from the source and W₁ the work done by the gas then from eqn (13)
       Q₁ = W₁ = nRT1 ln (V₂/V₁)     (19)
  as process is iso thermal.
  (ii) Step c→d is an adiabatic expension of gas from (P₂, V₂, T₁) to (P₃,V₃,T₂). Work done by gas in adiabatic esepansion is given by eqn (16)
       W₂ = nR (T₁-T₂)/(γ-1)               (20)
  (iii)Step d→a is iso-thermal compression of gas from (P₃,V₃,T₂) to (P₄,V₄,T₂). Heat Q₂ would be released by the gas to the at temperature T₂
• Work done on the gas by the environment is
       W₃ = Q₂
             = nRT₂ln(V₃/ V₄)               (21)
  (iv)Step a→b is adiabatic compression of gas from (P₄, V₄, T₂) to (P₁, V₁, T₁)
• Work done on the gas is
       W₄ =nR (T₁-T₂)/(γ-1)                    (22)
• Now total work done in one complete cycle is
       W = W₁ + W₂ - W₃ - W₄
        = nRT₁ln(V₂/V₁)-nRT₂ln(V₃/V₄)          (23)
  as W₂ = W₄
• Efficency of carnot engine
       η=W/Q₁ = 1-(Q₂/Q₁)
        = 1-(T₂/T₁)ln(V₃/V₄)/ln(V₂/V₁)          (24)
  or     η= 1-[T₂ln(V₃/V₄)/T₁ln(V₂/V₁)]          (25)
  Since points b and c lie on same iso thermal
  ⇒ P₁V₁=P₂V₂                    (26)
  also points c and d lie on same adiabatic
  ⇒     P₂(V₂)^γ=P₃(V₃)^γ                    (27)
  also points d and a lie on same iso thermal and points a and b on sum adiabatic thus,
        P₃V₃=P₄V₄                    (28)
       P₄(V₄)^γ=P₂(V₁)^γ                         (29)
  multiplying all the above four eqns me get
       V₃/₄ = V₂/V₁                    (30)
  Putting this in equation (25) we get
       η= 1-(T₂/T₁)                         (31)
  From above eqn we can draw following conclusions that efficency of Carnot engine is
  (i) independent of the nature of working substance
  (ii) depend on temperature of source and sink

15. Carnot Theorem

• Carnot Engine is a reversible engine.
• Carnot’s theorem consists of two parts
  (i) no engine working between two given temperatures can be more efficent than a reversible Carnot engine working between same source and sink.
  (ii) all reversible engines working between same source and sink (same limits or temperature) have the same efficiency irrespective the working substance.

Solved Examples

Question-1. What is true of Isothermal Process
a, ΔT >0
b, ΔU=0
c ΔQ=ΔW
d PV=constants

Solution-1:

In an Isothermal Process
Temperature remains constant ΔT =0
Since Internal energy depends on the temperature
ΔU=0

From first law of Thermodynamics
ΔU=ΔQ-ΔW
Since ΔU=0
ΔQ=ΔW

Also PV=nRT
As T is constant
PV= constant



Question-.2 Two absolute scales A and B have triple points of water defined as 200A and 350A. what is the relation between T_A and T_B



Solution-2

Given that on absolute scale
Triple point of water on scale A = 200 A
Triple point of water on scale B = 350 B
Also, triple point of water on Kelvin scale = 273.16 K
Now temperature on scale A and on scale B is equivalent to 273.16 K on absolute temperature scale.
Thus, value of one degree on absolute scale A = (273.16/200) K
Or, 
Value of temperature T_A on absolute scale A = (273.16XT_A)/200 
Similarly value of temperature T_B on absolute scale B = (273.16XT_B)/350
Since T_A and T_B represent the same temperature
273.16×T_A/200 = 273.16×T_B/350
Or, T_A = 200T_B/350 = 4T_B/ 7 

Question 3:A gas is contained in a cylinder with a moveable piston on which a heavy block is placed. Suppose the region outside the chamber is evacuated and the total mass of the block and the movable piston is 102 kg. When 2140 J of heat flows into the gas, the internal energy of the gas increases by 1580 J. What is the distance s through which the piston rises?

Solution:3

Total heat supplied =Workdone + Change in internal energy

So work done=2140-1580=560 J

Let s be the distance moved then
the workdone is given by =Fs
Fs=560
s=560/F
=560/102*10
s=.54 m


Question-4: At 27°C,two moles of an ideal monoatomic gas occupy a volume V.The gas is adiabatically expanded to a volume 2V.
Calculate the ratio of final pressure to the intial pressure
Calculate the final temperature
Change in internal energy
Calculate the molar specific heat capacity of the process

Solution-4
Given 
n=2 T=27°C=300 K ,V₁=V,V₂=2V

Now PV^y=constant
P₁V₁^y=P₂V₂^y
P₂/P₂=(V₁/V₂)^y
=.5^5/3

ALso
T₁V₁^y-1=T₂V₂^y-1
or T₂=300/2^5/3=189K

Change in internal energy=nC_vΔT
For monoatomic gas C_v=3R/2
Substituting all the values
Change in internal energy==-2764J

As in adiabatic process ΔQ=0,molar specific heat capacity=0


Question-5
An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C.It absorbs 6*10² cal of heat at the higher temperature.Calculate the amount of heat supplied to the engine from the source in each cycle 

Solutions-5:
T₁=227°C =500K
T₂=127°C =400K

Efficiency of the carnot cycle is given by
=1-(T₂/T₁)=1/5

Now also efficency =Heat supplied from source/Heat absorbed at high temperature
so Heat supplied from source=6*10²*(1/5)==1.2*10²cal

Physic 11th Kinetic Theory Of Gases

KINETIC THEORY OF GASES

1. Gas Laws

Gas laws are study of any two of quantities like pressure, volume and temperature, when the third is kept constant

A. Boyle's law


• "At constant temperature, the volume of a given mass of gas is invesely propertional to pressure." Thus
                 V ∝ 1/P
  Or,                PV = constant               (1)
• If P₁, V₁are initial pressure and volumes and P₂, V₂be final values then,            
       P₁V₁= P₂V₂                    
• Graph between P and V at temperature T₁ and T₂ such that T₁<T₂ are shown below,
  
   
• Graph above shows that Boyle's law is strictly not obeyed by gases at all values of P and T but it obeys this law only at low pressure and high temperature i.e., at law density

B. Charle's Law 

• Charle's Law is stated as follows :
  "When pressure of a gas is constant the volume of a given mass of gas is directly proportional to its absolute temperature".
                 V/T = Constant               (2)
• Graph between V and T is
  
  
  This graph shows that experimental graph deviates from straight line. Theoritical and enperimental graphs are in agreement at high temperature.

2. Ideal gas equation

• We can combine Boyle’s law eqn (1) and charle’s law eqn (2) in to a single eqn i.e.,
       PV/T = Constant          (3)
  If n moles is the mass of gas then we write
       PV = nRT               (4)
  where, n is number of moles of gas, R=N_AK_B is the universal constant known as gas constant and T is the absolute temperature.
• A gas satisfying eqn (4) at all values of presserves and temperatures is said to be an ideal gas
  now no of moles of gas
       n = m/M = N/N_A
  where
       m - mass of gas containing N molecules
       M - molar mass
       N_A - Avagadro’s number.
  From this,
       P = ρRT/M
       ρ - mass density of gas.

3. Moleular nature of matter

• We know that molecules which are made up of one or more atoms constitute matter.
• In solids these atoms and molecules are rigidly fixed and space between then is very less of the order of few angestrem and hance they can not move.
• In liquids these atoms and molecules can more enabeling liquids to flow.
• In gases atoms are free to travell without colliding for large distances such that if gases were not enclosed in an enclosure they would disappear.

4. Dalton's law of partial pressures :

• Consider a mixture of non-interacting ideal gases with n₁ moles of gas 1,n₂ of gas 2 and so on
• Gases are enclosed in an encloser with volume V, temperature T and pressure P.
• Equation of state of mixture
       PV = (n₁+ n₂)RT
  or     P = n₁RT/V + n₂RT/V + - - - -
        = P₁ + P₂ + - - --
  where,
       P₁ = n₁RT/V
  is pressure the gas 1 would exert at same V and T if no other gases were present in the enclosure. This is know as law of partial pressure of the gases.
• The total pressure of mixture of ideal gases is sum of partial pressures of individual gases of which mixture is made of.

5. Kinetic Theory of an ideal gas

Following are the fundamental assumptions of kinetic theory of gases.

• Gas is composed of large number of tiny invisible particles know as molecules
• These molecules are always in state of motion with varying velocities in all possible directions.
• Molecules traverse straight line path between any two collisions
• Size of molecule is infinitely small compared to the average distance traverse by the molecules between any two consecutive collisions.
• The time of collision is negligible as compared with the time taken to traverse the path.
• Molecules exert force on each other except when they collide and all of their molecular energy is kinetic. 
• Intermolecular distance in gas is much larger than that of solids and liquids and the molecules of gas are free to move in entire space free to them.

6. Pressure of gas

• Consider a cubical vessel with perfectly elastic walls containing large number of molecules say N let l be the dimension of each side of the cubical vessel.
• v_1x , v_1y, v_1z be the x, y, and z componenet of a molecule with velocity v.
• Consider the motion of molecule in the direction perpandicular to the face of cubical vessel.
• Molecule strikes the face A with a velocity v_1x and rebounds with the same velocity in the backward direction as the collisions are perfectly elastic.
  
   
• If m is the mass of molecule, the change in momentum during collision is
       mv_1x - (-mv_1x) = 2 mv_1x          (1)
• The distance travelled parallel to x-axis an is between A to A´ and when molecule rebounds from A´and travel towards A is 2L
• Time taken by molecule to go to face A´ and then comeback to A is
            Δt = 2l/v_1x
• Number of impacts of this molecule with A in unit time is
       n = I/Δt = v_1x/ 2l               (2)
  Rate of change of momentum is
       ΔF = ΔP/Δt
       =nΔP
  from (1) and (2)
       ΔF = mv_1x² / l
  this is the force exerted on wall A due to this movecule.
• Force on wall A due to all other molecules
       F = Σmv_1x²/L           (3)
• As all directions are equivalent
  Σv_1x²=Σv_1y²=Σv_1z²
  Σv_1x²= 1/3Σ((v_1x)² + (v_1y)² +( v_1z)² )
       = 1/3 Σv₁²
  Thus     F = (m/3L) Σv₁²
• N is total no. of molecules in the container so
       F = (mN/3L) (Σ(v₁)²/N)
• Pressure is force per unit area so
       P = F/L²
        =(M/3L³)(Σ(v₁)²/N)
  where ,M is the total mass of the gas and if ρ is the density of gas then
       P=ρΣ(v₁)²/3N
  since Σ(v₁)²/N is the average of squared speeds and is written as v_mq² known as mean square speed
  Thus, v_rms=√(Σ(v₁)²/N) is known as roon mean squared speed rms-speed and v_mq² = (v_rms)²
• Pressure thus becomes
       P = (1/3)ρv_mq²                     (4)
  or     PV = (1/3) Nmv_mq²                     (5)
  from equation (4) rms speed is given as
       v_rms = √(3P/ρ)
                 = √(3PV/M)                    (6)

7. Kinetic interpretation of temperature

• From equation (5) we have
       PV = (1/3)Nmv_mq²
  where N is the number of molecules in the sample. Above equation can also be written as
       PV = (2/3)N(1/2)Nmv_mq²                (7)
• The quantity (1/2)Nmv_mq² in equation (7) is the kinetic energy of molecules in the gas. Since the internal energy of an ideal gas is purely kinetic we have,
       E=(1/2)Nmv_mq²                     (8)
• Combining equation 7 and 8 we get
       PV=(2/3)E
• Comparing this result with the ideal gas equation (equation (4) ) we get
       E=(3/2)K_BNT
  or,     E/N=(1/2)mv_mq² =(3/2)K_BT          (9)
  Where, K_B is known as Boltzmann constant and its value is K_B=1.38 X 10^-23 J/K
• From equation (11) we conclude that the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas and is independent of the pressure , volume and nature of the gas.
• Hence average KE per molecule is
       (1/2)mv²¯=(3/2)K_BT
  from this since v²¯=(v_rms)², rms velocity of a molecule is
       v_rms=√(3K_BT/m)                    (10)
  This can also be written as
       v_rms=√(3K_BNT/Nm)
        =√(3RT/M)                    (11)
  where, M=mN is the molecular mass of the gas.
  
  Assignment
  (1)Gas laws (Boyle's and charle's law) and perfect gas equation can be derived using kinetic theory of gases. try to derive them.

7. Law of Equipartition of energy

• According to the principle of equipartition of energy, each velocity component has, on the average, an associated kinetic energy (1/2)KT.
• The number of velocity components needs to describe the motion of a molecule completely is called the number of degrees of freedom.
• For a mono atomic gas there are three degrees of freedom and the average total KE per molecule for any monotomic gas is 3/2 K_BT.
       

8. Specific Heat Capacity

(i) Monoatomic gases :

• Monoatemic gas moleules has three translational degrees of freedom.
• From law of equipartition of energy average energy of an molecule at temperature T is (3/2)K_BT
• Total internal energy of one mole of such gas is
       U= (3/2)K_BTN
        = (3/2) RT                    (12)
• If C_V is melar specific heat at constant volume then
       C_Vv = dU/dT
        = (3/2)R                         (13)
  now for an ideal gas
       C_P - C_V = R
  C_P - molar specific heat capacity at constant presseve
       C_P = 5/2 R                    (14)
  Thus for a monoatomic gas ratio of specific heats is
       γ_mono = C_P/C_V= 5/3               (15)

(ii) Diatomic gases :

• A diatomic gas molecule is treated as a rigid rotator like dumb-bell and has 5 degrees of freedom out of which three degrees of freedom are translatoinal and two degrees of freedom are rotational.
• Using law of equipartition of energy the total internal energy of one mole of diatomic gas is
       U= (5/2)K_BTN
       = (5/2) RT                    (16)
• Specific heats are thus
       C_V =(5/2)R
       γ_dia= 5/7          (rigid rotater)
• If diatomic molecule is not only rigid but also has an vibrational mode in addition, then
       U = (7/2) RT
       and C_V=(7/2)R
       C_P=(9/2)R
  and γ=C_P/C_V=9/7

9. Specific heat Capacity of Solids

• From law of equipartation of energy we can can also determine specific heats of solids.
• Consider that atoms in a solid are vibrating about their mean position at some temperature T.
• Oscillation in one dimension has average energy equals 2(1/2)K_BT=K_BT, as (1/2)K_BT is PE and (1/2)K_BT is KE of the atom.
• In three dimensions average kinetic energy is 3K_BT.
• For one mole of solid total energy is
       U= 3NK_BT
        = 3RT
• At constant pressure ΔQ =ΔU+PΔV=ΔU since for solids ΔV is negligible hence
       C=ΔQ/ΔT=ΔU/ΔT=3R
• This is Dulang and Petit law.
• Here we note that predictions of specific heats of solids on the basis of law of equipetation of energy are independent of temperature.
• As we go towards low temperatures T→0 there is a pronounced departure from the value of specific heat of solids as calculated.
• It is seen that specific heats of substance aproaches to zero as T→0.
• This result can further be explained using the principles of quantum mechanics which is beyond our scope.

10. Mean free Path

• On the basis of kinetic theory of gases, it is assumed that the molecules of a gas are continously collilding against each other.
• Molecules move in straight line with constant speeds between two successive collisions.
• Thus path of a single molecule is a series of zig-zag paths of different lengths as shown in fig -.
  
   
• These paths of different lengths are called free paths of the molecule
• Mean Free Path is the averege distance traversed by molecule between two successive collisions.
• If s is the Total path travelled in N_coll coilisions, then mean free path
       λ= s/N_coll
  Expression for mean free path :
• Consider a gas containing n molecules per unit volume.
• We assume that only one molecule which is under consideration is in motion while all others are at rest.
• If σ is the diameter of each molecule then the moving molecule will collide with all these molecules where centers lie within a distance from its centre as shown in fig
  
  < 
• If v is the velocity of the moving molecule then in one second it will collide with all moleculeswith in a distance σ between the centres.
• In one second it sweeps a volume πσ²v where any other molecule will collide with it.
• If n is the total number of molecules per unit volume, then nπσ²v is number of collisions a molecule suffers in one second.
• If v is the distance traversed by molecule in one second then mean free path is given by
       λ = total distance traversed in one second /no. of collision suffered by the molecules
        =v/πσ²vn
        =1/πσ²n
• This expression was derived with the assumption that all the molecules are at rest except the one which is colliding with the others.
• However this assumption does not represent actual state of affiar.
• More exact statement can be derived considering that all molecules are moving with all possible velocities in all possible directions.
• More exact relation found using distribution law of molecular speeds is
       λ=1/(√2)πσ²n
  its derivation is beyond our scope.

Solved examples

Question-1.Let A & B are two sample of ideal gases of equal mole .let T be the temperature of both the gas Let E_A and E_Bare there total energy respectively .Let M_A and M_B are these respective molecular mass .which of these is true
a,E_A > E_B
b,E_A < E_B
c,E_A =E_B
d,none of these

Solution:1
E_A = 3/2 nRT
E_B = 3/2 nRT
;E_A =E_B


Question-2. The velocities of the molecules are v, 2v, 3v, 4v & 5v. The rms speed will be
a,11v
b,v(11)^1/2
c, v
d, 3.3v

Solution:2
Vrms= (∑ V² / N)^1/2
= [(V² + 4V² + 9V² + 16V² + 25V²)/5]^1/2
=v(11)^1/2

Question-3:
An ideal gas A is there.Intial temperature is 27°C.The temperature of the gas is increased to 927°C.Find the ratio of final V_rmsto the initial V_rms 

Solution-9:

V_rms=√3RT/M

So it is proportional to Temperature
Now 
T₁=27°C =300K
T₂=927°C =1200K

So intial V_rms=k√300
Final V_rms=k√1200

ratio of final to intial=2:1


Question-.4 Two absolute scales A and B have triple points of water defined as 200A and 350A. what is the relation between T_A and T_B



Solution-4

Given that on absolute scale
Triple point of water on scale A = 200 A
Triple point of water on scale B = 350 B
Also, triple point of water on Kelvin scale = 273.16 K
Now temperature on scale A and on scale B is equivalent to 273.16 K on absolute temperature scale.
Thus, value of one degree on absolute scale A = (273.16/200) K
Or, 
Value of temperature T_A on absolute scale A = (273.16XT_A)/200 
Similarly value of temperature T_B on absolute scale B = (273.16XT_B)/350
Since T_A and T_B represent the same temperature
273.16×T_A/200 = 273.16×T_B/350
Or, T_A = 200T_B/350 = 4T_B/ 7 

Question-.5 . which one is not the assumption in kinetic theory of gases
a.the molecules of the gas are in continual random motion
b. The molecules interact during the collison
c. The molecules are tiny hard sphere undergoing inelastic collision
d. The collison are of short duration
Solution-5

Answer is c

Question-.6. 
There are two statement
A. Equal volumes of all gases at the same temperature T and pressure P contain an equal no of Molecules
B.the no of molecules in one mole of any gas is 6.0255 * 10²².
which one of the following is correct
a. A and B both
b. A only 
c B only
d. A and B both are incorrect
Solution-6
Answer is b

Question-.7. .There are two statement about Ideal gases
A. The V_rms of gas molecules depends on the mass of the gas molecule and the temperature
B. The V_rms is same for all the gases at the same temperature
which one of the following is correct
a. A and B both
b. A only 
c B only
d. A and B both are incorrect
Solution-7

Answer is b

Question-.8. .Choose the correct statement of the following
a. The pressure of the gas is equal to the total kinetic energy of the molecules in a unit volume of the gas
b. The product of pressure and volume of the gas is always constant
c. The average kinetic energy of molecule of the gas is proportional to its absolute temperature
d. The average kinetic energy of molecule of the gas is proportional to the square root of its absolute temperature
Solution-8

Answer is c

Question-.9...which one is true according Vander Waal gas equation
a. The attractive forces between the molecules is not negligible
b Volume of the molecules is negligible as compared to the volume occupied by the gas
c. Volume of the molecules is not negligible as compared to the volume occupied by the gas
d. The attractive forces between the molecules is negligible
Solution-9
Answer is a,c