Thermodynamics

 Thermodynamics system

$\text{Thermodynamics system:}$$\to$It is an assembly of an extremely large number of particles (atoms or molecules)$\\$having a certain pressure, volume and temperature.$\\$$\to$It is of two types:$\\$1)Isolated System: (no heat can enter or leave out the system)$\\$2)Closed system: (There can be heat exchange between the system and surrounding)$\\$

$\textbf{Thermal Equilibrium:}\\$A thermodynamics system is in thermal equilibrium if the temperature of all parts of it is same.

Zeroth law of Thermodynamics

$\text{Law:}$When two bodies A and B are in thermal equilibrium with third body C then the body A and B$\\$also must be in thermal equilibrium with each other.

Work Done

Points:$\\$

$\to$Work is said done when volume of gas changes.$\\$

$\to$Work done is positive if expansion takes place and is negative if compression takes place.$\\$

Work (W) = PdV$\\$

$\to$Area under PV curve between volume axis is equal to work done.$\\$

$\to$For a closed cycle, area of closed loop gives work done.$\\$

$\to$When P remains constant throughout the expansion, the work done by the gas is$\\$

W=P(V2-V1)$\\$

$\textbf{Internal Energy of a Gas}\\$The sum of energy due to molecular motion (KE) and due to molecular configuration (PE)$\\$is called internal energy of gas.$\\$:. Internal energy (U) = PE + KE$\\$For ideal gas intermolecular force of attraction is neglected so PE=0,$\\$so internal energy of ideal gas is KE which is only the function of temperature.

First law of thermodynamics

$\text{Intro:}$When heat energy is given to a system then some part of heat energy supplied is used to$\\$change the internal energy of system and rest of energy is used to do external work.$\\$∆Q=∆U+∆W$\\$

$\textbf{Note:}\\$For cyclic process, the change in internal energy of the system is zero because the system is$\\$brought back to the initial condition. Therefore, dU=0 and from the first law of thermodynamics,$\\$dQ= du + PdV =0+dW= dW$\\$

$\textbf{Molar Heat Capacities:}\\$1)Molar heat capacity at constant pressure ($C_p$)$\\$$\to$Heat required to rise the temperature of one mole of gas through 1 degree C at constant pressure.$\\$Its unit is J/(molK)$\\$$\to$Heat required (dQ)=n$C_p$dT$\\$2)Molar heat capacity at constant volume($C_v$):$\\$$\to$Heat required to rise the temperature of mole of gas through 1 degree C at constant volume.$\\$Its unit is J/(molK)$\\$$\to$Heat required (dU) = n$C_v$dT$\\$

$\textbf{Mayer`s Formula:}\\$$\to$ $C_p-C_v$=R

Specific heat capacities

$\text{Specific heat capacity at constant pressure :}$($c_p$)$\\$$\to$Heat required to rise the temperature of unit mass of gas through 1 degree C temperature$\\$at constant pressure. $\to$Heat required (dQ)=m$c_p$dT$\\$$\to C_p$=M$c_p \\$

$\textbf{Specific heat capacity at constant volume :}$($c_v$)$\\$$\to$Heat required to rise the temperature of unit mass gas through 1 degree C temperature$\\$at constant volume. $\to$Its unit is J/(kg K)$\\$$\to$Heat required (du)=m$c_v$dT$\\$$\to C_v$=M$c_v \\$

$\textbf{Note:}\\$$\to$Heat required to rise certain temperature at constant pressure is always greater than heat$\\$required to rise same temperature at constant volume. So gas has two types of heat capacities$\\$$\to$ i.e. $C_p$>$C_v \\$$\to$Because in constant pressure, internal energy and work done both is done.

Thermodynamical process

$\text{Isochoric Process:}$$\to$Volume remains constant$\\$$\to$Work done (dw)=0$\\$$\to$ Heat supplied = change in internal energy: dQ=dU$\\$$\to$nCvdT=dU$\\$

$\textbf{ Isobaric Process:}\\$$\to$Pressure remains constant$\\$$\to$dQ=CvdT+ PdV$\\$

$\textbf{Isothermal Process:}\\$$\to$Temperature remains constant. i.e. dT=0$\\$$\to$For this process cylinder with conducting wall is used and ideal gas filled inside is allowed to$\\$expand or is compressed very slowly.$\\$$\to$Eg: Melting process and boiling process$\\$$\to$Specific heat capacity during isothermal process is infinity$\\$$\to$Change in internal energy(du)= 0$\\$$\to$Gas Equation: $P_1 V_1$=$P_2 V_2 \\$$\to$Slope of curve (dP/dV)=-P/V$\\$$\to$Work Done (w) =nRT ln($V_2$/$V_1$)$\\$=$P_1 V_1$ ln($V_2$/$V_1$)=$P_1 V_1$ ln($P_1$/$P_2$)$\\$

$\textbf{Adiabatic Process: }\\$$\to$The process in which exchange of heat energy is zero i.e. dQ=0$\\$$\to$So, dW=-dU i.e. work is done by gas on the expense of internal energy so cooling is observed after$\\$adiabatic expansion$\\$$\to$Fast process in which wall of cylinder is perfectly insulator$\\$$\to$Specific Heat capacity of gas is 0.$\\$$\to$Eg: Propagation of sound wave, sudden bursting of tire, the compression stroke in an internal$\\$combustion engine.$\\$$\to$Slope of curve (dP/dV)=-γP/V$\\$$\to$Gas equation is$\\$$P_1 V_1^γ$=$P_2 V_2^γ \\$$T_1 V_1^{γ-1}$=$T_2 V_2^{γ-1} \\$$T_1^γ$/$P_1^{γ-1}$ =$T_2^γ$/$P_2^{γ-1}$$\\$$T_1^γ$/$T_2^γ$ =$P_1^{γ-1}$/$P_2^{γ-1}$$\\$

$\to$Work done (w) =nR[$T_1$-$T_2$]/(γ-1)=($P_1 V_1$-$P_2 V_2$)/(γ-1)$\\$$\to$For heat capacity: dQ=m$C_p$dT$\\$:.C=dQ/dt$\\$$\to$Heat capacity = 0

Different Process

Cyclic Process:

$\to$Complete a closed cycle$\\$$\to$Change in internal energy is zero$\\$$\to$Work done (w) = Area of closed loop in PV diagram$\\$

$\textbf{Reversible Process:}\\$$\to$An infinitesimally slow expansion and compression of an ideal gas at a constant pressure$\\$$\to$At mechanical processes take place under the action of conservative force. No loss of energy due to$\\$conduction or radiation during the cycle of operation.$\\$$\to$All thermal processes taking place at infinitesimally slow rate.$\\$

$\textbf{Irreversible Process:}\\$$\to$Cannot retrace in the opposite order by reversing the controlling factors$\\$$\to$Eg: Rusting of iron, dissolve of soap in water, decay of matter, flow of current through a conductor, etc.

Limitation of 1st law of thermodynamics

$\text{Points}$

1)Does not indicate the direction of heat transfer.$\\$2)Does not indicate as to why heat energy developed in the target cannot be converted back into mechanical$\\$energy of the bullet enabling it to fly back.$\\$3)Does not give to what extent the mechanical energy in obtained from the heat energy.$\\$4)1st law is silent about the efficiency of the heat engine.$\\$

$\textbf{Important notes:}\\$1)Work done:$\\$$\to$During expansion:$\\$W(isobaric) > W(isothermal) > W(adiabatic)$\\$

$\to$During compression:$\\$W (adiabatic) > W(isothermal) > W (isobaric)$\\$2)Pressure is constant during state change.$\\$3)Generally if no any information of gas is given we have to use γ=1.4 i.e. diatomic.$\\$4)The expansion of gas against zero external pressure is known as free expansion.$\\$5)The gas has the greatest internal energy whereas the solid has the least.$\\$6)Specific heat of saturated vapor pressure is negative.$\\$7)Work done during isothermal expansion is more than work done during adiabatic expansion$\\$if the initial and final volumes are same.$\\$8)Adiabatic curve is steeper than that of isothermal curve.$\\$9)In a cyclic process, the internal energy of a gas remains constant.$\\$10)In adiabatic process, the temperature of an isolated system changes$\\$11)When air of atmosphere rises up, it cools. Why? (Expansion occurs in the air, it becomes cool.)$\\$12)Internal energy of an ideal gas is wholly kinetic in nature and function of temperature.$\\$13)Two isothermal curves cannot intersect each other.$\\$14)When the cold air blowing across the mountain tops descends into the valley, it is adiabatically compressed.$\\$Consequently, the temperature in the valley is increased,

Second Law of Thermodynamic

$\text{Basics:}$1)Kelvin Planck statement:$\\$$\to$ It is impossible for engine to convert all the heat energy into work without rejecting some$\\$energy to sink i.e. no engine will have 100% efficiency.$\\$$\to$Presence of sink is essential for continuous conversion of heat into work.$\\$2)Clausius statement:$\\$It is impossible to absorb heat energy from cold body and reject to hot body without doing work$\\$on it i.e. self acting refrigerator is impossible.

Different Engines

$\text{Heat engine:}$$\to$ Any device which converts heat energy continuously into mechanical work$\\$

$\to$ Its main parts are:$\\$

i)Source: A hot body at a constant high temperature ($T_1$) from which the heat engine can draw heat (Q1).$\\$

ii)Sink: A cold body at a constant low temperature ($T_2$) to which any amount of heat can be rejected.$\\$

iii)Working Substance: Working substance is an ideal gas which on being supplied with heat performs$\\$mechanical work.$\\$

$\textbf{Efficiency of heat engine:}\\$$\to$ External work obtained to the heat energy absorbed by the working substance from the source.$\\$$\to$ Denoted by ɳ$\\$$\to$ ɳ=$\frac{Work done}{input}$×100%$\\$$\to$ ɳ=$\frac{W}{Q_1}$×100%$\\$$\to$ ɳ=(1-$\frac{Q_1}{Q_2}$)×100%$\\$$\to$ɳ=(1-$\frac{T_2}{T_1}$)×100%$\\$$\to$ When heat engines are placed in series then sink of 1st engine act as source for 2nd engine and so on.$\\$$\to$ Efficiency of heat engine always less than 1 or 100%$\\$

$\textbf{Carnot`s engine:}\\$$\to$ An ideal cycle of operation for a heat engine.$\\$$\to$ Essential parts:$\\$i)Source of heat: Hot body of infinite thermal capacity and maintained at a fixed temperature, from which the$\\$working substance draws heat without changing its temperature.$\\$ii)Cylinder: Cylinder is fitted with a perfectly non-conducting and frictionless piston enclosing an ideal gas.$\\$Its bottom is a perfect heat conductor whereas the walls are perfect heat insulator.$\\$iii)Sink of heat: Should be in fixed lower temperature and has the infinite thermal capacity.$\\$iv)Working substance: An ideal gas.$\\$

$\to$ A cycle of four operation consisting of two isothermal processes and two adiabatic process makes a complete Carnot`s cycle.$\\$$\to$ In adiabatic compression if ρ=$V_1$/$V_2 \\$ɳ=1-$\frac{1}{ρ})^{γ-1}$×100%

$\textbf{Note: }\\$$\to$Carnot engine is perfectly reversible because:$\\$i)NO friction between the cylinder and the piston.$\\$ii)The operations on the gas should be performed very slowly.$\\$iii)Lose of heat due to conduction is prevented.$\\$

$\to$Efficiency depends on the temperature of the source $T_1$ and that of the sink $T_2$$\\$but does not depends upon the nature of the working substance.$\\$$\to$As $T_2$ is always less than $T_1$, so efficiency of a heat engine is always less than one or$\\$efficiency cannot be 100%.$\\$

$\textbf{Refrigerator:}\\$$\to$Operates in a manner opposite to that of a heat engine.$\\$

$\to$Coefficient of Performance (β): Ratio of the amount of heat absorbed from the cold body to the$\\$work done is running the machine.$\\$

β=$\frac{T_2}{T_1-T_2} \\$

$\to Q_2$ heat is absorbed from cold reservoir on doing work ‘W’ on it then $Q_1$=$Q_2$+W heat is$\\$rejected to hot reservoir.

Entropy

$\text{Points:}$$\to$The measure of disorder of molecule of system is called entropy.$\\$$\to$On increasing the disorder, the entropy of system increases$\\$∴ ∆S= (Heat absorbed)/(Absolute temperature)$\\$∆Q=T∆S$\\$

$\to$The change in entropy during change in state of matter ∆S=±$\frac{ML}{T} \\$$\to$Where +ve sign indicates heat absorbed and –ve sign indicates heat evolved.$\\$$\to$When temperature of body changes from $T_i$ to $T_f$ then$\\$The change in entropy (∆S) =ms ln($T_f$/$T_i$).$\\$

$\textbf{Kelvin Temperature Scale:}\\$$\to$The ratio of any two temperatures on this scale is equal to the ratio of heats absorbed and rejected$\\$by a Carnot reversible engine working between these two temperatures.$\\$$\frac{Q_1}{Q_2}$=$\frac{T_1}{T_2} \\$$\to T_1$and $T_2$temperatures are measured in terms of work and hence this scale is also known as$\\$work scale of temperature.$\\$

$\textbf{Notes:}\\$

1. Efficiency of Carnot`s engine is independent of the nature of the working substance.$\\$
2. Efficiency of Carnot heat engine is 100% only if the temperature of the sink is zero Kelvin which is not possible.$\\$
3. The efficiency of heat engine is more in hilly areas than in the planes.$\\$
4. Atmosphere is the sink in a steam engine.$\\$
5. As the refrigerator works its coefficient of performance goes on decreasing.$\\$
6. It is possible to convert work into the energy completely but the reverse is not possible.$\\$
7. Total change is entropy of the whole system in a Carnot`s cycle is zero.$\\$
8. Energy while burning petrol or diesel engine is input power.$\\$
9. A diesel engine has higher efficiency than that petrol engine.