1. Basic Concept of Thermodynamics

aChapter One Thermodynamic Fundamental Concepts“The branch of science that deals with energy levels and the transfer of energy between systems and between different states of matter” is known as thermodynamics.Some Key terms• System:-Anything which are kept in consideration is called -System. • Surrounding:-Everything external to the system is known as Surrounding. • Boundary:-The thermodynamic system and surroundings are separated by an envelope called Boundary of a system. • Universe:-Thermodynamic system + surrounding = Universe

• State - Refers to the energy content of a given system. The state is defined by specifying certain variables such as temperature, pressure, volume and composition.• State Variables specifically refer to the change inherent if a reaction proceeds because of a change in state State variables are either extensive or intrinsic Extensive → variables which are proportional to the quantity of matter (such as volume) Intrinsic → variables which are independent of quantity, that instead describe the whole system (such as density, temperature, and concentration) Thermodynamic System (Process)Isothermal ProcessIso => constant, thermal=> temperatureA thermodynamic process in which pressure and volume of system of a gas change but temperature remains constant is called isothermal process. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. At a particular constant temperature, the change of a substance, object or system is known as Isothermal Process. Usually, there are two phenomena under which this process can take place. If a system is in contact with a thermal reservoir from outside, then, to maintain thermal equilibrium, the system slowly adjusts itself with the temperature of the reservoir through heat exchange. Example of isothermal process1. The melting and boiling process of substance at constant temperature.2. Slow compression of air in the tyre using pumpEquation of state in Isothermal processFrom Ideal gas equation, we have PV = nRT ……………….(i)Here For isotherma process , T = constantthe equation (i) becomes PV = Constant

Work-done during an Isothermal Process Work, is done by a gas when it is allowed to expand isothermally. Let the initial and final volumes of n moles of an ideal gas be V1 and V2 respectively. Then the work done for a very small change in volume dV :dW=P⋅dVWhen the gas expands from volume V1 to V2 at constant temperature T, then work done;W=v2∫V1PdVSince, expansion is isothermal (i.e., T= Constant), then from ideal gas equation:a

PV=nRT or, P=nRT V

∴ W=nRTV2∫V1dV V

=nRTlogeV2 V1… (i) [∵T=Constant]

Also, from isothermal equation, we have: P1V1=P2V2or, V2

V1=P1

P2… (ii)Then, from equation (i) and (ii), we getW=nRTloge(V2

V1)=nRTloge(P1

P2)This is the required expression for work done in isothemal proces. Adiabatic Process A thermodynamic process in which pressure, volume and temperature of the system of gas changes, but there in no change of heat between the system and its surrounding (i.e., dQ=0 ) is called adiabatic process. So, such process occurs suddenly and quickly in the system of gas. Two essential conditions for a perfect adiabatic process are:(i) The walls of the cylindrical vessel should be insulating/nonconducting.ii) The process of compression or expansion should be sudken and quick. Some examples of adiabatic process are:(i) brusting of a air tube of vehicle(ii) propagation of sound waves in air and other gases(iii) expansion of steam in the cylinder of steam engine, etc. The graphical plot variation of P and V for adiabatic process is shown in figure.

Here, the locus AB of two isotherms shows there is change id temperature without external supply of heat and represents the adiabatic expansion of gas. Adiabatic equation of a perfect gasLer's consider 'n' moles of ideal gas inside a cylinder with nonconducting wall and frictionless piston. Consider the gas have pressure P, volume V and temperature T at its initial state. Now considering gas which undergoes expansion so that its pressure, volume and temperature change dP,dV and dT. From first law thermodynamics, dQ=dU+PdV… (i) . In adiabatic prucess, there is no exchange of heat. So, dQ=0 Thus the equation (i) becoms 0=dU+PdVie, dU+PdV=0 ...........(ii)Also, if Cv is the molar specific heat capacity of the gas at constant volume, then heat required to raise temperature through dT for n moles of gas is nCndT. ∴ dU=nC.dT Using the value of dU from the equation (iii) in equation (ii), we gu nCvdT+PdV=0 … (iv)For 'n' moles of ideal gas, the perfect gas equation is PV= nRT Differentiating both side with respect to T, we get d(PV)

dT=d(nRT)

dTor, PdV

dT+VdP

dT=nRor, PdV+VdP

dT=nRor, PdV+VdP=nRdT∴ dT=PdV+VdP

nR... (v)From equation (iv) and ( v ) we get, nCV(PdV+VdP)

nR+PdV=0or, CV

RPdV+CVVdP

R+PdV=0or, PdV(CV

R+1)+CVVdP

R=0or, PdV(CV+R

R)+CVVdP

R=0[a

∵Cp-Cv=R

⇒Cp=Cv+R

]or, PdVCp+CVVdP=0Dividing this equation by PVCV, we get:PdVCP

PVCV+CVVdP

PVCv=0or. (CP

CV)dV

V+dP

P=0or, 𝛾dV

V+dP

P=0 [∵CP

CV=𝛾]Now, integrating this equation, we get𝛾∫dV

V+∫dP

P=const.or, 𝛾(lnV+lnP= constantor, lnV𝛾+lnP= constantor. ln(V𝛾P)= constant [∵lnx+lny=ln(xy) & alnx=lnxa]or. V𝛾P=econstanti.e. PV𝛾= constant .............................(vi) This is the adiabatic equation in terms of pressure and volume. If A(P1,V1,T1) and B(P2,V2,T2) be the initial and final state of gas in adiabatic equation, then we can write: P1VY1=P2VY2Case I: Adiabatic equation in terms of temperature and volume We have, from ideal gas equation,PV=nRT or, P=nRT

VSubstituting this value of 'P' in equation (vi), we get:nRT

VV𝛾= Constant or, TVY-1= ConstantIn general, T1VY-11=T2VY-12Case II: Adiabatic equation in terms of temperature and pressure We have, from ideal gas equation.PV=nRT or, V=nRT

PSubstituting this value of ' V ' in equation (vi), we get:a

P(nRT P)𝛾=Constant

or, P1-YTY(nR)𝛾=Constant

∴ P1-YTY=constantIn general.

P11-YTY1=P1-Y2T2𝛾

Isobaric Process(Pressure Constant)In thermodynamics, an isobaric process is a type of thermodynamic process in which volume and temperature change but the pressure of the system stays constant: ΔP = 0. e.g. heating / boiling of water at atmospheric pressure.In this process, working substance (i.e., gas) is taken within an panding chamber. As a result, gas either expands or contracts maintain a constant pressure.

If dQ is the amount of heat given to the system, then it is partly used in increasingthe temperature of the system by 'dT', and it is also partly used in doing external work 'dW' Now, from first law of thermodynamics, dQ=dU+dW or,

dQ=nCvdT+PdV

Further, for isobaric process: P is constant, then the gas expands from volume V1 to V2 so work done in the process becomes; dW=P[V2-V1] [∵PV=nRT] or, dW=nR(T2-T1) ∴ dQ = dU + nR(T2-T1)Isochoric Process:(Constant Volume)Any thermodynamic process, in which pressure and temperature of the system of gas changes, but its volume remains constant is called isochoric process. e.g. Gas heated in a tightly sealed container. In this process, the working substance is kept in a nonexpanding chamber. So, there is no change in volume and hence work-done in such process is zero. Now, from first law of thermodynamics, we have:dQ=dU+PdVor, dQ=dU (i)[∵dW=PdV=0 as dV=0 : for isochoric process]Thus the total amount of heat supplied to the system is used to increase the internal energy of the system in this process.

Thus , we have: dQ = nCvdT ..................... (ii)Where, Cv is molar heat capacity of a gas at constant volume. # Show that for an isobaric process, change in enthalpy is equal to the heat given to the system. → Enthalpy is the total heat content of the system mathematically, it is defined as, H=U+pVBy differentiation, dH=dU+d(pV)=(TdS-pdV)+pdU+VdpFor isobaric process, dp=0 So at constant persure (dp=0),dH=TdS=dQwhere dQ is the quantity of heat given from external source. Hence for an isobanic process, change of enthalpy is equal to the heat given of the system.# Explain quasi-static process in a thermo dynamical system. → The process in which there is small change or deviation in thermodynamic equilibrium is known as quasi-static. This process is only possible at slow rate. For e.g. It the piston is pushed very rapidly, the gas requires. k. e. and cannot establish thermodynamic equilibrium. But when the piston is pushed slowly, the system at all remains in thermodynamic equilibrium and process can be considered quasi-static. This process is an idealized concept and the conditions for it can never be rigorously satisfied in practice.Numericals# One gram of water becomes 1671 ce of steam when boiled at a pressure of one atm. The heat of vaporization at this temperature is 540cal/gm. Compute the external work and the increase in internal energy. Solution:a

Here, V1=1cc (∵𝜌w=lgm/cc and m=1gm)

V2=1671cc

∴ dV=(1671-1)cc=1670×10-6m3

P=1atm=1.01×105N/m2

Now, from first law of thermodynamicsdQ=dU+dWa

Here, dW	=PdV=1.01×105×1670×10-6=169.17J=169.17 4.2Cal
	=40.47Cal(:1cal=4.2~J)

and dQ=mL=lgm×540cal/gm=540cal.∴ dU=dQ-dWi.e., dU=499.53Cal# At normal temperature (0∘C) and normal pressure (1.013 105N/m2 ), when 1gm of water freezes, its volume increases by 0.091cm3. Calculate the change in internal energy. Solution:Amount of heat subtract from system isa

dQ	=mL=1gm×80cal/gm=80cal
	=80×4.2J=336J(∵1cal,=4.2J)

Again, dW=PdV=1.01×105×0.019×105 =0.0093J (∵dV=0.091cm3=0.091×106m3)from first law of thermodynamics, Now.a

dQ=dW+dU

dU=dQ-dW=(-336-0.0093)J

∴dU=-336.0093J Here, Q is taken - ve since heat is subtracted. The -ve sign shows that decrease in internal energy.# Calculate the work done when a gram molecule of a gas expands isothermally at 27∘C to double its original volume.Solution:We have, work done during an isothermal process:a

W	=nRTlogc(V2 V1)=2.3026nRTlog10(V2 V1)
or, W	=2.3026×8.3×300log102

Where, n=1: one gram moleculeT=27∘C=300Ka

Let V1=V , then V2=2V

∴ W=1725.8 Joule

# An ideal gas of volume 1 litre at pressure 8 atmosphere undergoes an adiabatic expansion until pressure drops to 1 atmosphere. Calculate the final volume and the work done during the expansion given 𝛾=1.4. Solution:Here, Initial volume of an ideal gas (V1)=1 litre =10-3m3 Initial pressure (P1)=8 atm=8×1.01×105N/m2Final pressure (P2)=1atm=1.0×105N/m2Workdone during the expansion (W)= ?We know,Now, work-done in adiabatic process =1

𝛾-1(P1V1-P2V2)=1

(1.4-1)(8×1.01×105×10-3-1.01×10-5×4.42×10-3)=1

0.4(8.08×102-4.46×102)=2.5×3.36×102=905JTherefore the work done in adiabatic expansion is 905J.# Air contained in a cylinder fitted with piston is compressed reversibly according to the equation pv1.25= const. The mass of air in the cylinder is 0.1kg. The initial pressure is 100kPa and the initial temperature 20∘C. The final volume is 1

8 of the initial volume. Determine the work. Sol": M=0.1kg=100gm,V1=100cc,P1=100kpa=100×103paT1=20∘c=(273+20)k=293kV2=1

8 of the initial volume =1

8×100=12.5ccWe have, T1 v𝛾-11=T2v2𝛾-1or, T2=T1(v1

v2)𝛾-1=293(100

12.5)14-1=293(8)0.4=673kwe know that work done is w=R

1-𝛾(T2-T1)=8.3

1-1.4(673-293)=-7885 JouleThe work done (w)=-7885 Joule