Heat Study Guide

HEAT Page # 1 HEAT 1. HEAT The energy that is being transferred between two bodies or between adjacent parts of a body as a result of temperature difference is called heat. Thus, heat is a form of energy. It is energy in transit whenever temperature differences exist. Once it is transferred, it becomes the internal energy of receiving body. If should be clearly understood that the word “heat” is meaningful only as long as the energy is being transferred. Thus, expressions like “heat in a body” or “heat of body” are meaningless. T1>T2 A T2 Heat B When we say that a body is heated it means that its molcules begin to move with greater kinetic energy. S.I. unit of heat energy is joule (J). Another common unit of heat energy is calorie (cal). 1.1 Mechanical Equivalent of Heat In early days heat was not recongnised as a form of energy. Heat was supposed to be something needed to raise the temperature of a body or to change its phase. Calorie was defined as the unit of heat. A number of experimets were performed to show that the temperature may also be increased by doing mechanical work on the system. These experiments established that heat is equivalent to mechanical energy and measured how much mechanical energy is equivalent to a calorie. If mechanical work W produces the same temperature change as heat H, we write, W = JH Where J is called mechanical equivalent of heat. J is expressed in joule/calories. The value of J gives how many joules of mechanical work is needed to raise the temperature of 1 g of water by 1°C. 1 calorie : The amount of heat needed to increase the temperature of 1 gm of water from 14.5 to 15.5 °C at one atmospheric pressure is 1 calorie. 1 calorie = 4.186 Joule 1.2 Specific Heat Specific heat of substances is equal to heat gain or released by that substance to raise or fall its temperature by 1°C for a unit mass of substance. When a body is heated, it gains heat. On the other hand, heat is lost when the body is cooled. The gain or loss of heat is directly proportional to : (a) the mass of the body Q  m (b) rise or fall of temperature of the body Q  T Q  m T or Q  m s T or dQ  m s d T or Q=m  s dT where s is a constant and is known as the specific heat of the body s = Q . S.I. unit of s is m T joule/kg-kelvin and C.G.S unit is cal/gm °C Specific heat of water : s = 4200 J/kg°C = 1000 cal/kg°C = 1 Kcal/kg°C = 1 cal/gm°C Specific heat of steam = half of specific heat of water = specific heat of ice Important Points : Q , if the substance undergoes the change of state which occurs at constant mT temperature (T = 0), the s = Q/0 = . Thus the specific heat of a substance when it melts or boils at constant temprature is infinite. (a) We know, s = Q = 0. mT Thus when liquid in the thermos flask is shaken, its temperature increases without the tranfer of heat and hence and the specific heat of liquid in the thermos flask is zero. (b) If the temperature of the substance changes without the transfer of heat (Q = 0) then s = HEAT Page # 2 specific heat (cal g–1C–1) (c) To raise the temperature of saturated water vapour, heat (Q) is withdrawn. Hence, specific heat of saturated water vapour is negative. (This is for your information only and not in the course) (d) The slight variation of specific heat of water with temperature is shown in the graph at 1 atmosphere pressure. Its variation is less than 1% over the interval form 0 to 100°C. 1.008 1.000 0 15 35 100 Temp(°C) 1.3 Heat capacity or Thermal capacity : Heat capacity of a body is defined as the amount of heat required to rasie the temperature of that body by 1°C. If ‘m’ is the mass and ‘s’ the specific heat of the body, then Heat capaicty = m s. Units of heat capacity in : CGS system is, cal °C–1 ; SI unit is, JK–1 1.4 Relation between Specific heat and Water equivalent : It is the amount of water which requires the same amount of heat for the same temperature rise as that of the object ms ms T = mw Sw T  mw = s w In calorie sw = 1  mw = ms mw is also represented by W so W = ms 2. LAW OF MIXTURE : When two substances at different temperatures are mixed together, the exchange of heat continues to take place till their temperatues become equal. This tempeature is then called final temperature of mixtue. Here, Heat taken by one substance = Heat given by another substance m1s1(T1 – Tm) = m2s2(Tm – T2) m2, s2, T2 m1, s1, T1 (T1>T2) Mixture Temperature = Tm 3. PHASE CHANGE : Heat required for the change of phase or state, Q = mL, L = latent heat. (a) Latent heat (L) : The heat supplied to a substance which changes its state at constant temperature is called latent heat of the body. (b) Latent heat of Fusion (Lf) : The heat supplied to a substance which changes it from solid to liquid state at its melting point and 1 atm. pressure is called latent heat of fusion. (c) Latent heat of vaporisation (Lv) : The heat supplied to a substance which changes it from liquid to vapour state at its boiling point and 1 atm. pressure is called latent heat of vaporization. If in question latent heat of water are not mentioned and to solve the problem it require to assume that we should consider following values. HEAT Page # 3 Latent heat of ice : L = 80 cal/gm = 80 Kcal/kg = 4200 × 80 J/kg Latent heat of steam : L = 540 cal/gm = 540 Kcal/kg = 4200 × 540 J/kg The given figure, represents the change of state by different lines T m ea St Tb ice Tm O er at ice+water w l1 water+Steam l2 Q Tm = Melting Temperature Tb = Boiling Temperature > l2 > l1 HEAT TRANSFER 4. INTRODUCTION Heat is energy in transit which flows due to temperature difference; from a body at higher temperature to a body at lower temperature. This transfer of heat from one body to the other takes place through three routes. (i) Conduction (ii) Convection (iii) Radiation (a) CONDUCTION (i) Requires Medium (ii) Energy is transmitted from one particle to another particle without displaced of particle (iii) No transfer of particle (b) CONVECTION (i) Requires Medium (ii) Enegy is transfer through movement of the particle of medium. (c) RADIATION (i) Does not requires any medium (ii) Enegy is transfer through Electromagnetic waves. Vaccum Radiation T1 Air T2 T1 Conduction 5. Convection + Radiation T2 T1 Air Convection + Radiation T2 Conduction CONDUCTION Figure shows a rod whose ends are in thermal contact with a hot reseroir at temperature Ti and a cold reservoir at temperature T2. The sides of the rod are covered with insulating medium, so the transport of heat is along the rod, not through the sides. The molecules at the hot reservoir have greater vibrational energy. This energy is transferred by collisions to the atoms at the end face of the rod. These atoms in turn transfer energy to their neighbours further along the rod. Such transfer of heat through a substance in which heat is transported without direct mass transport is called conduction. T1>T2 T1 (Hot) Q T2 (Cold) Most metals use another, more effective mechanism to conduct heat. The free electrons, which move throughout the metal can rapidly carry energy from the hotter to cooler regions, so metals are generally good donductors of heat. The presence of 'free' electrons also causes most metals to be good electrical conductors. A metal rod at 5°C feels colder than a piece of wood at 5°C because heat can flow more HEAT Page # 4 easily from your hand into the metal. Heat transfer occurs only between regions that are at different temperatures, and the rate of heat flow dQ is . This rate is also called the heat current, denoted by H. Experiments show that the heat current is dt dT proportional to the cross-section area A of the rod and to the temperature gradient , which is the dx rate of change of temperature with distance along the bar. In general dQ dT H  –kA dt dx dQ dT The negative sign is used to make a positive quantity since is negative. The constant k, called dt dt the thermal conductivity is a measure of the ability of a material to conduct heat. A substance with a large thermal conductivity k is a good heat conductor. The value of k depends on the temperature, increasing slightly with increasing temperature, but k can be taken to be practically constant throughout a substance if the temperature difference between its ends is not too great. Let us apply Eq. (i) to a rod of length L and constant cross sectional area A in which a steady state has been reached. In a steady state the temperature at each point is constant in time. Hence. – dT  T1 – T2 dt Therefore, the heat Q transferred in time t is  T – T2 Q  kA  1  L   t  Here, T = temperature difference (TD) and R  • 1. l = thermal resistance of the rod. kA Important Points in conduction Consider a section ab of a rod as shown in figure. Suppose Q1 heat enters into the section at 'a' and Q2 leaves at 'b', then Q2 < Q1 . Part of the energy Q1 – Q2 is utilized in raising the tempeature of section ab and the remaining is lost to atmosphere thorugh ab. If heat is continuously supplied from the left end of the rod, a stage comes when temperature of the section becomes constant. In that case, Q1 = Q2 if rod is insulated from the surroundings (or loss thorugh ab is zero). This is called the steady state condition. Thus, in steady state temperature of different sections of the rod becomes constant (but not same). Q2 Q1 a b Hence, in the figure : Q Q T1 T2 T3 T4 Insulated rod in steady state T1 = constant, T2 = constant etc. and T1 > T 2 > T 3 > T 4 Now, a natural question arises, why the temperature of whole rod not becomes equal when heat is being continuously supplied ? The answer is : there must be a temperature difference in the rod for the heat flow, same as we require a potential difference across a resistance for the current flow thorugh it. In steady state, the temperature varies linearly withd T T1 istance along the rod if it is insluated. 2. Comparing equation number (iii), i.e., heat current T4 x HEAT Page # 5 H l    where R   kA   dQ T  dt R with the equation, of current flow through a resistance, i l    where R   A   dq V  dt R We find the following similarities in heat flow through a rod and current flow through a resistance. Heat flow through a conducting rod Current flow thorugh a resistance dQ Heat current H = = rate of dt heat flow H Electric current i = i T TD  R R R dq = rate of charge flow dt V PD  R R R I kA k = thermal conductivity I A   electrical conductivity From the above table it is evident that flow of heat through rods in series and parallel is analogous to the flow of current through resistances in series and parallel. This analogy is of great importance in solving complicated problems of heat conduction. RADIATION 6. RADIATION The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation. The term radiation used here is another word for electromagnetic waves. These waves are formed due to the superposition of electric and magnetic fields perpendicular to each other and carry energy. Propoerties of Radiation : (a) All objects emit radiations simply because their temperature is above alsolute zero, and all objects absorb some of the radiation that falls on them from other objects. (b) Maxwell on the basis of his electromagnetic theory proved that all radiations are electromagnetic waves and their sources are vibrations of charged particles in atoms and molecules. (c) More radiations are emitted at higher temperature of a body and lesser at lower temperature. (d) The wavelength corresponding to maximum emission of radiations shifts from longer wavelength to shorter wavelength as the temperature increases. Due to this the colour of a body appears to be changing. Radiations from a body at NTP has predominantly infrared waves. (e) Thermal radiations travels with the speed of light and move in a straight line. (f) Radiations are electromagnetic waves and can also travel through vacuum. (g) Similar to light, thermal radiations can be reflected, refracted, diffracted and polarized. 1 (h) Radiation from a point source obeys inverse square law (intensity  2 ) r 7. STEFAN-BOLITZMANN'S LAW : Consider a hot body at temperature T placed in an environment at a lower temperature T0. The body emits more radiation than it absorbs and cools down while the surroundings absorb radiation from the body and warm up. The body is losing energy by emitting radiations and this rate. d d  T4 ,  A, dt dt d e  dt d  eAT 4 dt HEAT Page # 6 P1  eAT 4 and is receiving energy by absorbing radiations and this absorption rate d = P2  aAT04 dt Here, 'a' is a pure number between 0 and 1 indicating the relative ability of the surface to absorbs radiation from its surroundings. Note that this 'a' is different from the absorptive power 'a'. In thermal equilibrium, both the body and the surrounding have the same temperture (say Tc) and, P1 = P2 or eATc4  aATc4 or e=a Thus, when T > T0, the net rate of heat transfer from the body to the surroundings is, Net heat loss = or 8. dQ  eA ( T 4 – T04 ) dt  dT  ms    eA(T 4 – T04 )   dt  Rate of cooling  dT  eA 4 (T – T04 ) –   dt  mc dT  (T 4 – T04 ) dt or NEWTON'S LAW OF COOLING According to this law, if the temperature T of the body is not very different from that of the surroundings dT T0, then rate of cooling – is proportional to the temperature difference between them. To prove it let dt us assume that T = T0 + t  d 4  Ae ( T  T ) 4 – T0 dt   4 T  d  AeT04 1  – 1  4AT 3 T 0 dt T0   if the temperature difference is small. Thus, rate of colling dT d  T   or – dt dt or T = D – as 9. dT = d CONCEPT OF AN IDEAL GAS A gas has no shape and size and can be contained in a vessel of any size or shape. It expands indefinitely and uniformly to fill the available space. It exerts pressure on its surroundings. The gases whose molecules are point massses (mass without volume) and do not attract each other are called ideal or perfect gases. It is a hypothetical concept which can't exist in reality. The gases such as hydrogen, oxygen or helium which cannot be liquified easily are called permanent gases. An actual gas behaves as ideal gas most closely at low pressure and high temperature. 9.1 Ideal gas Equation According to this equation. m PV  nRT  RT M In this equation n = number of moles of the gas = m M HEAT Page # 7 m = total mass of the gas. M = molecular mass of the gas R = Universal gas constant = 8.31 J/mol-K = 2.0 cal/mol- K 10. KINETIC THEORY OF GASES Kinetic Theory of gases is based on the following basic assumptions. (a) A gas consists of very large number of molecules. These molecules are identical, perfectly elastic and hard spheres. They are so small that the volume of molecules is negligible as compared with the volume of the gas. (b) Molecules do not have any preferred direction of motion, motion is completely random. (c) These molecules travel in straight lines and in free motion most of the time. The time of the collision between any two molecules is very small. (d) The collision between molecules and the wall of the container is perfectly elastic. It means kinetic energy is conserved in each collision. (e) The path travelled by a molecule between two collisions is called free path and the mean of this distance travelled by a molecule is called mean free path. (f) The motion of molecules is governed by Newton’s law of motion (g) The effect of gravity on the motion of molecules is negligible. Note : At higher temperature and low pressure or at higher temperature and low density, a real gas behaves as an ideal gas. 11. EXPRESSION FOR PRESSURE OF A GAS : Let us suppose that a gas is enclosed in a cubical box having length  . Let there are ‘N’ identical molecules, each having mass ‘m’. Since the molecules are of same mass and perfectly elastic, so their mutual collisions result in the interchange of velocities only. Only collisions with the walls of the container contribute to the pressure by the gas molecules. Let us focus on a molecule having velocity v1 and components of velocity v X1 , v y1 , v z1 along x, y and z-axis as shown in figure. H G D v z1 v y1 Fm Y C v1 v x1 Z E A v 12  v 2x1  v 2y1  v 2z1  X  B The change in momentum of the molecule after one collision with wall BCHE = m v x1 – (– m v x1 ) = 2 m v x1 . The time taken between the successive impacts on the face BCHE = Time rate of change of momentum due to collision= 2 dis tan ce = v velocity x1 2mv x1 change in momentum mv 2 x1 = 2 / v = time taken x1  Hence the net force on the wall BCHE due to the impact of n N molecules of the gas is : Fx = mv 2x1  + mv 2x2  + mv 2x3  + ........ mv 2xn  =  m 2 v x1  v 2x2  v 2x3 ...... v 2xn  v  = mN  2 x  HEAT Page # 8 where < vx 2 > = mean square velocity in x-direction. Since molecules do not favour any particular direction therefore < vx 2 > = < vy 2 > = < vz 2 >. But = + +  = P= Fx 2 =  v2  . Pressure is equal to force divided by area. 3 M 3 3 = M . Pressure is independent of x, y, z directions 3V where  3 = volume of the container = V M = total mass of the gas, = mean square velocity of molecules 1  P =  3 from PV = nRT n= M Mass = (in kg/mole) M0 Molecular Weight M RT RT 1 2 P = M V RT = M  M = Vrms  Vrms = 3 0 0 0 Vrms = 3RT M0 = 3RT mN A = 3RT M0 3Kt m R K = Boltzman’s const. = N A 11.1 Co-ordinates of a Gas: (P, V, T) is the coordinate of the gas If initial condition of gas is given by (P1 V1 T1) and final condition of gas is given by (P2, V2 T2) such as (P1 V1 T1) (P2 V2 T2) Then (P, V, T) define situation of gas. When a gas changes from one coordinate system to another coordinate system, then we have to follow a process. 12. GAS LAWS Assuming permanent gases to be ideal, through experiments, it was established that gases irrespective of their nature obey the following laws : 12.1 Boyle's Law According to this law, for a given mass of a gas the volume of a gas at constant temperature (called isothermal process) is inversely proportional to its pressure, i.e., 1 (T = constant) P or PV = constant or PiVi = Pf Vf Thus, P – V graph in an isothermal process is a rectangular hyperbola. Or PV versus P or V graph is a straight line parallel to P or V axis. V P PV T = constant V T = constant P or V HEAT Page # 9 12.2 Charle's Law According to this law, for a given mass of a gas the volume of a gas at constant pressure (called) isobaric process) is directly proportional to its absolute temperature, i.e., VT or V = constant T or Vi V f  Ti Tf Thus, V – T graph in an isobaric process is a straight line passing through origin. Or V/T versus V or T graph is a straight line parallel to V or T axis. V V/T P = constant P = constant V or T T (in K) 12.3 Gay Lussac's Law or Pressure law According to this law, for a given mass of a gas the pressure of a gas at constant volume (called isochoric process) is directly proportional to its absolute temperature i.e., PT or P = constant T or Pi P f  Ti T f Thus, P - T graph in an isochoric process is a straight line passing through origin or P/T versus P or T graph is a straight line parallel to P or T axis. P P/T V = constant T (in K) P or T 12.4 Avogadro's Law Two gases at same volume pressure and temperature contain equal amount of moles (mass of gas may be different) or we can say contain equal no. of particle. 1 mole = 6.023 × 10–23 Particles 13. DEGREE OF FREEDOM Total number of independent co-ordinates which must be known to completely specify the position and configuration of dynamical system is known as “degree of freedom f”. Maximum possible translational 1 1 1 2 2 2 degrees of freedom are three i.e.  mVx  mVy  mVz  2 2 2 1 1 1 2 2 2 Maximum possible rotational degrees of freedom are three i.e.  Ix  x  I y  y  Iz  z  2 2 2 HEAT Page # 10 Vibrational degrees of freedom are two i.e. (Kinetic energy. of vibration and Potential energy of vibration) Monoatonic Eg : (all inrent gases, He, Ar, etc.) f=3 (translational) (Vx, Vy, VZ) Diatomic Eg : (gases like H2, N2, O2 etc) f=5 (3 translational + 2 rotational) y x (Vx, Vy, Vz ; Wx, Wy, Wz) z If temp < 70 K for diatomic molecules, then f=3 If temp is in between 250 K to 5000 K, then f=5 If temp is very high (> 5000K) f = 7 [3 translational +2 rotational +2 vibrational] Triatomic (Non - linear) y Vx , Vy , Vz , Wx , Wy , Wz     3 Trans. 3 Rotational x If linear (CO2) z Total D.O.f = 5 x Vx , Vy , Vz , Wy , Wz     3 Trans. 2 Rotational Maxwell’s law of equipartition of energy. Energy associated with each degree of freedom= 1 KT.. 2 1 KT 2 If degree of freedom of a molecule is f then of one pasticle is same and = total kinetic energy of that molecule = t KT 2 Monoatomic Energy of one particle = 3 3 3 KT, one mde = RT, n mole = nRT T 2 2 2 Diatomic Energy of one Barticle = D.O.f. = 6 5 5 5 KT, one mole = RT, n mole = nRT T 2 2 2 HEAT Page # 11 General Degree of Freedom. t t t Energy of one particle = KT, one mole = RT, n mole = nRT T 2 2 2 Internal energy of a gas only dipends on the temperature of the gas desn’t depend on the process taken by the gas to reach the tempreature. 14. INTERNAL ENERGY The internal energy of a system is the sun of kinetic and potential energies of the molecules of the system. It is denoted by U. Internal energy (U) of the system is the function of its absolute temperature (T) and its volume (V). i.e. U = f(T , V) In case of an ideal gas, intermolecular force is zero. Hence its potential energy is also zero. In this case, the internal energy is only due to kinetic energy. Which depends on the absolute temperature of the gas. f i.e. U = f(T). For an ideal gas internal energy U = nRT. 2 15. THERMODYNAMICS Thermodynamics is mainly the study of exchange of heat energy between bodies and conversion of the same into mechanical energy and vice versa. Thermodynamic System Collection of an extremely large number of atoms or molecules confined within certain boundaries such that it has a certain value of pressure (P),volume (V) and temperature (T) is called a thermodynamic system. Anything outside the thermodynamic system to which energy or matter is exchanged is called its surroundings. Taking into consideration the interaction between a system and its surroundings thermodynamic system is divided into three classes : (a) Open system : A system is said to be an open system if it can exchange both energy and matter with its surroundings. (b) Closed system : A system is said to be closed system if it can exchange only energy (not matter with its surrounding (c) Isolated system : A system is said to be isolated if it can neither exchange energy nor matter with its surroundings. Zeroth law of Thermodynamics : If two systems (B and C) are separately in thermal equilirbrium with a third one (A), then they the mselves are in thermal equilibrium with each other. adiabatic wall diathermic wall B C A Equation of State (for ideal gases) : The relation between the thermodynamic variables (P, V, T) of the system is called equation of state. The equation of state for an ideal gas of n moles is given by PV = nRT, Work done by a gas : Let P and V be the pressure and volume of the gas. If A be the area of the piston. then force exerted by gas on the piston is, F = P × A Let the piston move through a small distance dx during the expansion of the gas. Work done for a small displacement dx is dW = F dx = PA dx Since A dx = dV, increase in volume of the gas is dV  dW = P dV HEAT Page # 12 area enclosed P dx P,V,T vi vf V or W  d W   P dV Area enclosed under P-V curve gives work done during process DIFFERENT TYPES OF PROCESSES (a) Isothermal Process : T = constant [Boyle’s law applicable] PV = constant T P T V P V There is exchange of heat between system and surroundings. System should be compressed or expanded very slowly so that there is sufficient time for exchange of heat to keep the temperature constant. Slope of P-V curve in isothermal process : PV = constant = C dP P  dV V  Work done in isothermal process : Vf W = nRT ln V i If v f  vi then W is positive    If v f  vi then W is negative  V  W  2.303 nRT log10 f  Vi   P P Compression Expansion vi vf V vi vf V Internal energy in isothermal process : U = f (T) U = 0 (b) Iso-Choric Process (Isometric Process) : V = constant  Change in volume is zero  P is constant T P = const. (Galussac-law) T HEAT Page # 13 Work done in isochoric process : Since change in volume is zero therefore dW = p dV = 0 Indicator diagram of isochoric process : P V T  V T P Change in internal energy in isochoric process : f U  n R T 2 Heat given in isochoric process : f Q = U  n R T 2 (c) Isobaric Process : Pressure remains constant in isobaric process V  cons tan t P = constant  T Indicator diagram of isobaric process : P P P T T V (in Kelvin) Work done in isobaric process : W = P V = P (Vfinal – Vinitial) = nR(Tfinal – Tinitial) Change in internal energy in isobaric process : f R T 2 Heat given in isobaric process : Q = U + W U = n f f R T + P[Vf – Vi] = n R T + nR T 2 2 Above expression gives an idea that to increase temperature by T in isobaric process heat required is more than in isochoric process. Q = n (d) Cyclic Process : In the cyclic process initial and final states are same therefore initial state = final state Work done = Area enclosed under P-V diagram. Change in internal Energy U = 0 Q = U + W  Q = W If the process on P-V curve is clockwise, then net work done is (+ve) and vice-versa. The graphs shown below explains when work is positive and when it is negative (–)work P P P V V (–)work (+)work (–)work P V V HEAT Page # 14 16. SPECIFIC HEAT The specific heat capacity of a substance is defined as the heat supplied per unit mass of the substance per unit rise in the temperature. If an amount Q of heat is given to a mass m of the substance and its temperature rises by T, the specific heat capacity s is given by equation Q mT The molar heat capacities of a gas are defined as the heat given per mole of the gas per unit rise in the temperature. The molar heat capacity at constant volume, denoted by Cv, is : s= f  Q   R Cv =  nT   cons tan t volume 2 and the molar heat capacity at constant pressure, denoted by Cp is,  Q  CP =  nT   cons tan t Pr essure f     1 R 2  where n is the amount of the gas in number of moles and f is degree of freedom. Quite often, the term specific heat capacity or specific heat is used for molar heat capacity. It is advised that the unit be carefully noted to determine the actual meaning. The unit of specific heat capacity is J/kg-K whereas that of molar heat capacity is J/mol-K. MOLAR HEAT CAPACITY OF IDEAL GAS IN TERMS OF R : (i) For a monoatomic gas f = 3 CV = CP 3 5   = 5 = 1.67 R , CP = R  Cv 3 2 2 (ii) For a diatomic gas f = 5 CV = CP 5 7  1.4 R , CP = R  = C 2 2 V (iii) For a Triatomic gas f = 6 CV = 3R, CP = 4R CP 4 = C = = 1.33 3 V [Note for CO2 ; f = 5, it is linear] ln general if f is the degree of freedom of a molecule, then CV = CP 2 f   f  1   R , CP =   1  R ,  = 2 C f    2 V for any general process C = f R work done by gas  2 nT ADIABATIC PROCESS When no heat is supplied or extracted from the system the process is called adiabatic. Process is sudden so that there is no time for exchange of heat. If walls of a container are thermally insulated no heat can cross the boundary of the system and process is adiabatic. insulating wall Gas Equation of adiabatic process is given by HEAT Page # 15 PV = constant [Poission law] TP1– = constant T V – 1 = constant P Slope of P-V curve in adiabatic process : Since PV is constant   dP  P    –  dV  V dp P  –   dV V v Slope of P – T – curve in adiabatic process : Since T P1– is a constant   P ( ) P dV =– = (1 – ) T ( – 1) T dT P dP  P  dT (  – 1) T Slope of T-V-curve : P dV  1 V  –  dT  (  – 1) T  dV 1 V – dT ( – 1) T T T Work done in adiabatic Process : W = – U = nCv (Ti – Tf ) = = PV i i – Pf Vf ( – 1) nR(Ti – T f )  –1 work done by system is (+ve), if Ti > Tf (hence expansion) work done on the system is (–ve) if Ti < Tf (hence compression) Reversible and Irreversible Process A process is said to be reversible when the various stages of an operation in which it is subjected can be traversed the back in the opposite direction in such a way that substance passes through exactly the same conditions at every step in the reverse process as in the direct process. HEAT Page # 16 Comparison of slope of Iso-thermal and Adiabatic Curves Adiabatic P isobaric P Isothermal isothermal poly adiabatic di mono Isothermal V1 Adiabatic  adia V Wmono < Wdi < Wpoly < Wisothermal < Wisobaric V dP dV V2 dP dV isothermal In compression up to same final volume : Wadia  Wisothermal In Expansion up to same final volume : Wisothermal  Wadia Limitations of Ist Law of Thermodynamics : The first law of thermodynamics tells us that heat and mechanical work are interconvertible. However, this law fails to explain the following points : (i) It does not tell us about the direction of transfer of heat. (ii) It does not tell us about the conditions under which heat energy is converted into work. (iii) It does not tell us whether some process is possible or not. Mixture of non-reacting gases : (a) Molecular weight  n1M1  n2M2 n1  n2 M1 & M2 are molar masses. (b) Specific heat CV  n1C V1  n2CV2 n1  n2 , CP  n1CP1  n2CP2 CPmix n1CP1  n2CP2  ......  (c) for mixture,   C n1C V1  n2C V2  ...... Vmix (d) Degree of freedom for mixture f = n1f1  n2 f2 n1  n2 n1  n2