Thermodynamics Notes

THERMAL PROPERTIES OF MATTER Thermal equilibrium Branch of physics which deals with the inter–conversion between heat energy and any other form of energy is known as thermodynamics. In this branch of physics we deals with the processes involving heat, work and internal energy. In this branch of science the conversion of heat into mechanical work and vice versa is studied. Thermodynamical System The system which can be represented in of pressure (P), volume (V) and temperature (T), is known thermodynamic system. A specified portion of matter consisting of one or more substances on which the effects of variables such as temperature, volume and pressure are to be studied, is called a system. e.g. A gas enclosed in a cylinder fitted with a piston is a system. (i) Surroundings Anything outside the system, which exchanges energy with the system and which tends to change the properties of the system is called its surroundings. (ii) Heterogeneous System A system which is not uniform throughout is said to be heterogeneous. e.g. A system consisting of two or more immiscible liquids. (iii) Homogeneous System A system is said to be homogeneous if it is completely uniform throughout. e.g. Pure solid or liquid. (iv) Isolated System A system in which there can be no exchange of matter and energy with the surroundings is said to be an isolated system. (v) Universe The system and its surroundings are together known as the universe. Thermodynamic variables of the system (i) Composition () (ii) Temperature (T) (iii) Volume (V) (iv) Pressure (P) Thermodynamic state The state of a system can be described completely by composition, temperature, volume and pressure. If a system is homogeneous and has definite mass and composition, then the state of the system can be described by the remaining three variables namely temperature, pressure and volume. These variables are interrelated by equation PV = µRT The thermodynamic state of the system is its condition as identified by two independent thermodynamic variables (P, V or P, T or V, T). 1. ZEROTH LAW OF THERMODYNAMICS If objects A and B are separately in thermal equilibrium with a third object C (say thermometer), then objects A and B are in thermal equilibrium with each other. Zeroth law of thermodynamics introduce the concept of temperature. Two objects (or systems) are said to be in thermal equilibrium if their temperatures are the same. In measuring the temperature of a body, it is important that the thermometer be in the thermal equilibrium with the body whose temperature is to be measured. Thermal equilibrium is a situation in which two objects in thermal contact cease to exchange energy by the process of heat. Heat is the transfer of energy from one object to another object as a result of a difference in temperature between them. Internal Energy Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration. The energy due to molecular motion is called internal kinetic energy (Uk) and that due to molecular configuration is called internal potential energy (Up). dU = dUk + dUp If there no intermolecular forces, then dUp = 0 and dU = dUk = m cv dT cv = Specific heat at constant volume and dT = Infinitesimal change in temperature m = Mass of system M = Molecular weight Molar heat capacity Cv = Mcv For µ–moles of ideal gas dU  µC v dT  m C v dT M Internal energy in the absence of intermolecular forces is simply the function of temperature and state only, it is independent of path followed. U = Uf – Ui Ui = Internal energies in initial state and Uf = Internal energies in final state Thermodynamic Processes In the thermodynamic process pressure, volume, temperature and entropy of the system change with time. Thermodynamic process is said to take place if change occurs in the state of a thermodynamic system. Sign convention used for the study of thermodynamic processes Heat gained by a system Positive Heat lost by a system Negative The work done by a system Positive Work done on the system Negative Increase in the internal energy of system Positive Decrease in the internal energy of system Negative Indicator Diagram or P–V Diagram In the equation of state of a gas PV = µRT Two thermodynamic variables are sufficient to describe the behavior of a thermodynamic system. If any two of the three variables P, V and T are known then the third can be calculated. P–V diagram is a graph between the volume V and the pressure P of the system. The volume is plotted against X–axis while the pressure is plotted against Y–axis. The point A represents the initial stage of the system. Initial pressure of the system is P i and initial volume of the system Vi . The point B represents the final state of the system. Pf and Vf are the final pressure and final volume respectively of the system. The points between A and B represent the intermediate states of the system. With the help of the indicator diagram we calculate the amount of work done by the gas or on the gas during expansion or compression. P A (P i , Vi) B (P f, Vf) V THERMAL PROPERTIES OF MATTER (i) Cyclic process Cyclic process is that thermodynamic process in which the system returns to its initial stage after undergoing a series of changes. (ii) Non–cyclic process Non–cyclic process is that process in which the system does not return to its initial stage. (iii) Quasi–static or equilibrium process Quasi–static is a thermodynamic process which proceeds extremely slowly such that at every instant of time, the temperature and pressure are the same in all parts of the system. (iv) Reversible and Irreversible processes A reversible process is one in which the changes in heat and work of direct process from initial to a final state are exactly retraced in opposite sense in the reverse process and the system and surroundings are left in their initial states. The reversibility is an ideal concept and can not be realized in practice. The process which is not reversible is the irreversible process. In nature the processes are irreversible. WORK DONE BY THERMODYNAMIC SYSTEM dx One of the simple example of a thermodynamic system is a gas in a cylinder with a movable piston. (i) area of piston is A (P, V) If the gas expands against the piston F = P× A Gas exerts a force on the piston and displace it through a distance and does work on the piston. (ii) If the piston compresses the gas When piston moved inward, work is done on the gas. (iii) The work associated with volume changes P A If pressure of gas on the piston = P. (Pi , Vi) (Pf, Vf) B Then the force on the piston due to gas is F = PA When the piston is pushed outward an infinitesimal distance dx, P the work done by the gas is dW = F × dx = PA dx dV The change in volume of the gas is dV = Adx,  dW = PdV V For a finite change in volume from Vi to Vf, this equation is then integrated between Vi to Vf to find the net work done W   Vf Vi dW   Vf Vi PdV Hence the work done by a gas is equal to the area under P–V graph. Following different cases are possible. (i) Volume is constant P initial V initial final or final V V = constant and WAB = 0 P (ii) Volume is increasing P P Initial A Final B Initial A Final B or V V is increasing (iii) WAB > 0 V WAB = Shaded area Volume is decreasing P Final P B Final B Initial A Initial A or V V is decreasing (iv) WAB < 0 V WAB = – Shaded area Cyclic process P P V V Wclockwise cycle = + Shaded area Wanticlockwise cycle = – Shaded area WORK DONE IN CLOCKWISE CYCLE P P P B B = A B + A V A V V WAB I Positive Wcyclic = WAB I(positive) + WAB II WAB (negative) = II f a o th are d pa se clo Negative THERMAL PROPERTIES OF MATTER 2. FIRST LAW OF THERMODYNAMICS Let a gas in a cylinder with a moveable piston changes from an initial equilibrium state to a final equilibrium state. P initial a System change its state through path 'a' : b The heat absorbed by the system in this process = Qa c final The work done by the system = Wa V Again for path 'b' : Heat absorbed by the system = Qb , Work done by the system = Wb. It is experimental fact that the Qa – Wa = Qb –Wb Both Q and W depend on the thermodynamic path taken between two equilibrium states, but difference Q – W) does not depends on path in between two definite states of the system. So, there is a function (internal energy) of the thermodynamic coordinates (P, V and T) whose final value (U f) minus its initial value (Ui) equals the change Q – W in the process. dU =Q – W. This is the first law of thermodynamics. Heat supplied to the system and work done by the system are path dependent so they are denoted by Q and W respectively. Change in internal energy U = Uf – Ui does not depends on path it depends only on initial and final positions of the system. So, it is denoted by dU (or U) First Law of Thermodynamics If some quantity of heat is supplied to a system capable of doing external work, then the quantity of heat absorbed by the system is equal to the sum of the increase in the internal energy of the system and the external work done by the system. Q = dU + W or Q = W + U (i) This law is applicable to every process in nature (ii) The first law of thermodynamics introduces the concept of internal energy. (iii) The first law of thermodynamics is based on the law of conservation of energy. (iv) Q, dU and W must be expressed in the same units (either in units of work or in units of heat). (v) This law is applicable to all the three phases of matter, i.e., solid, liquid and gas. (vi) dU is a characteristic of the state of a system, it may be any type of internal energy–translational kinetic energy, rotational kinetic energy, binding energy etc. (i) Work done by the gas, (ii) Increase in the internal energy, (iii) Amount of heat supplied, (iv) Molar heat capacity of the gas R = 8.31 J mol–1 K–1 P1 = 4 × 105 Nm–2 P2 = 8 × 10+5 Nm–2, V1 = 0.2 m3, V2 = 0.5 m3 Work done by the gas = Area under P–V graph (Area ABCDEA) 8 B 2 (i) P  1 1 (AE +BD) × AC  (4 × 105 + 8 × 105) × (0.5–0.2) 2 2 1  × 12 × 105 × 0.3 = 1.8 × 105 J 2 5 Sol. The pressure in monoatomic gas increases linearly from 4 × 105 Nm–2 to 8 × 10+5 Nm–2 when its volume increases from 0.2m3 to 0.5 m3. Calculate. (10 × N/m ) Ex. 4 0 C A E 0.2 3 (m ) D 0.5 V (ii) Increase in internal energy U = CV (T2 – T1)  CV CV R(T2–T1)  (P2V2 – P1V1) R R For monoatomic gas CV  Ex. 3 3 3 R U  [(8 × 105 × 0.5) – (4 × 105 × 0.2)]  [4 × 105 – 0.8 × 105 ] = 4.8 × 105 J 2 2 2 (iii) Q = U + W = 4.8 × 105 + 1.8 × 105 = 6.6 × 105 J (iv) C QR Q QR 6 .6  10 5  8 .3 1    = 17.14 J/mole K (P2 V2  P1 V1 ) T RT 1  3 .2  10 5 P When a system is taken from state a to state b, in figure along the path a  c  b, 60 J of heat flow into the system, and 30 J of work is done : c b a d (i) How much heat flows into the system along the path a  d  b if the work is 10 J. V (ii) When the system is returned from b to a along the curved path, the work done by the system is –20 J. Does the system absorb or liberate heat, and how much? (iii) If, Ua = 0 and Ud = 22 J, find the heat absorbed in the process a  d and d  b. Sol. For the path a, c, b, U = Q – W = 60 – 30 = 30 J or Ub – Ua = 30 J (i) Along the path a, d, b, Q = U + W = 30 + 10 = 40 J (ii) Along the curved path b, a, Q = (Ua – Ub) + W = (–30) + (–20) = –50 J, heat flows out the system (iii) Qad = 32 J; Qdb = 8 J APPLICATION OF FIRST LAW OF THERMODYNAMICS (i) Melting Process : When a substance melts, the change in volume (dV) is very small and can, therefore, be neglected. The temperature of a substance remains unchanged during melting process. Let us consider the melting of a mass m of the solid. Let L be the latent heat of fusion i.e., the heat required L to change a unit mass of a solid to liquid phase at constant temperature. Heat absorbed during melting process, Q = mL By the first law of thermodynamics Q = U + W  mL = U [W=PV=P × 0=0] So, the internal energy increases by mL during the melting process. (ii) Boiling Process : When a liquid is heated, it changes into vapour at constant temperature (called boiling point) and pressure. When water is heated at normal atmospheric pressure, it boils at 100°C. The temperature remains unchanged during the boiling process. Let us consider the vaporisation of liquid of mass m. Let V and Vv be the volumes of the liquid and vapours respectively. The work done in expanding at constant temperature and pressure P, W = PV = P (Vv – V) Let the latent heat of vaporisation = L  Heat absorbed during boiling process, Q = mL Let U and Uv be the internal energies of the liquid and vapours respectively then U = Uv – U According to the first law of thermodynamics, Q = U + W  mL = (Uv – U) + P(Vv – V) THERMAL PROPERTIES OF MATTER Ex. 1 kg of water at 373 K is converted into steam at same temperature. Volume of 1 cm3 of water becomes 1671 cm3 on boiling. What is the change in the internal energy of the system if the latent heat of vaporisation of water is 5.4 × 10 5 cal kg–1? Sol. Volume of 1 kg of water = 1000 cm³ = 10–3 m³ , Volume of 1 kg of steam = 10³ × 1671 cm³ = 1.671 m³ Change in volume, V = (1.671 – 10–3) m³ = 1.670 m³ , Pressure, P = 1 atm. = 1.01 × 105 N m–2 In expansion work done, W = PV = 1.01 × 105 × 1.67 J = 1 .6 9 1  1 0 5 cal = 4.026 × 104 cal 4 .2 But U = Q – W (first law of thermodynamics) or U = (5.4 × 105 – 0.4026 × 105) cal = 4.9974 × 105 cal ISOMETRIC OR ISOCHORIC PROCESS Isochoric process is a thermodynamic process that takes place at constant volume of the system, but pressure and temperature varies for change in state of the system. Equation of state P = constant × T (P and T are variable, V is constant) Work done In this process volume remains constant V = 0 or dV = 0  W  Form of first Law Q = U It means whole of the heat supplied is utilized for change in internal energy of the system. Q = U = µ Cv T P Slope of the P–V curve dP = dV Specific heat at constant volume (CV)  Vf Vi PdV  0 V dP = dV V dV =0 dP P (c) The quantity of heat required to raise the temperature of 1 gram mole gas through 1 °C at constant volume is equal to the specific heat at constant volume. A gas enclosed in a cylinder having rigid walls and a fixed piston. When heat is added to the gas, there would be no change in the volume of the gas. When a substance melts, the change in volume is negligibly small. So, this may be regarded as a nearly isochoric process. Heating process in pressure cooker is an example of isometric process. Ex. An ideal gas has a specific heat at constant pressure CP = Sol. CV = CP – R = (a) (b) 5R . The gas is kept in a closed vessel of volume 0.0083 2 m3 at a temperature of 300K and a pressure of 1.6 × 106 Nm–2. An amount of 2.49 × 104 J of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas. 5R 3R R  , V = 0, T1 = 300 K, V = 0.0083 m3, P1 = 1.6 × 106 Nm–2 2 2 From first law of thermodynamics Q = U + PV U = Q = 2.49 × 104 J From gas equation n  PV 1 .6  1 0 6  0 .0 0 83 1 6   RT 8 .3  30 0 3 U 2 .49  1 0 4  6 U = nCVT  T  nC  3  8.3  16 = 375 K v Final temperature = 300 + 375 = 675K P2 T2 T2 1 .6  10 6  6 7 5 According to pressure law P T  P  T  P2  T  P1  = 3.6 × 106 Nm–2 3 00 1 1 1 Ex. 5 moles of oxygen is heated at constant volume from 10°C to 20°C. What will be change in the internal energy of the gas? The gram molecular specific heat of oxygen at constant pressure. CP = 8 cal/mole and R = 8.36 joule/mole °C. Sol. Q CV = CP – R = 8 – 2 = 6 cal/mole °C Heat absorbed by 5 moles of oxygen at constant volume Q = nCV T = 5 × 6 (20 – 10) = 30 × 10 = 300 cal At constant volume V = 0. W = 0 From first law of thermodynamics Q = U + W  300 = U + 0 U = 300 cal ISOBARIC PROCESS Isobaric process is a thermodynamic process that takes place at constant pressure, but volume and temperature varies for change in state of the system. Equation of state : V = constant × T or V T Work done In this process pressure remains constant P = 0 Vf Work done W   PdV  P(Vf  Vi ) Vi Form of first LawQ = U + P (Vf – Vi) µ Cp dT = µ Cv dT + P(Vf – Vi) It is clear that heat supplied to the system is utilized for : (i) Increasing internal energy and (ii) Work done against the surrounding atmosphere. P  dP  0 Slope of the PV curve :  dV  isobaric V dP =0 dV V dV = dP P Specific heat at constant pressure (CP) The quantity of heat required to raise the temperature of 1 gram mole gas through 1°C at constant pressure is equal to the specific heat. Heating of water at atmospheric pressure. • Melting of solids and boiling of liquids at atmospheric pressure. Ex. At normal pressure and 0°C temperature the volume of 1 kg of ice is reduced by 91 cm3 on melting. Latent heat of melting of ice is 3.4 × 105 J/kg. Calculate the change in the internal energy when 2kg of ice melts at normal pressure and 0°C. (P=1.01 × 105 Nm–2) Sol. Heat energy absorbed by 2 kg of ice for melting Q = mL = 2 × 3.4 × 105 = 6.8 × 105J Change in volume of 2 kg of ice = 2 × 91 = 182 cm3 = 182 × 10–6 m3 W = PV = 1.01 × 105 × (–182 × 10-6) = –18.4 J Since, work is done on ice so work W is taken –ve. Now from first law of thermodynamics Q = U + W U = Q – W = 6.8 × 105 – (–18.4) = (6.8 × 105 + 18.4)J Ex. Sol. What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise the temperature by 45°C at constant pressure. Molecular mass of N2 = 28, R = 8.3 J mol–1 K–1. 7 m 2  10 2 5 5 7   & C P  R Q = nC T =   8.3  4 5  933 .7 5 J Here m = 2 × 10 kg, n = P M 2 8  1 0 3 7 7 2 2 –2 THERMAL PROPERTIES OF MATTER ISOTHERMAL PROCESS In this process pressure and volume of system change but temperature remains constant. In an isothermal process, the exchange of heat between the system and the surroundings is allowed. Isothermal process is carried out by either supplying heat to the substance or by extracting heat from it. A process has to be extremely slow to be isothermal. Equation of state P V = constant (µRT) (T is constant) Work Done Consider µ moles of an ideal gas, enclosed in a cylinder, at absolute temperature T, fitted with a frictionless piston. Suppose that gas undergoes an isothermal expansion from the initial state (P 1, V 1) to the final state (P2, V2). Work done : W   V2 V1  V2  V2 dV µRT V  = µRT  log e V  V2 = µRT [loge V2 – loge V1] = µRT log e  dV = µRT  1 V1 V V  V1   P1   W = 2.303µRT log10   [P1V1 = P2V2]  P2  Form of First Law There is no change in temperature and internal energy of the system depends on temperature only So  V2  U = 0, Q = 2.303 µRT log10    V1  It is clear that Whole of the heat energy supplied to the system is utilized by the system in doing external work. There is no change in the internal energy of the system. Slope of the isothermal curve P Pi For an isothermal process, PV = constant Differentiating, PdV + VdP = 0 dP P   VdP = – PdV  dV V isotherm PV = constant hyperbola Pf P  dP  Slope of isothermal curve,    V  dV  isothermal Vi Vf V For a given system : (i) The product of the pressure and volume of a given mass of a perfect gas remains constant in an isothermal process. (ii) Boyle's law is obeyed in an isothermal process. (iii) A graph between pressure and volume of a given mass of a gas at constant temperature is known as isotherm or isothermal of the gas. (iv) Two isotherms for a given gas at two different temperatures (v) (vi) P T2 > T1 T1 and T2 are shown in figure. T2 The curves drawn for the same gas at different temperatures are mutually parallel and do not cut each other. If two isotherms intersect each other at a single point we get same value of P and V at intersection point. PV = µ RT1 for temperature T1 and PV = µ RT2 for temperature T2. It means T1 = T2 which is not possible. T1 V at const P V T so T2 > T1 (vii) An ideal gas enclosed in a conducting cylinder fitted with a conducting piston. Let the gas be allowed to expand very–very slowly. (viii) This shall cause a very slow cooling of the gas, but heat will be conducted into the cylinder from the surrounding. Hence the temperature of the gas remains constant. If the gas is compressed very–very slowly, heat will be produced, but this heat will be conducted to the surroundings and the temperature of the gas shall remain constant. (ix) The temperature of a substance remains constant during melting. So, the melting process is an isothermal process. (x) Boiling is an isothermal process, when a liquid boils, its temperature remains constant. (xi) If sudden changes are executed in a vessel of infinite conductivity then they will be isothermal. Ex. Two moles of a gas at 127°C expand isothermally until its volume is doubled. Calculate the amount of work done. V2 n = 2, T = 127 + 273 = 400K, V = 2 1 Sol. V2 From formula W = 2.3026 nRT log10 V = 2.3026 × 2 × 8.3 × 400 × log10 2 1 = 2.3026 × 2 × 8.3 × 400 × 0.3010  4.6 × 103J Ex. Figure shows a process ABCA performed on an ideal gas. Find the net heat given to the system during the process. Sol. Since the process is cyclic, hence the change in internal energy is zero. The heat given to the system is then equal to the work done by it. V V2 V1 C A T1 B T2 T The work done in part AB is W1 = 0 (the volume remains constant). The part BC represents an isothermal process V2 so that the work done by the gas during this part is W2 = nRT2 n V 1 During the part CA So, V/T is constant and hence, P  VT nRT is constant V The work done by the gas during the part CA is W3 = P(V1 – V2) = nRT1 – nRT2 = – nR (T2 – T1).   V2  (T2  T1 ) The net work done by the gas in the process ABCA is W = W1 + W2 + W3 = nR  T2 l n V1   The same amount of heat is given to the gas. ADIABATIC PROCESS It is that thermodynamic process in which pressure, volume and temperature of the system do change but there is no exchange of heat between the system and the surroundings. A sudden and quick process will be adiabatic since there is no sufficient time available for exchange of heat so process adiabatic. Equation of state : PV = µRT  Equation for adiabatic process PV = constant THERMAL PROPERTIES OF MATTER Work done Let initial state of system is (P1, V1, T1) and after adiabatic change final state of system is (P2, V2, T2) then we can   write P1 V1 = P2 V2 = K (here K is const.) So W  V2 V1 PdV  K  V2 V1 V2  V   1  K  K V dV     1    (  1) V1   V2    1  V1   1   Q K  P1 V1  P2 V2  1 1     P1 V1 V1 .V1  P2 V2 V2 .V2  = W   [P V – P V ]  (  1)  (  1) 1 1 2 2 µR  T1  T2   W  (  1) ( PV = µRT) Form of first law : dU = – W It means the work done by an ideal gas during adiabatic expansion (or compression) is on the cost of change in internal energy proportional to the fall (or rise) in the temperature of the gas. If the gas expands adiabatically, work is done by the gas. So, Wadia is positive. The gas cools during adiabatic expansion and T 1 > T 2. If the gas is compressed adiabatically, work is done on the gas. So, Wadia is negative. The gas heats up during adiabatic compression and T1 < T2. (Adiabatic expansion of gas) Slope of the adiabatic curve  For an adiabatic process, PV = constant  –1  Differentiating, P V dV + V dP = 0   V dP = –  PV –1 dV  P dP PV  1 P  P          V dV V V mon oato mic = 1.33 d i a t om po ic lya to = 1.4 m ic = 1.67 V P  dP   Slope of adiabatic curve,   dV V   adiabatic Slope of adiabatic is greater than the slope of isotherm slope of adiabatic changes  dP   P  dP   dV      V     dV   slope of isothermal changes =    adia     iso Since  is always greater than one so an adiabatic is steeper than an isotherm Examples of adiabatic process (expansion of gas) P is o th a di ab at ic e rm al V (i) A gas enclosed in a thermally insulated cylinder fitted with a non–conducting piston. If the gas is compressed suddenly by moving the piston downwards, some heat is produced. This heat cannot escape the cylinder. Consequently, there will be an increase in the temperature of the gas. (ii) If a gas is suddenly expanded by moving the piston outwards, there will be a decrease in the temperature of the gas. (iii) Bursting of a cycle tube. (iv) Propagation of sound waves in a gas. (v) In diesel engines burning of diesel without spark plug is done due to adiabatic compression of diesel vapour and air mixture Ex. Why it is cooler at the top of a mountain than at sea level? Sol. Pressure decreases with height. Therefore if hot air rises, it suffers adiabatic expansion. From first law of thermodynamics Q = U + W U = – W [Q Q = 0] This causes a decrease in internal energy and hence a fall of temperature. Ex. 2m3 volume of a gas at a pressure of 4 × 105 Nm–2 is compressed adiabatically so that its volume becomes 0.5m3. Find the new pressure. Compare this with the pressure that would result if the compression was isothermal. Calculate work done in each process.  Sol. V1 = 2m3, P1 = 4 × 105 Nm–2, V2 = 0.5m3 1.4 In adiabatic process  2  P1 V = P2 V2 P2 = 4 × 105  0.5    In isothermal process P1 V1 4  10 5  2 P1V1 = P2V2  P2  V  = 1.6 × 106 Nm–2. 0.5 2  1 Now work done in adiabatic process W  = 4 × 105 (4)1.4 = 2.8 × 106 Nm–2 (2 .8  1 0 6  0 .5 )  (4  1 0 5  2 ) P2 V2  P1 V1  = 1.48 × 106 J. 1 .4  1  1 Work done in isothermal process W = 2.3026RT log V2 V1 = 2.3026P1V1 log V2 V1  0 .5  1 = 2.3026 × 4 × 105 × 2 × log  = 2.3026 × 4 × 105 × 2log   = –1.1 × 106 J  4  2 .0  Ex. Two samples of a gas initially at same temperature and pressure are compressed from a volume V to V . One 2 sample is compressed isothermally and the other adiabatically. In which sample is the pressure greater? Sol. Let initial volume, V1 = V and pressure, P1 = P , final volume, V2 = For isothermal compression P2V2 = P1V1 or P2   V and final pressure, P2 = ? 2 P1 V1 PV   2P V V2 2   V1   V      For adiabatic compression P2' = P1    P2 '  P    2 P  P2´ = 2 P > 1 2 > 2 and P2' > P2 V / 2   V2  Pressure during adiabatic compression is greater than the pressure during isothermal compression. Ex. Calculate the work done when 1 mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are 105 N/m2 and 6 litre respectively. The final volume of the gas is 2 liters. Molar specific heat of the gas at constant volume is 3R 2 [(3)5/3 = 6.19] THERMAL PROPERTIES OF MATTER Sol.   For an adiabatic change PV = constant P1V1 = P2V2  As molar specific heat of gas at constant volume C v  3 2 R 5 R 3 5 C 5 CP = CV + R = R  R  R    P  2  2 2 CV 3 3 R 2  5/ 3  V1  6   10 5 = (3)5/3 × 105 = 6.19 × 105 N/m2  P2    P1 =   V 2    2 Work done W  1 1 [P2 V2  P1 V1 ] = 1 1  (5 / 3 ) [6 .19  10 5  2  10 3  10 5  6  10 3 ]  2  10 2  3    (6.19  3 ) = – 3 × 102 × 3.19 = – 957 joules 2   – ive sign shows external work done on the gas Ex. A motor tyre pumped to a pressure of 3 atm. It suddenly bursts. Calculate the fall in temperature due to adiabatic expansion. The temperature of air before expansion is 27°C. Given =1.4.  Sol.  We know that T2 P2 1.4  T2     300  1    3  1 –  T1 P1  T2   P1       T  1  P2  1 – 1  1 .4  T2      300  1 1.4 3     1  0.4 T2 = 219.2 K  T1 – T2 = (300 – 219.2) K = 80.8 K ETOOS KEY POINTS (i) P When a gas expands its volume increases, then final pressure is less for adiabatic expansion. But, when a gas compresses its volume decreases, then the final pressure is more in case of adiabatic compression. iso th er m al adiabatic P P final P isobaric W IB > W IT > W AD P IB > P IT > P AD PAD > PIT > PIB IB IT isobaric iso the adiabatic AD rm a l V P V 2V V WIB > WIT > WAD PAD > PIT > PIB Pfinal P AD IT PAD > PIT > PIB IB V/2 V V First Law of Thermodynamics Applied to Different Processes Process Q U W Cyclic W 0 Area of the closed curve Isochoric U µCvT (µ mole of gas) 0 Isothermal W 0 V  P  RT loge  f  = RTlog e  i   Vi   Pf  Adiabatic 0 –W µR(Tf  Ti ) 1 Isobaric Ex. µCPT P (Vf – V i) = µR(Tf – Ti ) CVT Plot P – V , V – T graph corresponding to the P–T graph for an ideal gas shown in figure. Explain your answers. D A V Sol. C B (P - V curve) T (V - T curve) For process AB T = constant so P  1 For process BC V For process CD T = constant so V  1 For process DA P = constant so V  T P P = constant so V  T FREE EXPANSION Take a thermally insulated bottle with ideal gas at some temperature T 1 and, by means of a pipe with a stopcock, connect this to another insulated bottle which is evacuated. If we suddenly open the stopcock, the gas will rush from the first bottle into the second until the pressures are equalized. free expansion of a gas Experimentally, we find that this process of free expansion does not change the temperature of the gas – when the gas attains equilibrium and stops flowing, the final temperature of both botles are equal to the initial temperature T1. This process is called a free expansion. The change in the internal energy of the gas can be calculated by applying the first law of thermodynamics to the free–expansion process. The process is adiabatic because of the insulation, So Q = 0. THERMAL PROPERTIES OF MATTER No part of the surroundings moves so the system does no work on its surroundings. (i) For ideal gas (W)ext. = Work done against external atmosphere = P dV = 0 (because P = 0) (W)int. = Work done against internal molecular forces = 0 Q = dU + W The internal energy does not change dU = 0 0 = dU + 0 So U and T are constant. The initial and final states of this gas have the same internal energy. Which implies that the internal energy of an ideal gas does not depend on the volume at all. The free–expansion process has led us to the following conclusion : The internal energy U(T) of an ideal gas depends only on the temperature. (ii) For real gas In free expansion of real gases, measurements show that the temperature changes slightly in a free expansion. Which implies that the internal energy of a real gas depends on the volume also. Q =0 (W)int.  0 Q (W)ext. = 0 (P = 0) (Intermolecular forces are present in real gases) = dU + W 0 = dU + (W)int. dU = – (W)int. U decreases. So T decreases. RELATION BETWEEN DEGREE OF FREEDOM AND SPECIFIC HEAT OF GAS Energy related with each degree of freedom = 1 f kT , Energy related with all degree of freedom = kT 2 2 Internal energy of one mole of ideal gas (total K.E.) U  By first law of thermodynamics CV  dU f R  R dT 2  1 CV  Q = W + dU f RT for Isometric process (volume constant) W = 0 2  R f  . CP  CV  R    1 R   1 2  R  1 , CP  R  1 CV dT = dU   and  1  2 f R R   1 1  x For isobaric process P = constant so x=0  C  CP  For isothermal process, PV = constant so x=1  C= so x=  C=0  For adiabatic process PV = constant dU dT CP 2 1 CV f and General expression for C (CP or CV) in the process PVx = constant C  CV = R  R  CV  R  1 Values of f, U, CV, CP and for different gases are shown in table below. Atomicity of gas Cv f Monoatomic 3 Diatomic 5 Triatomic and Triatomic linear (at high temperature) 7 Poly atomic Triangular Non-linear 6 CP  3 R 2 5 R 2 5  1 .67 3 5 7 7 2 R 2 7 R 2 6 R  3R 2 R 9 R 2 8 R  4R 2 5 9 7  1 .4  1.28 4  1.33 3 (i) 1 < < 2 (ii) If atomicity of gases is same U, CP, CV and  is same for gas mixture. (iii) If in a gas mixture gases are of different atomicity, then for gas mixture  changes according to following condition. Diatomic 1 < mixture < 2 mono atomic where 1 < 2 (iv) If 'f' is the degree of freedom per molecule for a gas, then Total energy of each molecule = fkT 2 f f kT  RT 2 2 According to kinetic theory of gases, the molecule are not interacting with each other. So potential energy is zero and internal energy of gas molecules is only their kinetic energy. Total energy per mole of gas = N 0 (v) (vi) For '' mole of a gas : Internal energy at temperature T is U  (vii) Change in internal energy is given by dU  fR T  C V T 2 fR (dT)  C V dT 2 This change is process independent. CP is greater than CV If a gas is heated at constant volume, the gas does no work against external pressure. In this case, the whole of the heat energy supplied to the gas is spend in raising the temperature of the gas. If a gas is heated at constant pressure, its volume increases. In this case, heat energy is required for the following two purpose : (i) To increase the volume of the gas against external pressure. (ii) To increase the temperature of 1 mole of gas through 1 K. Thus, more heat energy is required to raise the temperature of 1 mole of gas through 1 K when it is heated at constant pressure than when it is heated at constant volume.  CP > CV The difference between CP and CV is equal to thermal equivalent of the work done by the gas in expanding against external pressure. Mayor's formula : CP – CV = R  At constant pressure dQ = µCPdT, dU = µCVdT & dW = PdV = µRdT Now from first law of thermodynamics dQ = dW + dU  µCPdT = µRdT + µCVdT CP = R + CV CP – CV = R THERMAL PROPERTIES OF MATTER Ex. Calculate the difference between two specific heats of 1 g of helium gas at NTP. Molecular weight of helium = 4 and J = 4.186 × 107 erg cal–1. Sol. Gas constant for 1 g of helium, r  CP – CV = 7 6  1 3.6  9 81  2 24 00 R PV   = 2.08 × 107 erg g–1 K–1 2 73  4 Mw T  M w r 2.08  1 0 7  = 0.5 cal g–1 K–1 J 4.186  1 0 7 Ex. Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 cal mol–1 K–1 and density of hydrogen = 0.0899 g cm–3. One mole of gas = 2.016 g, J = 4.2 × 107 erg cal–1 and 1 atmosphere = 10 6 dyne cm–2. Sol. Since the density of hydrogen is 0.0899 g cm–3 therefore volume occupied by 0.0899 g of hydrogen at NTP is 1000 cm3. So, volume of 1 mole (2.016 g) of gas, V  CP – CV = 1 00 0  2.01 6 cm3 0.0 89 9 R PV 10 6  1 0 00  2 .0 16   = 1.96 cal mol–1 K–1 J T  J 0 .0 8 9 9  2 73  4 .2  10 7  CV = CP – 1.96 = (6.85 – 1.96) = 4.89 cal mol–1 K–1 Ex. The specific heat of argon at constant volume is 0.075 kcal/kg K. Calculate its atomic weight, [R = 2cal/mol K] Sol. As argon is monoatomic, its molar specific heat at constant volume will be CV  So 3 3 R   2  3 cal/mol K, CV =Mw cV and 2 2 3 = Mw × 0.075  Mw = cV = 0.075 cal/g K 3 = 40 gram/mole 0 .0 7 5 Effeciency of a cycle () :  total Mechanical work done by the gas in the whole process Heat absorbed by the gas (only  ve) = Ex. area under the cycle in P-V curve Heat injected into the system n moles of a diatomic gas has undergone a cyclic process ABC as shown in figure. Temperature at a is T 0. Find (i) Volume at C. P (ii) Maximum temperature. (iii) Total heat given to gas. (iv) Is heat rejected by the gas, if yes how much heat is rejected. (v) Find out the efficiency. Sol. (i) Since triangle OAV0 and OCV are similar therefore B 2P0 2 P0 P  0  V = 2V0 V V0 P0 C A V0 V V Since process AB is isochoric hence P A PB  TB = 2T0  TA TB Since process BC is isobaric therefore T TB  C  TC = 2TB = 4 T0 VB VC (iii) Since process is cyclic therefore Q = W = area under the cycle = 1 PV . 2 0 0 (iv) Since (ii) U and W both are negative in process CA  Q is negative in process CA and heat is rejected in process CA QCA = WCA + UCA = 1 5 [P0 + 2P0] V0 – nR (Tc – Ta ) 2 2 =  4 P0 V0 P0 V0  1 5 [P0 + 2P0] V0 – nR  nR  nR  2 2 = –9P0V0 = Heat injected. (v)  = efficiency of the cycle = work done by the gas P0 V0 / 2  heat injected Q injected × 100 5  5  19 where Qinj = QAB + QBC =  nR(2 T0  T0 )   nR(2 T0 )  2 P0 (2 V0  V0 )  PV. 2  2  2 0 0 Therefore = 100 % 19 Carnot cycle Cannot devised an ideal engine which is based on a reversible cycle of four operations in succession : (i) Isothermal expansion, A B (ii) Adiabatic expansion, B C (iii) Isothermal compression C D (iv) Adiabatic compression. D A MAIN PARTS OF CARNOT'S ENGINE ARE (i) Source of heat : It is a hot body of very large heat capacity kept at a constant high temperature T 1. Its upper surface is perfectly conducting so that working substance can take heat from it. THERMAL PROPERTIES OF MATTER (ii) Mechanical arrangements and working substance : It is a hollow cylinder whose walls are perfectly non–conducting and its base is perfectly conducting fitted with non–conducting piston. This piston move without any friction. Ideal gas enclosed in cylinder as a working substance. (iii) Heat sink : It is a cold body at low temperature T2. It is a body of large heat capacity and its upper surface is highly conducting so that working substance can reject heat to it. (iv) Stand : It is made by perfectly insulating material. When cylinder placed on it working substance can expended or compressed adiabatic. (v) Working : A set of reversible processes through them working substance is taken back to initial condition to get maximum work from this type of ideal engine. PROCESSES OF CARNOT'S CYCLE CAN BE DENOTED BY AN INDICATOR DIAGRAM. (a) Isothermal expansion A  B Initially the cylinder is taken to be in thermal equilibrium with the high temperature T1, this is initial state of working substance denoted by point A (P1, V1, T1). After that the piston is allowed to move outward slowly. With the movement of the piston. The process is very slow so that it is isothermal. Heat from reservoir flows through the base of cylinder into the gas so temperature of the gas remains T1. Gas expand and receive heat Q1 from source and gets state B(P2, V2, T1) Q source This heat input Q1 to the gas from path A to B is utilized for doing work W1. By path A to B the heat input to the gas = the work done against the external pressure. W1  Q 1  (b)  V2 V1 PdV  Adiabatic expansion B  C  V2 V1 µRT1 V dV  µRT1 l n 2 V V1 Now cylinder is put in contact with a non–conducting stand and piston is allowed to move outward, because no heat can enter in or leave out so the expansion of gas is adiabatic. The temperature falls to T 2K and gas describes the adiabatic from B to point C (P3, V3, T2). During this expansion more work is done (W2) at the expense of the internal energy. Work done in adiabatic path BC is W2  (c) Isothermal compression C  D Q=0 µR  T  T2   1 1 Now the gas cylinder is placed in contact with sink at temperature T 2. The piston is moved slowly inward so that heat produced during compression passes to the sink. The gas is isothermally compressed from C to point D (P4, V4, T2). The heat rejected Q2 to the cold reservoir (sink) at T 2 occurs over this path. Amount of work done on gas W3 = amount of heat rejected to the sink V  V  Q2 = W3 = RT2n  4   Q2 = µRT T2n  4  V  3  V3  Q2 sink (d) Adiabatic compression D  A The cylinder is removed from the sink and is put in contact with insulating stand now piston moves inward. Heat is not allowed to go out and it increases the internal energy of the system. Now work is done on the gas during adiabatic compression from state D to initial point A (P1, V1, T1). No heat exchanges occur over the adiabatic path. Work done on the system W4 = µR (T  T1 )  1 2 This cycle of operations is called a Carnot cycle. Q=0 In first two steps work is done by engine W1 and W2 are positive In last two steps work is done on gas W3 and W4 are negative The work done in complete cycle W = the area of the closed part of the P–V cycle. W = W 1 + W2 + W3 + W4  W  µRT1 l n V2 µR V µR V V  (T1  T2 )  µRT2 l n 4  (T2  T1 )  µRT1 l n 2  µRT2 l n 4 V1   1 V3   1 V1 V3 Efficiency of Carnot Engine,   W Q1 µRT1 l n    V2 V1  µRT2 l n µRT1 l n B to C and D to A are adiabatic paths so T1V2     ( – 1) = T 2 V3 ( – 1) and T1V1 ( – 1) = T 2 V4 ( – 1)  V2 V4 V3 V1 V2 V3  V1 V4 Q2 T2 T1  T2 Q  Q2 = 1  1  Q 1  T T1 Q1 1 1 Q1 Q 2  T1 T2 T1  T2 Q  Q2  10 0%    1  1 0 0% T1 Q1 The efficiency for the Carnot engine is the best that can be obtained for any heat engine. The efficiency of a Carnot engine is never 100% because it is 100% only if temperature of sink T 2 = 0 which is impossible. CARNOT THEOREM No irreversible engine (I) can have efficiency greater than Carnot reversible engine (R) working between same hot and cold reservoirs. R  I  1 – T2 Q >1– 2 T1 Q1 HEAT ENGINE Heat engine is a device which converts heat into work. Three parts of a heat engine: (i) Source of high temperature reservoir at temperature T1 (ii) Sink of low temperature reservoir at temperature T2 (iii) Working substance. source hot reservoir T1 Q1 working substance W Q2 sink cold reservoir T2 THERMAL PROPERTIES OF MATTER In a cycle of heat engine the working substance extracts heat Q1 from source, does some work W and rejects remaining heat Q2 to the sink. Efficiency of heat engine   T  T2 Q 1  Q 2 work done (W)  1 = Q1 heat taken from source (Q 1 ) T1 TYPES OF HEAT ENGINE (i) External combustion engine Steam engine is the example of external combustion engine. Its efficiency is 10 to 20% (small). (ii) Internal combustion engine The examples are petrol engine and diesel engine. There are four strokes in internal combustion engine. (i) Intake stroke or Charging stroke (ii) Compression stroke (iii) Working (or power) stroke of Expansion stroke (iv) Exhaust stroke The useful work is done in third stroke called work stroke or power stroke. The efficiency of internal combustion engine is 40 to 60%. Ex. A carnot engine working between 400 K and 800 K has a work output of 1200 J per cycle. What is the amount of heat energy supplied to the engine from source per cycle? Sol. W = 1200 J, T1 = 800 K, T2 = 400 K = 1 – Heat energy supplied by source Q 1  400 120 0 1200 T2 W   1  80 0  Q  0.5 = Q T1 Q 1 1 1 120 0 = 2400 joule per cycle 0.5 Ex. The temperatures T1 and T2 of the two heat reservoirs in an ideal carnot engine are 1500°C and 500°C respectively. Which of the following : increasing T1 by 100°C or decreasing T2 by 100°C would result in a greater improvement in the efficiency of the engine? Sol. T1 = 1500°C = 1500 + 273 = 1773 K and T2 = 500°C = 500 + 273 = 773 K. The efficiency of a carnot's engine   1  T2 T1 When the temperature of the source is increased by 100°C, keeping T2 unchanged, the new temperature of the source is T´1 = 1500 + 100 = 1600°C = 1873 K. T2 77 3 The efficiency becomes ´ 1  T '  1  18 73  0 .59 1 On the other hand, if the temperature of the sink is decreased by 100°C, keeping T 1 unchanged, the new temperature of the sink is T´2 = 500 – 100 = 400°C = 673 K. The efficiency now becomes ´´ 1  T´2 67 3 1   0 .6 2 T1 17 73 Since ´´ is greater than ´, decreasing the temperature of the sink by 100°C results in a greater efficiency than increasing the temperature of the source by 100°C. Ex. Sol. A heat engine operates between a cold reservoir at temperature T2 = 300 K and a hot reservoir at temperature T1. It takes 200 J of heat from the hot reservoir and delivers 120 J of heat to the cold reservoir in a cycle. What could be the minimum temperature of hot reservoir? W 80   0 .4 Work done by the engine in a cycle is W = 200 – 120 = 80 J.  = Q 20 0 From carnot's Theorem 0.4  1  3 00 T2 30 0 3 00   T1  500 1  0 .6  T1  0.6 T1 T1 T1 REFRIGERATOR It is inverse of heat engine. It extracts heat (Q2) from a cold reservoir, Hot reservoir T1 Q1 = Q2 + W same external work W is done on it and rejects heat (Q1) to hot reservoir. The coefficient of performance of a refrigerator.  H eat extracted from cold reservoir  Work done on refrigerator For Carnot reversible refrigerator W Q2 Q2 1   Q1 W Q1  Q 2 1 Q2 Q2 Cold reservoir T2 T2 Q 1 T1 Q 1 1      2  = Q 2 T2 W  Q1   T1  T1  T2  Q  1  T  1  2   2  Ex. A carnot engine works as a refrigrator between 250 K and 300 K. If it receives 750 cal of heat from the reservoir at the lower temperature. Calculate the amount of heat rejected at the higher temperature. Sol. T1 = 300 K Q1 Q2  T2 = 250 K T1 Q2 = 750 Q1 = ? Q1  T2 30 0  750 = 900 cal 2 50 Ex. The temperature inside and outside of refrigerator are 260 K and 315 K respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to surroundings for every joule of work done. Sol. T2 = 260 K, T1 = 315 K ; W = 1 joule Coefficient of performance of Carnot refrigerator    Q2 T2  W T1  T2 Q2 2 60 2 60 26 0    Q2  = 6.19 J 1 31 5  2 60 42 42 Ex. A refrigerator takes heat from water at 0°C and transfer it to room at 27°C. If 100 kg of water is converted in ice at 0°C then calculate the work done. (Latent heat of ice is 3.4 × 105 J/kg) Sol. Coefficient of performance (COP) = W T2 27 3 27 3   T1  T2 30 0  2 73 27 Q2 mL 100  3.4  10 5 1 0 0  3 .4  1 0 5  27    = 3.36 × 107 J COP COP 273 / 27 273