Thermodynamics

THERMODYNAMICS 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 deal 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.  Thermodynamic System The system which can be represented in terms 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 thermodynamic 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.  Surrounding Anything outside the system, which exchanges energy with the system and which tends to change the properties of the system is called its surrounding.  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.  Homogeneous System A system is said to be homogeneous if it is completely uniform throughout. e.g. Pure solid or liquid.  Isolated System A system in which no exchange of matter and energy with the surrounding take place, is said to be an isolated system.  Universe The system and Its surrounding are together known as the universe.  Thermodynamic variables of the system (i) Composition () (ii) Temperature (T) (iv) Pressure (P) (v) Mass (iii) Volume (V)  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 state of the system can be described by the remaining three variables namely temperature, pressure and volume. These variables are inter related by equation PV = RT. The thermodynamic state of the system is its condition as identified by two independent thermodynamic variables (P, V or P, T or V, T).  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 [ideal gas] cv = Specific heat at constant volume and dT = Infinitesimal change in temperature m = Mass of system M = Molecular weight m Molar heat capacity Cv = Mcv For -moles of ideal gas dU = CvdT = CvdT M Internal energy in the absence of inter-molecular 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  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 [ideal gas] i cv = Specific heat at constant volume and dT = Infinitesimal change in temperature m = Mass of system M = Molecular weight m molar heat capacity Cv = Mcv For -moles of ideal gas dU = CvdT = CvdT M Internal energy in the absence of inter-molecular 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 Positive Heat lost by a system Negative 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 = RT 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.   Cyclic process Cyclic process is a thermodynamic process in which the system returns to its initial stage after undergoing a series of changes.  Non-cyclic process Non-cyclic process is a process in which the system does not return to its initial stage.  Quasi-static or equilibrium process Quasi-static is a thermodynamic process which proceeds extremely slowly such that a every instant of time, the temperature and pressure are the same in all parts of the system.  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 realized in practice. The process which is not reversible is the irreversible process. In nature the processes are irreversible. Work done by thermodynamic system One of the simple example of a thermodynamic system is gas in a cylinder with a movable piston. • If the gas expands against the piston Gas exerts a force on the piston and displace it through a distance and does work on the piston. • If the piston compresses the gas When piston moved inward, work is done on the gas. • The work associated with volume changes If pressure of gas on the piston = P. Then the force on the piston due to gas is F = PA When the piston is pushed outward an infinitesimal distance dx, the work done by the gas is dW = F x dx = PA dx The change in volume of the gas is dV = Adx,  dW = PdV 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 Vf dW   PdV Vi 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 (ii) Volume is increasing (iii) Volume is decreasing (iv) Cyclic process Work Done in Clockwise Cycle Wcyclic = WAB (positive) + WAB (negative) = First law of thermodynamics Let a gas in a cylinder with a movable piston changes froth an initial equilibrium state to a final equilibrium state. System change its state through path 'a' : The heat absorbed by the system in this process = Qa The work done by the system = Wa 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 (Uf) 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 • This law is applicable to every process in nature • The first law of thermodynamics introduces the concept of internal energy. • The first law of thermodynamics is based on the law of conservation of energy. • Q, dU and W must be expressed in the same units (either in units of work or in units of heat). • This law is applicable to all the three phases of matter, i.e., solid, liquid and gas. • dU is a characteristic of the state of a system, it may be any type of internal energytranslational kinetic energy, vibrational, rotational kinetic energy, binding energy etc.   Application of First Law of Thermodynamics 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 = U + PV, where V is volume change during phase change. 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 100ºC. 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) Illustrations Illustration 53. The pressure of one mole monoatomic gas increases linearly from 4 × 105 Nm–2 to 8 × 10+5 Nm–2 when its volume increases from 0.2 m3 to 0.5 m3. Calculate. (i) Work done by the gas, (ii) Increase in the internal energy, (iii) Amount of heat supplied, (iv) Molar heat capacity of the gas –1 –1 R = 8.31 J mor K Solution: P1 = 4 × 105 Nm–2 P2 = 8 × 10+5 Nm–2, V1 = 0.2 m3, V2 = 0.5 m3 (i) Work done by the gas = Area under P–V graph (Area ABCDEA) 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 C (ii) Increase in internal energy U = CV (T2 – T1) = V (P2V2 – P1V1) R 3 For monoatomic gas CV = R 2 3 3  U = [(8 × 105 × 0.5) – (4× 105 × 0.2)] = [4 × 105 – 0.8 × 105] = 4.8 × 105 J 2 2 (iii) Q = U + W = 4.8 × 105 + 1.8 × 105 = 6.6 × 105 J QR 6.6 105  8.31 Q QR   (iv) C= = = 17.14 J/mol K (P2 V2  P1V1 ) 1  3.2 10 5 T RT Illustration 54. As shown in figure when a system is taken from state a to state b, along the path A  c  b, 60 J of heat flow into the system and 30 J of work is done : (i) How much heat flows into the system along the path a  d  b if the work is 10 J. (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. Solution: For the path abc U = Q – W = 60 – 30 = 30 J or Ub – Ua = 30 J (i) Along the path adb Q = U + W = 30 + 10 = 40 J (ii) Along the curved path ba Q = (Ua – Ub) + W = (–30) + (–20) = –50, heat liberates from system (iii) Qad = Ud – Ua + Wad but Wad = Wadb – Wdb = 10 – 0 = 10 Hence and Qdb = Ub – Ud + Wdb = 30 – 22 + 0 = 8 J Qad = 22 – 0 + 10 = 32 J Illustration 55. 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 vaporization of water is 5.4 × 105 cal kg–1? Solution: 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 1.686 155 In expansion work done, W = PV = 1.01 × 105 × 1.67 J = cal = 4.015 × 104 cal 4.2 But U = Q – W (first law of thermodynamics) or U = (5.4 × 105 – 0.4015 × 105) cal = 4.9985 × 105 cal 1. BEGINNER’S BOX - 7 One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as shown in fig. calculate (a) (b) (c) The work done by the gas. The heat rejected by the gas in the path CA and heat absorbed in the path AB. The net heat absorbed by the gas in the path BC. 2. Calculate the heat supplied in figure shown for complete cycle. 3. Given figure shows a P–V graph of the thermodynamic behaviour of an ideal gas. Find out work in the processes A  B, B  C, C  D and D  A form this graph. 4. At 8 atm pressure a monoatomic, gas expands from 2 litre to 5 litre then calculate the following (a) Work done by the gas (b) Increase in internal energy (c) Amount of heat supplied 5. A gas is taken through a cyclic process ABCA as shown in figure. If 2.4 cal of heat is given in the process. Calculate the value of J. 6. 6.1 DIFFERENT PROCESSES Isometric or lsochoric 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 /T = constant (P and T are variable, V is constant) Work done In this process volume remains constant so dV = 0  W =  vf vi PdV = 0 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 =  Cv T dP Slope of the P–V curve =  dV Specific heat or molar heat capacity at constant volume (CV) The quantity of heat required to raise the temperature of 1 gram mole of 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.   6.2 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.  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 /T = Constant or V  T  Work done In this process pressure remain constant Work done W =  Vf Vi PdV = P(Vf – Vi)  Fraction of Heat given which – (i) is converted into internal energy (ii)  dP = 0 U 1  Q  W 1 does work against external pressure  1 Q   Form of first Law  Slope of the PV curve :  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. Ex. Heating of water at atmospheric pressure. • Melting of solids and boiling of liquids at atmospheric pressure. 6.3 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 (RT) (T is constant)  Q = U + P (Vf – Vi)  CP (Tf – Ti) =  Cv (Tf – Ti) + 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.  dP  =0    dV isobaric  Work Done Consider  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 (P1, V1) to the final state v2 RT (P2, V2). Work done : W =  PdV  PV = RT then P = v1 V v 2 RT v 2 dV V  v  W = = RT  log e V v2 = RT [loge V2 – loge V1] = RT loge  2  dV = RT  1 v1 v1 V V  V1  P   W = 2.303 RT log10  1  [  P1V1 = P2V2 ]  P2  Form of First Law There is no change in temperature and internal energy of the system depends on temperature only V  So U = 0, Q = 2.303 mRT log10  2   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 For an isothermal process, PV = constant Differentiating, PdV + VdP = 0  VdP = –PdV  dP P  dV V P  dP  Slope of isothermal curve,    V  dV  isothermal        For a given system : The product of the pressure and volume of a given mass of a perfect gas remains constant in an isothermal process. Boyle's law is obeyed in an isothermal process . A graph between pressure and volume of a given mass of a gas at constant temperature is known as isotherm or isothermal curve of the gas. Two isotherms for a given gas at two different temperatures T1 and T2 are shown in figure. 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 =  RT1 for temperature T1 and PV =  RT2 for temperature T2. It means T1 = T2 which is not possible. An ideal gas enclosed in a conducting cylinder fitted with a conducting piston.     6.4 Let the gas be allowed to expand very-very slowly. 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. The temperature of a substance remains constant during melting. So, the melting process is an isothermal process. Boiling is an isothermal process, when a liquid boils, its temperature remains constant. If sudden changes are executed in a vessel of infinite conductivity then they will be isothermal. 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 = RT Equation for adiabatic process PV = constant, TV–l = constant, T p1– = constant 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.) V2  V 1  K  V21  V11  Q K  P1V1  P2 V2  So W =  PdV  K  V dV  K   = V1 V1    1 V1 (  1) 1 1  P1V1 V1 .V1  P2 V2 V2 .V2  = W = [P1V1 – P2V2] (   1) (   1) R W = (T1 –T2) ( PV = RT) (   1) V2 V2  From 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 T1 > T2. 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. Slope of the adiabatic curve For an adiabatic process, PV = constant Differentiating,  P V–1 dV + V dP = 0 dP PV 1 P  P   V dP = –  PV–1 dV  = – =      V dV V  V P  dP   Slope of adiabatic curve,   V  dV  adiabatic Magnitude of slope of adiabatic curve is greater then the slope of isotherm dP P dP slope of adiabatic changes     slope of isothermal change dV adia V dV iso Since  is always greater than one so an adiabatic curve is steeper than an isothermal curve      Examples of adiabatic process 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 from the cylinder. Consequently, there will be an increase in the temperature of the gas. If a gas is suddenly expanded by moving the piston outwards, there will be a decrease in the temperature of the gas. Bursting of a cycle tube. Propagation of sound waves in a gaseous medium. In diesel engines burning of diesel without spark plug is done due to adiabatic compression of diesel vapour and air mixture Free Expansion Take a thermally insulated bottle with ideal gas at some temperature T1 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. 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 apply the first law of thermodynamics to the free-expansion process. The process is adiabatic because of the insulation, So Q = 0. No part of the surroundings moves so the system does no work on its surroundings.  For ideal gas (W)ext. = Work done against external atmosphere = P dV = 0 ( P = 0)  because vacuum (W)int. = Work done against internal molecular forces = 0 Q = dU + W 0 = dU + 0 The internal energy does not change 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. Relation between degrees of freedom and specific heat of gas Energy related with each degree of freedom for a gas molecule = 1/2 kT, Energy related with all degree of freedom for a gas molecule = f/2 kT Internal energy of one mole of ideal gas (total K.E.) U = f/2 kT for Isometric process (volume constant) W= 0 dU By first law of thermodynamics Q = Q + dU  CV dT = dU  CV = dT C R R 2 dU f f  CV = = R= , CP = CV + R =   1 R = and  = P  1  CV f dT 2  1  1 2  R R 2 CV = , CP = and  = 1 + f  1  1 For isobaric process P = constant so For isothermal process, PV = constant so R R +  1 1  x R x = 0  C = CP = + R = CV + R  1 x = 1  C =  For adiabatic process PV = constant so x= General expression for C(CP or CV) in the process PVx = constant C = C=0 Values of f, U, CV, CP and  for different gases are shown in table below. Atomicity of gas Monoatomic Diatomic, Triatomic and Triatomic linear (at normal temperature) Poly atomic Triangular Non-linear       f 3 5 6 CV CP 3 R 2 5 R 2 6 R = 3R 2 5 R 2 7 R 2 8 R = 4R 2  5 = 1.67 3 7 = 1.4 5 4 = 1.33 3 1<<2 If atomicity of gases is same U, CP, CV and  is same for gas mixture. 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. If 'f' is the degree of freedom per molecule for a gas, then fkT * Total energy of each molecule = 2 f f * Total energy per mole of gas = N0 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. fRT For '' mole of a gas : Internal energy at temperature T is U = = CVT 2  Change in internal energy is given by dU = f R (dT) = CvdT. This change is process 2 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. Mayer’s formula : CP – CV = R  At constant pressure dQ = CPdT, dU = CVdT & dW = PdV = RdT Now from first law of thermodynamics dQ = dW + dU  CPdT = RdT + CVdT  CP = R + CV  CP – CV = R  GOLDEN KEY POINTS 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 . First Law of Thermodynamics Applied to Different Processes Illustrations Illustration 56. An ideal gas has a specific heat or molar heat capacity at constant pressure C P = 5R/2. The gas is kept in a closed vessel of volume 0.0083 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. Solution: 5R 3R CV = CP – R = –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 PV 1.6 106  0.0083 16   From gas equation n = RT 8.3  300 3 4 2.49 10  6 U  U = nCVT  T = = = 375 K C v 3  8.3 16 Final temperature = 300 + 375 = 675 K 1.6 106  675 T According to pressure law P  T  P2 = 2 × P1 = = 3.6 × 106 Nm–2 t1 300 Illustration 57. 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. Solution:  CV = CP – R = 8 – 2 = 6 cal/mole °C  Heat absorbed by 5 moles of oxygen at constant volume  = CV  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 Illustration 58. 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– K–1. Solution: M 2 10 2 5 7   & Cp = R Here m = 2 × 10 kg,  = 3 M W 28 10 7 2 5 7 Q = CpT =  × 8.3 × 45 = 933.75 J 7 2 –2 Illustration 59. Two moles of a gas at 127°C expand isothermally until its volume is doubled. Calculate the amount of work done . Solution: V  = 2, T = 127 + 273 = 400K, 2 = 2 V1 V From formula W = 2.3026 RT log10 2 = 2.3026 × 2 × 8.3 × 400 × log10 2 V1 = 2.3026 × 2 × 8.3 × 400 × 0.3010  4.6 × 103 J Illustration 60. Figure shows a process ABCA performed on an ideal gas. Find the net heat given to the system during the process. Solution: 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. The work done in part AB is W1 = 0 (the volume remains constant). The part BC represents an V isothermal process so that the work done by the gas during this part is W2 = RT2 n 2 V1 RT During the part CA : V  T So, V/T is constant and hence, P = is constant V The work done by the gas during the part CA is W3 = P(V1 – V2) = RT1 – RT2 = – R (T2 – T1).   V The net work done by the gas in the process ABCA is W=W1+W2+W3=R T2 l n 2  (T2  T1 )  V1   The same amount of heat is given to the gas. Illustration 61. During an experiment an ideal gas is found to obey an additional law VP2 = constant. The gas is initially at temperature T and volume V. What will be the temperature of the gas when it expands to a volume 2V. Solution: By gas equation PV = RT and VP2 = constant on eliminating P  A    V = RT   V V R T A  T   1  V2  T2  V1 V T   T' = 2V T '  2 T Illustration 62. 2m3 volume of a gas at a pressure of 4 × 105 Nm–2 is compressed adiabatically so that its volume becomes 0.5 m3. Find the new pressure. Compare this with the pressure that would result if the compression was isothermal. Calculate work done in each process.  = 1.4 Solution: V1 = 2m3, P1 = 4 × 105 Nm–2, V2 = 0.5 m3 1.4  2  P1V1  P2 V2  P2 = 4 × 105  = 4 × 105 (4)1.4 = 2.8 × 106 Nm–2   0.5  P V 4 105  2 In isothermal process P1V1 = P2V2  P2 = 1 1  = 1.6 × 106 Nm–2. V2 0.5 Now work done in adiabatic process (4 105  2  2.8 106  0.5) PV P V W= 1 1 2 2 = = –1.48 × 106 J 1.4  1  1 V V Work done in isothermal process W = 2.303 RT log 2 = 2.3026 P1V1 log 2 V1 V1 In adiabatic process  0.5  1 = 2.3026 × 4 × 105 × 2 × log   = 2.3026 × 4 × 105 × 2 log   = –1.1 × 106 J  2.0  4 Illustration 63. Two samples of a gas initially at same temperature and pressure are compressed from a V volume V to . One sample is compressed isothermally and the other adiabatically. In which 2 sample is the pressure greater? Solution: V Let initial volume, V1=V and pressure, P1 = P , final volume, V2 = and final pressure, P2= ? 2 P1V1 PV For isothermal compression P2V2 = P1V1 or P2 = = 2P  V V2 2 For adiabatic compression   V   V  P = P1  1   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. ' 2 Illustration 64. Plot P – V , V – T graph corresponding to the P – T graph for an ideal gas shown in figure. Explain your answers. Solution: 1 V 1 For process CD ; T = constant so V  P For process AB ; T = constant so P  For process BC P = constant so V  T For process DA P = constant so V  T Illustration 65. The specific heat of argon at constant volume is 0.075 kcal/kg K. Calculate its atomic weight, [R = 2cal/mol K] Solution: As argon is monoatomic, its molar specific heat at constant volume will be 3 3 CV = R = × 2 = 3 cal/mol K, CV = Mw cV and cV = 0.075 cal/g K 2 2 3 So 3 = Mw × 0.075  Mw = = 40 gram/mole 0.075 Illustration 66. Two moles of helium gas ( = 5/3), assumed ideal are initially at 27°C and occupy a volume of 20 litres. The gas is first expanded at constant pressure till its volume is doubled. It then undergoes an adiabatic change until the temperature return to its initial value. Determine the final pressure and volume of the gas. Also, calculate the work done under isobaric and adiabatic processes. [R = 8.3 J mor–1 K–1] [AIPMT (Mains) 2005] Solution: For process AB For process BC  (P = constant) TV–1 = constant T2 = 2T1 = 600K ( V2 = 2V1) Therefore (2T1) (V2)5/3–1 = (T1) (V3)5/3–1 V3 = (V2) (2)3/2 = 2 2 V2 = 80 2 × 10–3 m3 = 80 2 litres RT3 (2)(8.3)(300)  P3 = = 0.44 × 105 N/m2 3 V3 80 2  10 Work done under isobaric process = PV = P1(V2 – V1) RT1 2  8.3  300 = (V2 – V1) = (40 × 10–3 – 20 × 10–3) = 4980 J V1 20  10 3 Work done under adiabatic process = 1. R(T2  T3 ) 2  8.3  (300) = 7470 J  (   1) (5 / 3  1) BENGINNER’S BOX - 8 32 gram of oxygen gas at temperature 27°C is compressed adiabatically to 1/3 of its initial volume. Calculate the change in internal energy ? (y = 1.5 for oxygen) 2. How much heat energy should be added to the gaseous mixture consisting of one gram of hydrogen and one gram helium to raise its temperature from 0°C to 50°C. (a) at constant volume and. (b) at constant pressure, assuming both the to be ideal. R = 2 cal/mol-K,  H 2 = 1.41,  He = 1.67 3. Match the column-I with column-II 4. Match the column-I with column-II Column-I Column-II (A) Isobaric process (p) No heat exchange (B) Isothermal process (q) Constant pressure (C) Isoentropic process (r) constant internal energy (D) Isochoric process (s) Work done is zero 5. Match the column-I with column-II Column-I Column-II (A) Adiabatic expansion (p) Q = 0 (B) Adiabatic free expansion (q) U = 0 6. (C) Isobaric expansion (r) W = 0 (D) Isothermal expansion (s) Q = W + U (E) Isochoric cooling An ideal gas expands isothermally along AB and does 700 J of work. (a) How much heat does the gas exchange along AB. (b) The gas expands adiabatically along BC and does 400 J of work. When the gas returns to A along CA, it exhusts 100 J of heat to its surroundings. How much work is done on the gas along this path. 7. For one mole of an ideal gas at 300 K occupies a volume of 0.36 m3 at a pressure of 2 atm. The gas expands adiabatically until its volume becomes 1.44 m3. Next gas is compressed isobarically to its original volume. Finally the pressure is increased isochorically until the gas return to its. initial state. (see figure). (a) Determine the temp. at the point B and C (b) The work done during the process A to B J (Assume  = 1.5, R = 8 ) mol  K 8. Following figure is given for 1 mole gas. Then find out net work done in the cyclic process. EFFECIENCY OF A CYCLE () : Total Mechanical work done by the gas in the whol e process = Heat absorbed by the gas (only  ve) Area of the P  V cycle = Heat injected into the system HEAT ENGINE Heat engine is a device which converts heat into work. Three parts of a heat engine: (l) Source of high temperature reservoir at temperature T1 (ii) Sink of low temperature reservoir at temperature T2 (iii) Working substance . In a cycle of heat engine the working substance extracts heat Q1 from source, does some work and rejects remaining heat Q2 to the sink. Efficiency of heat engine = T  T Q  Q2 woek done (W)  1 2  1 heat taken from source (Q) T1 Q1 REFRIGERATOR It is inverse of heat engine. It extracts heat (Q2) from a cold reservoir, some external work W is done on it and rejects heat (Q1) to hot reservoir. The coefficient of performance of a refrigerator (C.O.P. or ). Heat extracted from cold reservoir Q Q2 1  = = 2  Work done on refrigerator W Q1  Q 2 Q1  1 Q2 Q1 T1 Q 1 1  For Carnot reversible refrigerator = 2   Q 2 T2 W  Q1   T1   Q  1  T  1  2   2  T2  = T1  T2 W Relation between  and  :  = [Efficiency of heat engine] Q1 Q (Refrigerator and heat engine = 2 [C.O.P. of refrigerator] W Are maintained at same thermal reservoirs) and Q1 = Q2 + W 1 then  =  1  SECOND LAW OF THERMODYNAMICS The first law of thermodynamics is the principle of conservation of energy. Common experience shows that there are many conceivable processes that are perfectly allowed by the First law and yet are never observed. For example, nobody has ever seen a book lying on a table jumping to a height by itself. Would be possible if the principle of conservation of energy were the only restriction. This principle, which disallows many phenomena consistent with the first law of thermodynamics is known as the second law of thermodynamics. The Second Law of Thermodynamics gives a fundamental limitation to the efficiency of a heat engine and the co-efficient of performance of refrigerator. Second Law of Thermodynamics Kelvin Planck Statement No process is possible whose sole result is the absorption of heat from a reversoir and the complete conversion of the heat into work. Claussius Statement No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. It can be proved that the two statements above are completely equivalent. The second law of thermodynamics is a general principle which places constrains upon the direction of heat transfer and the attainable efficiencies of heat engines. Entropy Statement of the second Law of Thermodynamics (i) The entropy of the Universe never decrease. It either increases (for irreversible processes) or remains the same (for reversible processes) • REVERSIBLE AND IRREVERSIBLE PROCESS Reversible Process A reversible process is defined as a process that can be reversed without leaving any trace on the surroundings. Both the system and the surroundings are returned to their initial status at the end of the reverse process. Conditions for a Reversible process (a) The process should take place very slowly, i.e. quasi-statically, i.e. seemingly static (from the Latin word qasi meaning 'as if') so that it satisfies the following requirement at each stage of the process. • The system should be in mechanical equilibrium. • The System be in thermal equilibrium. • The system should be in chemical equilibrium, i.e., no new products should be formed. (b) There should be no friction losses etc. It should be remembered that a complete reversible process or cycle of operations is only an ideal case. In actual practice, there is always a loss of heat due to friction, conduction and radiation. However, it is possible to approximate reversible processes through carefully controlled procedures, but they can never be achieved. Examples • Frictionless pendulum. • Qusai-equilibrium expansion and compression of a gas. (ii) Irreversible Process A process that is not reversible is called an irreversible process. The spontaneous processes occuring in nature are irreversible. In fact irreversibility is a rule rather than an exception in nature. Examples (i) Sudden unbalanced expansion of a gas. (ii) Heat produced by friction. (iii) Heat generated when a current flows through an electric resistance in any direction. (iv) Cooking gas leaking from a cylinder. (v) Diffusion of liquids or gases. (vi) Breaking of an egg. (vii) The growth of a plant. A system can be restored to its initial state following a process, regardless of whether the process is reversible or irreversible. But for reversible process, this restoration is made without leaving any net change on the surroundings whereas for irreversible processes, the surroundings usually do some work on the system and therefore will not return to their original state. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS To have a better understanding of thermodynamic process and the second and third laws of thermodynamics, let us introduce the concept of entropy. It was introduced by clausius in 1865. A quantity that denotes the amount of disorder is called entropy and is denoted by S. The total energy always conserved in any process but the total entropy always increases or remains the same in any process (ie., the disorder increases or remain constant) dQ dQ  S =  T T For reversible processes, S = 0. For irreversible process S > 0 A process where S < 0 is not possible. dS = 3.  Carnot cycle Carnot 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 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.  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.  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.  Stand : It is made by perfectly insulating material. When cylinder placed on it working substance can expended or compressed adiabatic.  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.  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) 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 = Q1 = V 100  3.4 105  27 dV  RT1l n 2 V1 273  Adiabatic expansion B  C 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 T2K 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. R Work done in adiabatic path BC is W2 = (T1 –T2)  1  Isothermal compression C  D Now the gas cylinder is placed in contact with sink at temperature T2.The piston is moved slowly inward so that heat produced during compression passes lo the sink. The gas is isothermally compressed from C to point D (P4, V4, T2). The heat reject Q2 to the cold reservoir (sink) at T2 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 = RT2n  4   V3   V3   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 R on the system W4 = (T2 – T1).  1 This cycle of operations is called a Carnot cycle. 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 = W1 + W2 + W3 + W4 V V V V R R  W = RT1n 2  (T1 – T2) + RT2n 4  (T2 – T1) = RT1n 2 +RT2n 4 V1   1 V3   1 V1 V3 Efficiency of Carnot Engine,  = W h= Q1 B to C and D to A are adiabatic paths V2 V  RT2 l n 4 V1 V3 V RT1l n 2 V1 RT1l n so T1V2( 1)  T2 V3( 1) and T1V1( 1)  T2 V4( 1)  = = V2 V3  V1 V4 T1  T2 Q1  Q 2 Q T Q Q    = 1 – 2 = 1 – 2 1  2 T1 Q1 Q1 T1 T1 T2 T1  T2 × 100% T1  = Q1  Q 2 × 100% 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 T2 = 0 which is impossible. Carnot Theorem No irreversible engine (I) can have efficiency greater than Carnot reversible engine (R) T Q working between same hot and cold reservoirs. R > I  1  2  1  2 T1 Q1 Illustrations Illustration 67. 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 supped to the engine from source per cycle ? Solution: T W 400 1200  W = 1200 J, T1 = 800 K, T2 = 400 K   = 1 – 2  1– T1 Q1 800 Q1 1200  0.5 = Q1 1200 Heat energy supplied by source = Q1 = = 2400 joule per cycle 0.5 Illustration 68. The temperatures T1 and T2 of the two beat 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? Solution: T1 = 1500°C = 1500 + 273 = 1773 K and T2 = 500°C = 500 + 273 = 773 K. T The efficiency of a Carnot’s engine  = 1 – 2 T1 When the temperature of the source is increases by 100°C, keeping T2 unchanged the new temperature of the source is T1' = 1500 + 100 = 1600°C = 1873 K. T 773 The efficiency becomes ' = 1 – 2' = 1 – = 0.59 T1 1873 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 T2' = 500 – 100 = 400°C = 673 K. The efficiency now becomes T2' 673 =1– = 0.62 T1 1773 Since '' is greater then ', decreasing the temperature of the sink by 100°C results in a greater efficiency than increasing the temperature of the source by 100°C. '' = 1 – Illustration 69. 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? Solution: W 80 Work done by the engine in a cycle is W = 200 – 120 = 80 J. = = 0.4  Q 200 T 300 300 300  From Carnot’s Theorem 0.4  1 – 2 = 1 –  0.6  T1   T1  500 T1 T1 T1 0.6 Illustration 70. A Carnot engine works as refrigerator 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. Solution T1 = 300 K T2 = 250 K Q2 = 750 Q1 = ? Q1 T1 300  Q1 = × 750 = 900 cal Q 2 T2 250 Illustration 71. 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. Solution: T2 = 260 K, T1 = 315 K ; W = 1 joule Q T2 Coefficient of performance of Carnot refrigerator  = 2  W T1  T2 Q 260 260 260   2   Q2 = = 4.73 J So Q1 = Q2 + W = 5.73 J 1 315  260 55 55 Illustration 72. 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) Solution: T2 273 273   Coefficient of performance (COP) = T1  T2 300  273 27 W= 100  3.4 10 5 100  3.4 105  27 Q2 mL  = = = 3.36 × 106 J COP COP 273 / 27 273 BEGINNER’S BOX - 9 1. An ideal engine which works between 27°C & 227°C takes 100 calorie of heat in one cycle. Calculate the work done in one cycle ? 2. The coefficient of performance of a Carnot refrigerator is 5/3. If the refrigerator operating at a intake temperature –23°C then what will be the exhaust temperature ? 3. A Carnot engine absorbs 1000 J of heat from a reservoir at 127°C and reject 600 J of heat during each cycle. Calculate (a) The efficiency of engine .. (b) The temperature of the sink (c) The amount of the useful work done during each cycle. 4. Two Carnot engine A and B use one after another. Engine A absorb heat from source at T1 = 800 K and release heat to sink of temp T2K. Second engine B absorb heat which release by first (at T2K) and release heat at another sink at temperature T3 = 300 K. If work done by two engine is same than calculate value of T2. 5. A Carnot engine, efficiency 40% and temperature of sink 300 K to increase efficiency up to 60% calculate change in temperature of source ? 6. An ideal refrigerator is working between temperature 27°C and 127°C. If it expels 120 calorie of heat in one second then calculate its wattage ?