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Conducted by Efa Mai Inanignsih Objective to be able to formulate work, heat and internal energy based on the principal law of thermodynamics and apply it in problem solving to be able to apply general equation of ideal gases in isothermal, isochoric and isobaric processes to be able to analyze ideal gases process based on pressure against volume graph to be able to describe the principle of Carnot engine The phenomenon What do you think of Balloon explosion?? This is it!! If a balloon is inflated and kept in a hot place, it will finally explode, because the particles of gas in the balloon continue to expand and press the balloon wall so that when the wall is unable to stand the gas pressure, the balloon will explode Introduction Thermodynamics discuss the relationship between heat and mechanical work also the basic knowledge about temperature and heat, the influence of temperature and heat to the characteristics of substances, and kinetic theory of gases System and Environment is the is the everything everything becoming our which is observation outside the object system External Work p The figure is a tube containing ideal gas closed with a piston ∆s piston If the gas in is heated at a F constant pressure, then gas tube the gas will expand and push the piston with force F so that the piston will shift as far as s External Work (continued) W = F.s Because F = pA Then W = p.A.s Where W = mechanical work of gas (J) p = gas pressure A = section area of tube (m2) s = displacement of piston (m) Because A s = V, then the work equation above becomes W = p.V Where V is change of volume (m3) Work done by the gas is directly proportional to pressure (p) and change of volume (V) of the gas. Sample Problem A gas is compressed Solution so that the volume V1 = 5.0 L decreases from 5.0 L V2 = 3.5 L to 3.5 L at a constant p = 1.0 x 105 Pa pressure of 1.0 x 105 Pa. Calculate the because the gas is compressed, then external work W = p V applied to the gas! = p (V2 – V1) = 1.0 x 105 Pa (3.5 L – 0.5 L) = 1.0 x 105 Pa (-1.5 x 10-3 m3) = -150 J Thus, the external work applied to the gas is -150 J Exercises A number of ideal gases are heated at a constant pressure of 2 x 105 N/m2 so that the volume changes from 20 liters to 30 liters. Calculate the external work done by the gas during expansion! A gas is compressed at constant pressure 2.00 x 105 Pa from a volume of 2.00 m3 to a volume of 0.500 m3. What is the work done to the gas? If the temperature initially was 40oC, what is the final temperature of the gas? Thermodynamics Processes Isothermal Process Isobaric Process Isochoric Process Adiabatic Process Isothermal Process Isothermal process is the change process of gas state at constant temperature From the ideal gas state equation, pV = nRT obtained pV = constant, because nRT has a constant value. So pV = constant complies with then p1V1 = p2V2 the Boyle’s law p Where p1 = initial pressure V1 = initial volume p2 = final pressure V2 = final volume V External work in Isothermal Process The external work done by gas in the isothermal process can be determined from the equation pV = nRT and W = p V. W pΔV nRT W ΔV V ΔV W nRT V p W nRT ln V V2 p1 1 V1 W nRT ln V2 - ln V1 2 p2 V W nRT ln 2 V1 V1 V2 V Isobaric Process Isobaric process is the change process of gas state at constant pressure pV nRT V nR T p because p is constant and nR is always constant, then V1 V1 V2 constant or T1 T1 T2 the Gay-Lussac’s law Isobaric Process(continued) Graph of isobaric process p V External work in Isobaric Process The external work done by gas in isobaric process can be determined by equation W = p V = p (V1 – V2) it can also be determined by the area under the p – V graph (shaded region) p 1 2 p1 = p2 V1 V1 V Isochoric Process Isochoric process is the change process of gas state at constant volume pV nRT p nR T V because V is constant and nR is always constant, then p p p1 p2 constant or T T1 T2 This process complies the Gay- Lussac’s law. V graph of isochoric process External work in Isochoric Process The external work done by gas in isochoric process: W p ΔV The area under the p – V W 0 graph for isochoric process is zero In other words, in isochoric process p (constant volume), the gas does not do any external work. So the area under the p p 2 – V graph only forms a point. p1 V1 = V2 V Adiabatic Process Adiabatic process is the process change of a gas state which does not experience any transfer of heat or there is no heat entering or coming out of the system (gas) This process complies with the Poisson’s formula p V constant or p1V1 p2 V2 Where = Laplace constant = Cp/Cv Cp = specific heat of gas at constant pressure CV = specific heat of gas at constant volume Adiabatic Process(contd.) The equation above can also be expressed in another equation as follows γ γ Curvature of p – V adiabatic p1V1 p2V2 graph is steeper than isothermal nRT1 γ nRT2 γ V V1 V V2 p 1 2 adiabatic γ-1 γ-1 T1V1 T2V2 isothermal V External work in Adiabatic Process p The external work done by gas in p1 adiabatic process is expressed as follows. p2 1 W (p1V1 - p2V2 ) γ -1 V1 V2 V The external work done 3 by gas is equal to the or W nR (T1 - T2 ) area of shaded region 2 under the p – V graph Sample Problem Two moles of gas is compressed at Solution a constant Given temperature of - n = 2 mol 23oC so that its R = 8.31 J/mol K volume becomes T = (-23 + 273) K = 250 K half of the initial. Calculate the V2 = ½ V1 so V2/V1 = 1/2 external work The external work done in isothermal process done by the gas! W = V (R = 8.31 J/mol K, nRT ln 2 V1 ln 1 = 0, ln 2 = = (2 mol) (8.31 J/mol K) (250 K) (ln ) 0.69) = 4155 J (ln 1 – ln 2) = 4155 J (0 – 0.69) = -2866.95 J Thus, the external work done is -2866.95 J (-) sign indicates that at the gas is applied work. Sample Problem A monoatomic ideal Solution gas ( = 5/3) is compressed Because = 5/3 and V2/V1 = ½ adiabatically and the then volume decreases to its p1V1 = p2V2 half. Determine the ratio of the final p2 /p1 = V1 /V2 = 2 (5/3) pressure to the initial 3 pressure! =2 4 Thus, the ratio of the final to the initial pressure is 23 4 Exercise A gas occupying a room of 40 cm3 is heated at a constant pressure so that the volume becomes twice the initial. The gas pressure is 105 Pa. Calculate the external work done by the gas! Two moles of ideal gas initially has a temperature of 27oC, a volume V1 and pressure p1 = 6.0 atm. The gas expands isothermally to V2 volume and pressure p2 = 3.0 atm. Calculate the external work done by the gas! (R = 8.31 J/mole K) The First Law of Thermodynamic If an external work is applied on a system, then the temperature of the system will increase. This happens because the system receives energy from the environment. This increase of temperature relates to the increase of the internal energy. In adiabatic process, the external work applied on the ideal gas will be equal to the change of internal energy of the ideal gas. Where W ΔU or W - ΔU 0 W = external work (J) U = the change of internal energy (J) The First Law in non-adiabatic process, the gas will not only receive external work but also heat. Q ΔU W or ΔU Q - W Where Q = heat (J) This is it!!.....The first law of thermodynamics That states “Though heat energy has turned into the change of internal energy and external work, the amount of all energy is always constant”. The First Law If a system receives (absorbs) heat from the environment Q = U + (+W) = U + W If a system receives heat from the environment Q = U + (-W) = U – W environment + system Q + W - The rules of W and - Q Q values W The First Law in isothermal process In isothermal process (constant temperature), the change of internal energy U = 0, because the change of temperature T = 0. So that the first law of thermodynamics becomes ΔU Q - W 0 Q -W V2 Q W nRT ln V1 The First Law in isochoric process In isochoric process (constant volume), the work applied by gas W = 0 because the change of volume U = 0. So that the first law of thermodynamics becomes ΔU Q - W ΔU Q - 0 ΔU Q The First Law in isobaric process In isobaric process (constant pressure), the work done by gas W = p V = p (V2 – V1). So, the first law of thermodynamics becomes ΔU Q - W ΔU Q - p (V2 - V1 ) The First Law in adiabatic process In adiabatic process, the system does not receive heat or release heat, so Q = 0. Therefore, the first law of thermodynamics becomes ΔU Q -W ΔU 0 - W ΔU - W Heat Capacity Heat capacity of gas is the amount of heat energy needed to increase gas temperature by one Kelvin (1 K) or one degree Celsius (1oC). Q C Where ΔT C = heat capacity (J/K) Q = absorbed heat (J) T = the change of temperature (K) QV CV or QV CV ΔT for isochoric process ΔT Qp Cp or Qp C p ΔT for isobaric process ΔT Derivation The first law of thermodynamics can be derived as Q ΔU W Q ΔU W ΔT ΔT ΔT ΔU W C ΔT ΔT For isobaric process (p = constant) W p ΔV nRΔR 3 3 U nRT ΔU nRΔR 2 2 Monatomic gas: Diatomic gas: 3 nRΔR 7 Cp 2 nRΔR C p nR ΔT ΔT 2 3 C p nR nR 2 5 C p nR 2 For isochoric process (V = constant). Monatomic gas: Diatomic gas: W 0 5 C V nR 3 2 nRΔR 2 0 CV ΔT ΔT 3 CV nR 2 Relationship between Cp and CV : 7 5 C p - CV nR - nR 2 2 C p - CV nR Other parameters Q Molar heat capacity of gas Cm n ΔT QV CV,m or QV n CV ,m ΔT isochoric process n ΔT Qp C p,m or Q p n C p,m ΔT isobaric process n ΔT Q Specific heat of gas C m ΔT QV CV or QV mCV T for isochoric process m T 3 R 5 R CV for monatomic gas Cp 2 M 2M 5 R 7 R CV for diatomic gas Cp 2 M 2M Qp Cp or Qp mC p ΔT for isobaric process m ΔT Sample Problem One mole of gas is Solution compressed at a Because constant temperature n = 1 mol of -23oC so that its T = (-23 + 273) K = 250 K volume decreases to V2 = ½ V1 R = 8.31 J/mol K half of its initial volume. Calculate the Then work done by the gas! W = nRT ln V2/V1 = 1 x 8.31 x 250 ln ½ (R = 8.31 J/mole K; ln 1 = -1.4 x 103 joule = 0, ln 2 = 0.69) Thus, the work done by the gas is -1.4 x 103 joule. Exercise Two moles of ideal gas initially has temperature of 27oC, volume V1 and pressure p1 = 6.0 atm. The gas expands in isothermic process and reaches volume of V2 and pressure p2 = 3.0 atm. Calculate the external work done by the gas! (R = 8.3 J/mole K) (Answer : 11.5 J) 2.5 m3 of neon gas with temperature 52oC is heated in isobaric process to 91oC. If the pressure of the gas is 4.0 x 105 N/m2, determine the work done by the gas! 56 x 10-3 kg of nitrogen is heated from -3oC to 27oC. If it is heated in a free expanding vessel, then required heat of 2.33 kJ. If the nitrogen is heated in a stiff vessel (cannot expand), then the heat required is 1.66 kJ. If the relative mass of nitrogen molecules is 28 g/mole, calculated (a) the heat capacity of nitrogen, (b) the general gas constant! Thermodynamic p a Cycle Cycle means the process which c runs from the initial state and b returns to that initial state after gas does work V1 V2 V Random cycle in p – V diagram In the process a – b, the gas expands in adiabatic process and the work done by the gas is the area of plane abV2V1, its value is negative. In process b – c the compressed gas in isothermal process is the area of plane bcV1V2, its value is positive. In process c – a the gas does not do any work because its volume is constant. The process c – a is an isochoric process which is done to make the gas returns to its initial state. The total external work done by the gas in one cycle a – b – c – a is the area of abca. Wabca =Wab + Wbc + Wca area of abV2V1 + (-area of bcV1V2) + 0 Wabca =area of abca A thermodynamics cycle can occurs in a heat engine, such as otto engine (Otto cycle), diesel engine (diesel cycle), steam engine (Rankine cycle), and carnot engine Thermodynamic Cycle : Carnot engine High temperature reservoir T1 Carnot engine is assumed as an ideal heat engine which works cyclically and reversible Q1 between two temperatures without any loss of energy. Within one cycle, the gas returns to Heat engine W = Q1 - Q2 its initial state, so there is no change of internal energy (U = 0). Q = U + W Q1 Q1 – Q2 = 0 + W W = Q1 – Q2 Low temperature reservoir T2 imaginary Carnot engine Carnot cycle Work process of Carnot engine to produce Carnot cycle Entire of process in the Carnot can be represented in pressure (P) against volume (V) graph: Efficiency Efficiency of engine W η 100% Q1 In the Carnot engine, holds W = Q1 – Q2 Q1 - Q2 η 100% Q1 Q2 T2 η 1 100% Q 1 100% 1 T 1 p Sample Problem 90 C 60 The figure below Pressure (N/m 3 ) indicates the 30 A B thermodynamics change V O D E of system from iitial Volume (m 3 ) state A to B and C and back to A. If VA = 0, VB Determine = 30 joule and heat a. the system internal energy given to the system in in state C, process B C = 50 J b. the heat given to the system in A B process, and c. the heat given to the system or taken from in C A process. Solution a. The heat which is given away to the system in B C process, is QBC = +50 joule. QBC = +50 joule WBC = 0, because B C process is isochoric Use the first law of thermodynamics QBC = UBC + WBC UBC = QBC – WBC UBC = 50 – 0 = 50 J UBC = UC – UB UC = UBC + UB UC = 50 + 30 = 80 J Thus, the system internal energy in state C is 80 J. b. Process from A B WAB = area of ABED = AB x BE = 2 x 30 = 60 J UAB = UB – UA = 30 – 0 = 30 J QAB is calculated by using the first law of thermodynamics QAB = UAB + WAB QAB = 30 + 60 = 90 J Thus, the heat given in A B process is 90 J. c. Process from C A WCA = -area of ACED = -(area of ABED + area ABC) AB BC 2 60 60 60 2 2 = -120 J UCA = UA – UC = 0 – 80 = 80 J QCA is calculated by using the first law of thermodynamics. QCA = UCA + WCA = -80 + (-120) = -200 J Thus, the energy taken in C A process is -200 J. The Second Law of Thermodynamics is a restriction of the first law of thermodynamics which expresses the energy conservation It is states: “energy cannot be created or destroyed but can only change from one form to another” Rudolf Clausius: Heat flows spontaneously from an object of high temperature to an object of lower temperature and it does not flow spontaneously in the opposite direction without external work. Entropy The total entropy of the universe does not change when a reversible process occurs (Suniverse = 0) and increase when the irreversible process occurs (Suniverse > 0). entropy is a measurement of the amount of energy or heat which cannot changed into work Q ΔS with S = the change of entropy (J/K) T the total change of entropy of the Carnot engine is Q2 Q1 ΔS 2 ΔS1 - 0 T2 T1 Kelvin and Planck formulate the second law of thermodynamics about heat engine that it is impossible to make an engine with 100% efficiency Principle of cooler engines: is flowing heat from the cool reservoir T2 to the hot reservoir T1 by exerting external effort on the system. The magnitude of external work needed in a cooler engine is formulated as W Q1 - Q2 Where : Q1 = heat absorbed from low temperature Q2 = heat given at high temperature Sample Problem A motor operates a cooler engine for producing ice. Q2 heat is taken from a cooling room which contains an amount of water at 0oC and Q1 heat is given away to the air around it at 15oC. Suppose the cooler engine has a coefficient of performance of 20% of the coefficient of performance of an ideal cooler engine. a. Calculate the work done by the motor to make 1 kg of ice? (ice latent heat is 3.4 x 105 J/kg) b. What is time required to make 1 kg of ice if the power of the motor is 50 W? Solution The scheme of the cooler engine T1 = 15 + 273 = 288 K m = 1 kg Lice = 3.4 x 105 J/kg Q1 T1 = 15 + 273 = 298 K T2 = 0 + 273 = 273 K Cooler W from motor Cp engine = 20% x Cp ideal p = 50 W Q2 T2 T2 = 0 + 273 = 273 K Cp ideal T1 - T2 273 K 273 288 K - 273 K 15 20 273 91 Cp engine = 20% x Cp ideal 100 15 25 Q2 The work done by electric motor (W) is Cp engine W Q2 W m Lice / Cp engine 9.3 x 104 J C p engine Thus, the work done by the electric motor (W) is 9.3x104 J. b. The time required to make 1 kg of ice is work W t 1860 seconds 31 minutes power P Thus, the time required to make 1 kg of ice is 31 minutes. Exercises An ideal refrigerator has coefficient of performance of 5.0. If the room temperature outside the refrigerator is 27oC, what is the lowest temperature in the refrigerator which can be obtained? The coefficient of performance of a refrigerator is 4.0. What is the electric energy used to transfer 4000 joule of heat from the food in the refrigerator? That’s all!!! Thanks, ……bye bye

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posted: | 8/16/2011 |

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