The laws of Thermodynamics. Work in thermodynamic processes
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1 The laws of Thermodynamics ork in thermodynamic processes The work done on a gas in a cylinder is directly proportional to the force and the displacement. = F y = PA y It can be also expressed in terms of the pressure excerted by the piston and the variation in volume. = P V Compression: work is positive Expansion: work is negative 1
2 ork: questions A monoatomic gas is expanded from an initial state with volume V=1 L to a final state 4V. This process happens at constant pressure P=1 atm. Next, the pressure on the gas is increased at constant volume 4V to 2 atm. 1) The work done during the first process is: a) positive, b) negative c) zero 2) The work done during the second process is: a) positive, b) negative c) zero 3) hat is the net work and sketch the processes in a graph of P versus V. PV diagrams The work done on a gas when taking it from an initial thermodynamic state (I) to a final thermodynamic state (f) can be determined by the area below the PV diagram. 2
3 PV diagrams Isobaric process: constant pressure Isovolumetric process: constant volume The work depends not only on the initial and final state but also on the path taken. PV diagrams: question time In which case if there more work done on the system: a) A b) B c) Both the same 3
4 First law of thermodynamics The change in internal energy of the system is the sum of the heat transferred to the system and the work done on the system U U f Ui = Q + First law of Thermodynamics Isolated systems: =0 and Q=0 U f = U i Cyclic processes: if the process ends up in the same thermodynamics state as it started there will be no change in its internal energy U f = U i = Q 4
5 First law of Thermodynamics Isothermal processes: if the temperature is constant over the process 3 U = nrt a U 2 env f = U V f = nrt ln V i = Q i Adiabatic processes (Q=0): if the energy transferred by heat in a process is zero then U = First law of thermodynamics: question hich of the following processes correspond to 1. Isobaric 2. Isothermal 3. Isovolumetric 4. Adiabatic 5
6 First law of thermodynamics: example An ideal mono-atomic gas is confined in a cylinder by a movable piston. The gas starts at P=1.00atm, V=5.00L and T=300K. An isovolumetric process raises the pressure to 3 atm. Then an isothermal expansion brings the system back to 1atm. Finally an isobaric compression at 1atm completes the cycle and return the gas to its original state. (R=8.31 J/mol/K= atm L/mol/K) 1) Find the number of moles, the temperature B, and the volume of the gas at C. First law of Thermodynamics: example 2) Find the internal energy of the gas (in kj), at A, B, and C. List P,V,T, U for the points A,B and C. 3) Consider the process A to B, B to C and C to A. For each case determine the sign of and Q. 4) Calculate Q, and U for each transition 5) Tabulate,Q and U for each transition and calculate the net effect of each. 6
7 Heat engines The heat engine is a device that converts internal energy into other useful forms of energy (electrical, mechanical, etc) In general, it carries a working substance (ex: water) through cycles: 1) Energy is transferred from a hot reservoir 2) ork is done by the engine 3) Energy is expelled into a cold reservoir Heat engines Because the substance goes through a cycle its initial internal energy is the same as the final: engine engine = Q net = Qhot Qcold The work done by an engine for a cyclic process is the area enclosed in the curve of a PV diagram 7
8 Thermal efficiency of Heat engines Thermal efficiency of the heat engine is the ratio of the work done by the engine to the energy absorbed from the hot reservoir in a cycle ε = Q Q ε = eng hot hot Q Q hot cold Q = 1 Q cold hot Second law of thermodynamics: It is impossible to construct a heat engine with 100% efficiency! Reversible and Irreversible processes Reversible process: the system is taken through a path of states in equilibrium and can be returned to the original state along the same path. Most natural processes are Irreversible! 8
9 Carnot engine Carnot s theorem No real engine operating between two energy reservoirs can be made more efficient than a Carnot engine operating between the same reservoirs ε =1 T T cold hot All real engines operate irreversibly (friction and cycles are short) 9
10 Carnot engine: question time Three engines operate between reservoirs separated in temperature by 300 K. The reservoir temperatures are: Engine A operates between 1000K and 700K Engine B operates between 800K and 500K Engine C operates between 600K and 300K 1) hich one has the highest efficiency? 2) hich one has the lowest efficiency? Heat engines: example A car engine delivers 8.2 KJ of work per cycle. a) Before tune-up the efficiency is 25%. Calculate, per cycle, the heat absorbed from combustion of fuel and the energy lost by the engine b) After a tune-up the efficiency is 31%. hat are the new values of the quantities calculated in a) when 8.2 KJ of work is delivered per cycle. 10
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