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Thermodynamics

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Published in: AIEEE | IIT JEE Mains | Physics
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First Law of Thermodynamics.

Navin S / Hyderabad

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Qualification: B.Tech/B.E. (Bit Sindri , Sindri - 2011)

Teaches: Algebra, Geography, Mathematics, Physics, All Subjects, Biology, Science, Chemistry, Bank PO, PSU Exam, SBI Exam, AIEEE, BITSAT, Drawing, Mechanical, Production

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  1. The First Law of Thermodynamics : — The first law of thermodynamics states that, " If the quantity of heat supplied to a system is capable of doing 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 it. " Mathematical Expression of First Law:- Consider some gas enclosed in a barrel having insulating walls and conducting bottom. Let an amount of heat 'Q' be added to the system through the bottom. If 'UI' is the initial energy of the system, then, Total energy of the system in the beginning = UI+Q After gaining heat the gas tends to expand, pushing the piston from A to B as shown in below figure. As a result of this, some work 'W' is done by the gas. The work is external work, since the system undergoes a displacement. If 'IJ2' is final internal energy of the system, then, Total energy of the system at end = IJ2+W In accordance to the law of conservation of energy, total energy of the system in the beginning will be equal to total energy of the system at end. SO, U 1 + Q = I—J2+W It may be noted that 'UI', 'U2','Q' and 'W' all are being taken in same units. So, Q = - UI) + W When infinitesimal amount of heat 'dQ' is added to the system, corresponding changes in internal energy 'dQ' and external work done 'dW' are so small. Then, dQ = dlJ + dW Or, dQ = dlJ+ pdV This is mathematical statement of the first law of thermodynamics. Therefore, first law of thermodynamics signifies that, "energy can neither be created nor destroyed, but it can only be transformed from one form to another". Thus, dU+dW If work is done by the surroundings on the system (as during the compression of a gas), W is taken as positive so that dQ = dlJ+W. if however work is done by the system on the surroundings (as during the expansion of a gas), W is taken as negative so that dQ = dlJ - dW.
  2. Now,since you have gone through the basics of first law of thermodynamics,we would try now to understand with the help of example: Problem 1 : Consider that 214 J of work done on a system, and 293 J of heat are extracted from the system. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of (a) W, (b) Q, and (c) ?E2 Concept:- For a thermodynamic system, in which internal energy is the only type of energy the system may have, the law of conservation of energy may be expressed as Q+ AEint Here Q is the energy transferred (as heat) between the system and environment, Wis the work done on (or by) the system and AEint is the change in the internal energy of the system. By convention we have chosen Q to be positive when heat is transferred into the system and W to be positive when work is done on the system. Solution : (a) Since work is done on the system, therefore algebraic sign of the work done will be positive and the magnitude of work done is 214 J. Thus, +214 J. (b) Since heat is extracted from the system, the magnitude of heat is 293 J. Thus, -293 J. (c) To obtain internal energy of the system, AEint = Q+ W = (-293 J) + (+214 J) = -79.0 J therefore algebraic sign of the heat will be negative and substitute —293 J for Q and +214 J for W in the equation From the above observation we conclude that, the internal energy of the system would be -79.0 J. Problem 2 : 1 mole of ideal monoatomic gas at 270 C expands adiabatically against a constant external pressure of 1.5 atm from a volume of 4dm3 to 16 drn3. Calculate (i) q (ii) w and (iii) AU Solution : (i) Since process is adiabatic q (ii) As the gas expands against the constant external pressure. W = -PAV = -1.6(V2-Vl) = -1.5 (16-4) - 18 atm dm3 (iii) = q +w=O+ (-18) -- - 18 atm dm3 Problem 3 : A thermometer of mass 0.055 kg and heat capacity 46.1 J/K reads 15.00 C. It is then completely immersed in 0.300 kg of water and it comes to the same final temperature as the water. If the thermometer reads 44.40 C,what was the temperature of the water before insertion of the thermometer, neglecting other heat losses? Concept : — In accordance to the law of conservation of energy, for a thermodynamic system, in which internal is the only type of energy the system may have, the law of conservation of energy may be expressed as, Q+ AEint
  3. Here Q is the energy transferred between the system and its environment, W is the work done on or by the system and AEint is the change in the internal energy of the system. The heat capacity C of a body as the ratio of amount of heat energy Q transferred to a body in any process to its corresponding temperature change AT. C=Q/AT so, CAT The heat capacity per unit mass of a body, called specific heat capacity or usually just specific heat, is characteristic of the material of which the body is composed. c = C/m = Q/mAT so, cmAT Solution : The heat transfers for the water Qw is, Q = mwcw (T -T) Here, mass of water is mw, specific heat capacity of water is cw, final temperature is T and initial temperature is T. The heat transfers for the thermometer Q is Q = CAT Here, heat capacity of thermometer is C and AT is the temperature difference. As the internal energy of the system is zero and there is no work is done, therefore substitute AEint = 0 and W = 0 in the equation Q + W = AE int, Q+ AEint so, or, Qw+Q=O mwcw (T -T)+ CAT = O so, T = (mwcw T + CATO/ mwcw Here 44.4 15.0 cc = 29.4 oc To obtain the temperature of the water before insertion T of the thermometer, substitute 0.3 kg for mw, 4190 J/kg.m for cw, 44.4 0 C for T, 46.1 J/K for C and 29.4 0 C for AT in the equation T = (mwcw Tf+ CAT )/ mwcw = (mwcw T + CATt mwcw = [(0.3 kg) (4190 J/kg.m) (44.4 C) + (46.1 J/K) (29.4 C)] kg) (4190 J/kg.m)] =45.5 c From the above observation we conclude that, the temperature of the water before insertion of the thermometer was 45.5 0 C. Problem 4 : Air that occupies 0.142 m3 at 103 kPa gauge pressure is expanded isothermally to zero gauge pressure and then cooled at constant pressure until it reaches its initial volume. Compute the work done on the gas. Concept:- In an isothermal process the work done WI is defined as, WI = -nRTIn WV As from ideal gas equation, PI VI = P2V2 = nRT, thus WI = -nRTIn WV
  4. = - PI VI In PI/P2 Here nis the number of moles, Ris the gas constant, Tis the temperature, VI is the initial volume, is the final volume, PI is the initial pressure and p-z is the final pressure. The work done W2 at constant pressure, = -P2( V- V2) (Since, v-V2) = -p VI (I-p/p) - VI (p-P2) Solution : The total work done Won the gas will be equal to the sum of work done WI by isothermal process and work done at constant pressure during cooling. SO, WI + = (- p VI In p/P2) + (V (p-P2)) To find out the work done Won the gas, substitute Pa for PI, 0.142 rn3 for VI, 101 x 103 Pa for in the equation W = (- PI VI Inpl/p2) + (VI (p-p2)), W = (- PI VI In PI/P2) + (VI (p-p2)) = [-(204x 103 Pa) (0.142 m3) In 101 x103 Pa) ] + [(0.142 m3) ((204x 103 Pa) - (1 01 Pa))] = -5.74x103 Pa. m3 = (-5.74x103 Pa. n) (1 J/l Pa.m3) From the above observation we conclude that, work done Won the gas would be -5.74x1 03 J. Problem 5: — Calculate the work done by an external agent in compressing 1.12 mol of oxygen from a volume of 22.4 L and 1.32 atm pressure to 15.3 L at the same temperature. Concept : — In an isothermal process the work done by an external agent in compressing gas from its initial volume is, -nRTln V/ v = -nRTIn WV. In accordance to ideal gas equation, PI/ = nRT VI ) to its final volume (V =V2) Here Pis the pressure of the gas, Vis the volume of the gas, n is the number of moles, ris the gas constant and Tis the temperature of the gas. At constant temperature if the gas compress from its initial pressure R, intial volume VI to its final pressure B, final volume V2, then PI VI = R nRT To find work Win terms of R and VI, substitute R VI for nRTin the equation W = -nRTln 1/2/ VI , W = -nRTln W VI =-PI VI In WV Solution : To obtain work done, substitute 1.32 atm for pressure R, 22.4 L for initial volume VI and 15.3 L for initial volume 1/2 in the equation W = - PI VI In w VI,
  5. W=-R VI In WV = -(1.32 atm) (1.01 x105 Pa/ 1 atm) (22.4 L) (lc m3/ 1 L) In ((15.3 L) (1 C m3/1 L) /(22.4 L) (103 n / 1 L)) = Pa. m3 = (1.14x103 Pa. m3) ( 1 J/l Pa. m3) From the above observation we conclude that, the work done by an external agent in compressing the oxygen molecule would be 1.14x103J. Hope this note must be useful for all: Thank you and all the best!!!!!!

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