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المحاضرة الخامسة Ideal Gas Equation Boyle's Law When gas is kept at constant temperature its pressure is inversely proportional to the volume.

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Presentation on theme: "المحاضرة الخامسة Ideal Gas Equation Boyle's Law When gas is kept at constant temperature its pressure is inversely proportional to the volume."— Presentation transcript:

1 المحاضرة الخامسة Ideal Gas Equation Boyle's Law When gas is kept at constant temperature its pressure is inversely proportional to the volume.

2 Charle's Law When the pressure of the gas kept constant the volume directly proportional to the temperature. V  T at constant pressure. result can be summarized in one equation called the equation of state for an ideal gas.

3 Where n is the number of moles, R is a constant for a specific gas, which can be determined experimentally, and T is the absolute temperature in Kelvin When the pressure goes to zero then the quantity PV/nT become the same value of R for all gasses, therefore R called the universal gas constant R = 8.31 J/mole.K The ideal gas law can be expressed in terms of the total number of molecules N where N = nN A where N A is the Avogadro's number =  molecules/mole

4 where K is called Boltzmann's constant, which has the value R/N A K = R/N A = 1.38  J/K ¯ One mole of substance is that mass of the substance that contains Avogadro's number of molecules Example (1) An ideal gas occupies a volume of 100cm 3 at 20 o C and a pressure of 100Pa. Determine the number of moles of gas in the container. Solution PV = nRT What is the number of molecules in the container?

5 Example (2) Pure helium gas is admitted into a tank containing a movable piston. The initial volume, pressure and temperature of the gas are 15  m 3, 200kPa and 300K respectively. If the volume is decreased to 12  m 3 and the pressure is increased to 350KPa, find the final temperature of the gas. Solution Since the gas can not escape from the tank then the number of moles is constant, therefore, PV = nRT at the initial and final points of the process

6 Exercise Pure helium gas is admitted into a leak-proof cylinder containing a movable piston. The initial volume, pressure, and temperature of the gas are 15 L, 2.0 atm, and 300 K. If the volume is decreased to 12 L and the pressure increased to 3.5 atm, find the final temperature of the gas. (Assume helium behaves as an ideal gas.) T=420  K

7 Exercise An ideal gas undergoes the process shown in the figure. Find V 2, T a, T d and T b. where n=44kmole and R=8.314J/mole.K

8 Heat and the first law of thermodynamics The word of ''heat flow'' is an energy transfer that take place as a consequence of temperature difference only. temperature is defined as the derivative of the internal energy with respect to the entropy.internal energyentropy Unit of Heat The unit of heat is ''calorie'' which is defined as the amount of heat (energy) required to raise the temperature of 1g of water from 14.5 o C to 15.5 o C. 1cal = 4.186J or 1J = cal

9 Temperature is a physical property of matter that quantitatively expresses the common notions of hot and cold. Objects of low temperature are cold, while various degrees of higher temperatures are referred to as warm or hot. Heat spontaneously flows from bodies of a higher temperature to bodies of lower temperature, at a rate that increases with the temperature difference and the thermal conductivity. No heat will be exchanged between bodies of the same temperature; such bodies are said to be in "thermal equilibrium".physical propertyhotcoldHeat spontaneouslythermal conductivitythermal equilibrium The temperature of a substance typically varies with the average speed of the particles that it contains, raised to the second power; that is, it is proportional to the mean kinetic energy of its constituent particles. Formally,kinetic energy

10 Example A student eats a dinner rated at 2000 (food) Calories. He wishes to do an equivalent amount of work in the gymnasium by lifting 50Kg mass. How many times must he raise the weight to expend this much energy? Assume that he raises the weight a distance of 2m each time and no work is done when the weight is dropped to the floor. Solution Since 1 (food) Calories = 1000 cal then the work required is 2x10 6 cal. Converting this to joule, then the work required is W = 2x10 6 cal x 4.186J/cal = 8.37x10 6 J الشغل المبذول لرفع الاثقال لمسافة h يساوي mgh والشغل الكلي لرفع الأثقال عدة مرات يعطي بالعلاقة nmgh وعليه فإن عدد مرات رفع الاثقال هو W = nmgh = 8.37x10 6 J Since m = 50 Kg, and h = 2m n = 8.54x10 3 times

11 Heat capacity and specific heat The heat capacity is defined as the amount of heat energy needed to raise the temperature of a sample by 1 degree Celsius. J/C o

12 The specific heat capacity is defined as the amount of heat energy needed to raise 1kg of sample by 1 degree Celsius. where c is called the specific heat capacity or specific heat.

13 Al 900J/kg.C o wood 1700J/kg.C o Cu387J/kg.C o glass 837J/kg.C o Ag129J/kg.C o water 4186J/kg.C o Pb128J/kg.C o ice 2090J/kg.C o

14 Example A 0.05kg of metal is heated to 200 o C and then dropped into a beaker containing 0.4kg of water initially at 20 o C. If the final equilibrium temperature of the mixed system is 22.4 o C find the specific heat of the metal. What is the total heat transferred to water in cooling the metal? Solution Heat lost by the metal = heat gained by water m x c x (T i -T f ) = m w c w (T f -T i ) (0.05Kg) c x (200 o C-22.4 o C) = (0.4kg)(4186J/kg.C o )(22.4 o C-20 o C) c x = 453J/kg.C o (b) Q = m c (T i -T f ) = 0.05 ´ 453 ´ ( ) = 4020J

15 Example A man fires a silver bullet of mass 2g with a velocity of 200m/sec into a wall. What is the temperature change of the bullet? The kinetic energy of the bullet E k = 1/2 m v 2 = 40J Q = m c  T where c for silver is 234Jkg.C o  T = Q/mc = 85.5C o Solution

16 Latent heat Solid ---- Liquid melting Liquid gas boiling The heat or energy required to change the phase ( تغير حالة المادة ) of a given mass m of a substance is given by where L is called the latent heat (hidden heat) of substance

17 Latent heat of fusion L f melting (Solid º > Liquid) Latent heat of vaporization L v boiling) Liquid º > gas) L v > L f

18 Work and heat in thermodynamic processes dF = P dA dF dy = P dA dy dW = P dV dV=0 W = 0(Isochoric) Process Isothermal process Isobaric process Isochoric process Adiabatic process

19 The first law of thermodynamics change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work performed by the system on its surroundings. The law can also be stated: The energy of an isolated system is constant.internal energyclosed systemheat If a system (gas) in its initial state at P i, V i change to P f, V f by adding quantities of heat and applying work on the system. If the quantity Q-W is measured for various paths from i to f.

20 we find Q-W is always constant. Q-W is called the change in the internal energy of the system  U Q and W depend on the path but Q-W is independent of the path  U = U f - U i = Q - W The first law of thermodynamics for small changes dU = dQ - dW

21 Special cases In isolated system there is no heat flow and work is zero the change in internal energy is zero, i.e.  U=0 If the process is done on a system taken through a cycle, the change in the internal energy is zero, i.e.  U=0 and Q = W

22 Example A thermodynamic process is shown in Figure. In process ab, 600J of heat are added, and in process bd 200J of heat are added. Find 1) the internal energy change in process ab 2) the internal energy change in process abd 3) the total heat added in process acd

23 Solution (a) in ab W = 0 and  U = Q = 600J (b) in bd pressure is constant W = P (V 2 - V 1 ) = 8x10 4 pa (5x x10 -3 ) = 240J W abd = Q abd = 800J  U = = 560J (c) in acd  U = 560J as well W = 3x10 4 pa (5x x10 -3 ) = 90J Q =  U + W = = 650J

24 Example One gram of water occupies a volume of 1cm 3 at atmospheric pressure. When this amount of water is boiled, it becomes 1671cm 3 of steam. Calculate the change in internal energy for this process. Where the latent energy of water is (2.26x10 6 J/kg) Solution  U = Q - W Q = m L v = (1x10 -3 kg) x (2.26x10 6 J/kg) = 2260J W = P (V f - V i ) = (1.013x10 5 ) x [(1671-1)10 -6 ] = 169J  U = Q - W = 2260J - 169J = 2091J

25 Example A 1kg bar of copper is heated at atmospheric pressure(1.013x10 5 N/m 2 ). If its temperature increases from 20 o C to 50 o C, (a) find the work done by the copper. (b) What quantity of heat is transferred to the copper? (c) What is the increase in internal energy of the copper? Where intial volume of copper is (1kg/8.92x10 3 kg/m 3 ) the volumetric expansion coefficient β= 5.1x10 -5 C o-1 (a) the work done by the copper

26  V = [5.1x10 -5 C o-1 ](50 o C- 20 o C)x(1kg/ 8.92x10 3 kg/m 3 )  V = 1.7x10 -7 m 3 W = P  V = (1.013x10 5 N/m 2 ) x (1.7x10 -7 m 3 ) = 1.9x10-2J (b) What quantity of heat is transferred to the copper? Q = m C  T = (1kg) x (387J/kg.C o ) x (30C o ) = 1.16 x 10 4 J (c) What is the increase in internal energy of the copper?  U = Q - W = 1.16 x 10 4 J - 1.9x10-2J


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