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

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

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

where NA is the Avogadro's number = 6.022´10-23molecules/mole
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 = nNA where NA is the Avogadro's number = 6.022´10-23molecules/mole

where K is called Boltzmann's constant, which has the value R/NA
K = R/NA = 1.38´10-23J/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 100cm3 at 20oC 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?

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´10-3m3, 200kPa and 300K respectively. If the volume is decreased to 12´10-3m3 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

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

Exercise An ideal gas undergoes the process shown in the figure.  Find V2, Ta, Td and Tb. where n=44kmole and R=8.314J/mole.K

Heat Temperature 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 temperature is defined as the derivative of the internal energy with respect to the entropy. 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 14.5oC to 15.5oC. 1cal = 4.186J      or     1J = cal

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". 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,

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 2x106cal. Converting this to joule, then the work required is W = 2x106cal x 4.186J/cal = 8.37x106J 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 = 8.37x106J Since m = 50 Kg, and h = 2m n = 8.54x103 times

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/Co

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.

Al 900J/kg.Co wood 1700J/kg.Co Cu 387J/kg.Co glass 837J/kg.Co Ag
wood 1700J/kg.Co Cu 387J/kg.Co glass 837J/kg.Co Ag 129J/kg.Co water 4186J/kg.Co Pb 128J/kg.Co ice 2090J/kg.Co

mx cx (Ti-Tf) = mw cw (Tf-Ti)
Example A 0.05kg of metal is heated to 200oC and then dropped into a beaker containing 0.4kg of water initially at 20oC. If the final equilibrium temperature of the mixed system is 22.4oC 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 mx cx (Ti-Tf) = mw cw (Tf-Ti) (0.05Kg) cx (200oC-22.4oC) = (0.4kg)(4186J/kg.Co)(22.4oC-20oC) cx = 453J/kg.Co (b) Q = m c (Ti-Tf) = 0.05 ´ 453 ´ ( ) = 4020J

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? Solution The kinetic energy of the bullet Ek = 1/2 m v2 = 40J Q = m c DT where c for silver is 234Jkg.Co DT = Q/mc = 85.5Co

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 Solid ----  Liquid        melting            Liquid    gas           boiling

Lv > Lf Latent heat of fusion Lf melting (Solid º> Liquid)
Latent heat of vaporization Lv boiling)  Liquid º> gas) Lv > Lf

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

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. If a system (gas) in its initial state at Pi, Vi change to Pf, Vf 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.

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

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

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

Solution (a) in ab W = 0 and DU = Q = 600J
(b) in bd pressure is constant       W = P (V2 - V1) = 8x104 pa (5x x10-3) = 240J       W abd =       Q abd = 800J       DU = = 560J (c) in acd DU = 560J as well      W = 3x104 pa (5x x10-3) = 90J      Q = DU + W = = 650J

Example One gram of water occupies a volume of 1cm3 at atmospheric pressure. When this amount of water is boiled, it becomes 1671cm3 of steam. Calculate the change in internal energy for this process. Where the latent energy of water is (2.26x106J/kg) Solution DU =  Q - W Q = m Lv = (1x10-3kg) x (2.26x106J/kg) = 2260J W = P (Vf - Vi) = (1.013x105) x [(1671-1)10-6] = 169J DU =  Q - W = 2260J - 169J = 2091J

(a) the work done by the copper
Example A 1kg bar of copper is heated at atmospheric pressure(1.013x105N/m2) . If its temperature increases from 20oC to 50oC, (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/8.92x103kg/m3) the volumetric expansion coefficient β= 5.1x10-5Co-1 (a) the work done by the copper

Q = m C DT = (1kg) x (387J/kg.Co) x (30Co) = 1.16 x 104J
DV = [5.1x10-5Co-1](50oC-20oC)x(1kg/8.92x103kg/m3) DV = 1.7x10-7 m3 W = P DV = (1.013x105N/m2) x (1.7x10-7 m3) = 1.9x10-2J (b) What quantity of heat is transferred to the copper? Q = m C DT  = (1kg) x (387J/kg.Co) x (30Co) = 1.16 x 104J (c) What is the increase in internal energy of the copper? DU =  Q - W = 1.16 x 104J- 1.9x10-2J

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