الخامسة المحاضرة

Ideal Gas EquationBoyle's LawWhen gas is kept at constant temperature its pressure is inversely proportional to the volume.

Charle's LawWhen 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 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.KThe 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.SolutionPV = nRTWhat 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.SolutionSince 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

ExercisePure 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=420K

ExerciseAn 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 and the first law of thermodynamicsHeatThe 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 HeatThe 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 = 0.2389cal

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,

ExampleA 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.

SolutionSince 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 والشغل الكلي mgh يساوي hالشغل المبذول لرفع االثقال لمسافة

وعليه فإن عدد مرات nmghلرفع األثقال عدة مرات يعطي بالعالقة رفع االثقال هو

W = nmgh = 8.37x106J Since m = 50 Kg, and h = 2mn = 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 129J/kg.Co   water 4186J/kg.Co

Pb 128J/kg.Co ice 2090J/kg.Co

ExampleA 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?

SolutionHeat 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 ´ (200-22.4) = 4020J

ExampleA 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 Ek = 1/2 m v2 = 40J

Q = m c DTwhere c for silver is 234Jkg.Co

DT = Q/mc = 85.5Co

Solution

Latent heat

Solid ----  Liquid        melting           Liquid  ------  gas           boiling

The heat or energy required to change the phase of a given mass m of a substance is (تغير حالة المادة)given by

where L is called the latent heat (hidden heat) of substance

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

dF = P dA

dF dy = P dA dydW = P dV

dV=0 W =

0(Isochoric)

ProcessIsothermal processIsobaric processIsochoric processAdiabatic process

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 .

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 thermodynamicsfor small changes dU = dQ - dW

Special casesIn 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

ExampleA 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 ab2) the internal energy change in process abd3) 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 (5x10-3 - 2x10-3) = 240J      W abd = 240 + 0      Q abd = 800J      DU = 800 - 240 = 560J(c) in acd DU = 560J as well     W = 3x104 pa (5x10-3 - 2x10-3) = 90J     Q = DU + W = 560 + 90 = 650J

ExampleOne 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 - WQ = m Lv = (1x10-3kg) x (2.26x106J/kg) = 2260JW = P (Vf - Vi) = (1.013x105) x [(1671-1)10-6] =

169JDU = Q - W = 2260J - 169J = 2091J

ExampleA 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

 

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