prev

next

out of 3

Published on

17-Jan-2016View

27Download

1

DESCRIPTION

problemas resueltos de termodinamica

Transcript

Solved Examples ThermodinamicsQuestion-1. What is true of Isothermal Processa, T >0b, U=0c Q=Wd PV=constantsSolution-1:In an Isothermal ProcessTemperature remains constant T =0Since Internal energy depends on the temperatureU=0From first law of ThermodynamicsU=Q-WSince U=0Q=WAlso PV=nRTAs T is constantPV= constantQuestion-.2 Two absolute scales A and B have triple points of water defined as 200A and 350A.what is the relation between TA and TBSolution-2Given that on absolute scaleTriple point of water on scale A = 200 ATriple point of water on scale B = 350 BAlso, triple point of water on Kelvin scale = 273.16 KNow temperature on scale A and on scale B is equivalent to 273.16 K on absolute temperature scale.Thus, value of one degree on absolute scale A = (273.16/200) KOr,Value of temperature TA on absolute scale A = (273.16XTA)/200Similarly value of temperature TB on absolute scale B = (273.16XTB)/350Since TA and TB represent the same temperature273.16TA/200 = 273.16TB/350Or, TA = 200TB/350 = 4TB/ 7Question 3:A gas is contained in a cylinder with a moveable piston on which a heavy block is placed.Suppose the region outside the chamber is evacuated and the total mass of the block and themovable piston is 102 kg. When 2140 J of heat flows into the gas, the internal energy of the gasincreases by 1580 J. What is the distance s through which the piston rises?Solution:3Total heat supplied =Workdone + Change in internal energySo work done=2140-1580=560 JLet s be the distance moved thenthe workdone is given by =FsFs=560s=560/F=560/102*10s=.54 mQuestion-4: At 27C,two moles of an ideal monoatomic gas occupy a volume V.The gas isadiabatically expanded to a volume 2V.Calculate the ratio of final pressure to the intial pressureCalculate the final temperatureChange in internal energyCalculate the molar specific heat capacity of the processSolution-4Givenn=2 T=27C=300 K ,V1=V,V2=2VNow PVy=constantP1V1y=P2V2yP2/P2=(V1/V2)y=.55/3ALsoT1V1y-1=T2V2y-1or T2=300/25/3=189KChange in internal energy=nCvTFor monoatomic gas Cv=3R/2Substituting all the valuesChange in internal energy==-2764JAs in adiabatic process Q=0,molar specific heat capacity=0Question-5An ideal gas heat engine operates in Carnot cycle between 227C and 127C.It absorbs 6*102 cal ofheat at the higher temperature.Calculate the amount of heat supplied to the engine from the sourcein each cycleSolutions-5:T1=227C =500KT2=127C =400KEfficiency of the carnot cycle is given by=1-(T2/T1)=1/5Now also efficency =Heat supplied from source/Heat absorbed at high temperatureso Heat supplied from source=6*102*(1/5)==1.2*102cal