Thermal conductivity of the glass is 065wmk solution

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it conduct in one hour. [ thermal conductivity of the glass is 0.65W/mK] solution = ௞஺௱ఏ ݇ = 0.65W/mK A = 2 x 1.5 = 3.0m 2 θ = 23 – ( 20) = 43 0 C ߡ = 8mm = 0 008m T = 3600sec Φ = ௞஺௱ఏ ݔ ݐ = ଴.଺ହ ௫ ଷ ௫ ସଷ ଴.଴଴଼ X 3600 = 3773.25J Exercise (i)The control rod in a thermometer is a thin metal bar 150mm long held at one End and free to expand at the other end. The thermostat is assembled at 18 0 C with a gap 6mm between the free end of the rod and the mechanism operating a shut off valve. Calculate the temperature at which the valve acts. [Coefficient of linear expansion of the metal is 20x10 6 /K] (ii)Calculate the heat loss per second through a rectangular 8mm glass pane of dimensions 1.2m x 0.4m for a temperature difference of 15K. [The thermal conductivity of glass is 0.65W/K}
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Prepared By: ChilesheMwenyaMporokoso Page 72 UNIT 2.1: GAS LAWS Gases are relatively compressive and pressure, volume and temperature can change substantially. The pressure, volume and temperature of a fixed mass of a gas can be linked by simple laws and reduced to simple equation of state BOYLE’S LAW States that for a fixed mass of a gas at constant temperature, pressure is inversely proportional to volume. If T is constant, then P PV = K (constant) P 1 V 1 = P 2 V 2 CHARLE’S LAW : States that, for a fixed mass of a gas at constant pressure, volume is directly proportional absolute temperature. V T = K i.e. = THE IDEAL GAS EQUATION The ideal gas equation is obtained by combining Boyle’s law and Charles’s laws. P 1 V 1 = KT 1 P 2 V 2 = KT 2 K = = మೇ K is usually denoted by R PV = RT for one more of a gas PV = nRTfor n moles of a gas Where R = Universal Gas constant for 1 mole of a gas One mole of any gas occupies 22.4 liters at STP R = 8.31Jmol 1 K 1
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Prepared By: ChilesheMwenyaMporokoso Page 73 At STP P = 1.013 x 10 5 Pa V = 22.4 x 10 2 m 3 T = 273.15K In all the problems involving gas laws it is the absolute pressure and not the gauge pressure that is used Absolute pressure = gauge pressure + Atmospheric pressure. EXAMPLES 1. When heated in a garage at room temperature of 24 0 C, a motor tyre is found to have a pressure of 120KN/m 2 . If the volume of the air inside remains constant, what is the pressure after the tyre has been allowed to stand in the sun at 40 0 C (take atmospheric pressure to be 100KN/m 2 ) = మೇ but V 1 = V 2 P 1 = 120 KN/m 2 T 2 = 40+ 273 = 313K T 1 = 24 + 273= 297K = ,P 2 = భ೅ = ଶଶ଴ ଴଴଴ ௫ ଷଵଷ ଶଽ଻ = 231 850 Gauge pressure = 231 850 100 000 = 131 850N = 131.850KN 2.A given gas of mass of an ideal gas occupies 38ml at 20 0 C. If its pressure is held constant, what volume does it occupy at a temperature of 45 0 C? 3.After 1.5 litres of gas at STP is subjected to a chemical experiment, the temperature of the gas rises to 20 0 C and the pressure rises to 1.2 x 10 5 Pa. How many moles of gas were evolved? At STP V 1 = 1.5lt P 1 = 1.013 x 10 5 Pa
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Prepared By: ChilesheMwenyaMporokoso Page 74 T 1 = 273K P 2 = 1.2 x 10 5 Pa T 2 = 20 + 273 = 293K = మೇ V 2 = భೇ భ೅ మ೅ = ଵ.ହ ௑ ଵ.଴ଵଷ௑ଵ଴ ௑ଶଽଷ ଵ.ଶ ௑ଵ଴ ௑ ଶ଻ଷ = 1.35 litres 22.4 litres = 1 mol 1.65litres = x X = ଵ.଺ହ ௫ ଵ ଶଶ.ସ = 7.4 x 10 2 moles 1.65litres has 7.4 x 10 2 mole 3.A cylinder of nitrogen gas has an initial gauge pressure of 800Kpa. If the
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