Chapter 10 Exercise Key 1Chapter 10 Exercise Key Exercise 10.1 – Using the Ideal Gas Equation: Krypton gas does a better job than argon of slowing the evaporation of the tungsten filament in an incandescent light bulb. Because of its higher cost, however, krypton is used only when longer life is considered to be worth the extra expense. a.How many moles of krypton gas must be added to a 175-mL incandescent light bulb to yield a gas pressure of 117 kPa at 21.6 °C? ()3117 kPa 175 mLPV1 LPV = nRT n = = 8.3145 LkPaRT10mL294.8 K Kmol⎛⎞⎜⎟•⎝⎠•= 8.35 × 10−3mol Krb.What is the volume of an incandescent light bulb that contains 1.196 g of Kr at a pressure of 1.70 atm and a temperature of 97 °C?V = ? g = 1.196 g T = 97 °C + 273.15 = 370 K P = 1.70 atmgPV = RTM0.082058 L • atm1.196 g Kr 370 KgRTK • moleV = = gPM1.70 atm 83.80 mole⎛⎞⎜⎟⎝⎠⎛⎞⎜⎟⎝⎠= 0.255 L Kr c.What is the density of krypton gas at 18.2 °C and 762 mmHg?gV= ? P = 762 mmHg T = 18.2 °C + 273.15 = 291.4 K gPV = RTM()g762 mmHg 83.80 gPM1 atmmole= = L atmVRT760 mmHg0.082058 291.4 KK mole⎛⎞⎜⎟⎛⎞⎝⎠⎜⎝⎠⎟= 3.51 g/LCopyright 2004 Mark Bishop
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