# Gases - GASES Objectives 1 How was the equation of state...

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Gases 1 GASES Objectives. 1. How was the equation of state for ideal gases derived? The absolute temperature . 2. How can one describe real gases or mixture of gases? Virial coefficients , van der Waals gases , Boyle temperature 3. Which are the intensive properties of gases? critical region, isothermal compressibility , thermal expansion 4. Which properties are due to the fact that the gas particles are moving? pressure , diffusion , effusion , viscosity , thermal conduction 5. What are the effects of intermolecular forces?

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Gases 2 IDEAL GASES Volume = h V x Area Pressure = P atm + P h extra-P Add mercury. Record h V and h extra-P . h extra-P h V Boyle’s law (1660): V of air at P atm mercury P atm
Gases 3 g h P Area tant cons nal gravitatio x Area x height x density P Area tant cons nal gravitatio x volume x density P Area on accelerati x mass Area Force P P Volume = Area x h V Volume is proportional to h V . Pressure: Boyle noticed that the product of h V and h P was a constant: At a given temperature T, PV = constant P P extra atm P extra atm P extra atm total h g ) h h ( g g h g h g h P P Pressure on area is proportional to h P .

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Gases 4 Note: This relationship corresponds to the P exerted by a column of height h above an area. But if we want to know the pressure at a certain atmospheric height: Pressure decreases as height increases Barometric formula h RT gM 0 0 0 P P h 0 h e P P h RT gM P ln P ln h RT gM P ln P ln dh RT gM P dP dh g RT MP dP RT MP : gas ideal an for V M V nM V m dh g dP 0 Average molar mass for air: 1 2 2 2 mol g 97 . 28 ) CO ( 00 . 44 x 0003 . ) Ar ( 95 . 39 x 0093 . 0 ) O ( 2 x 00 . 16 x 2095 . ) N ( 2 x 01 . 14 x 7808 . 0 M
Gases 5 Example: Calculate the atmospheric pressure at an altitude of 1 km and 15 o C. bar 898 . 0 ) 118 . 0 exp( x bar 013 . 1 P m 1000 K 288 x mol K J 314 . 8 mol kg 10 x 8 . 28 s m 80665 . 9 exp x bar 013 . 1 P mol g 8 . 28 mol g 00 . 32 x 2 . 0 mol g 02 . 28 x 8 . 0 M 1 1 1 3 2 1 1 1 air h RT gM 0 e P P

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Gases 6 UNITS OF PRESSURE Hg = 13.5951 g cm -3 at 0 o C = 13.6 x 10 -3 kg x 10 6 m -3 = 13.6 x 10 3 kg m -3 g = 9.80665 m s -2 When P=1 atm, and the area is 1 cm 2 , the height of column = 76.000 cm. 1 atm = 13.5951 x 10 3 kg m -3 x 0.76000 m x 9.80665 m s -2 = 101,325 N m -2 or Pa = 1.01325 bar 1 atm = 760 mm Hg = 760 torr 1 bar = 750 torr Pressure Volume high temperature low temperature Boyle’s isotherms: Pressure 1/Volume high temperature low temperature P = constant x (1/V) V = constant x (1/P) 10 5 Pa = 1 bar 1 atm = 1.01 bar
Gases 7 Gay-Lussac or Charles’ Law (1802) 150 years later because a way of measuring temperature had to be invented: At constant pressure, the volume of a gas varies linearly with temperature. Volume temperature ( o C) low pressure high pressure -273 o C V = slope x (temperature + 273 o C) when t = -273 o C, V=0 Move y axis to the right by defining Kelvin scale ( absolute temperature ): T t + 273 o C V = constant x T

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Gases 8 Avogadro’s hypothesis (1811): Equal volumes of gases at the same T and P contain equal numbers of molecules. V = constant x (number of molecules)
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Gases - GASES Objectives 1 How was the equation of state...

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