MidtermKey

# MidtermKey - Chemistry 20B Midterm Summer Session Answer...

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Unformatted text preview: Chemistry 20B Midterm August 24, 2007 Summer Session Answer Key 1) The TauTona mine in South Africa was the deepest mine in the world at 3580. m (13) deep. In order to allow people to work in it, the mine is kept at 20.ºC. Suppose the mine is filled with UF 6(g) ( M = 352 g/mol). (This is a huge assumption as UF 6 is a solid at room temperature and 1 atm. Just go with it.) a) The pressure at the bottom of the mine can be estimated by P = ρ gh = 6.44 atm. (3) However, this estimation is grossly inaccurate as the density of a gas varies dramatically with pressure. To determine the pressure more accurately, the barometric formula can be used to calculate the pressure at height h. Assume the pressure at the top of the mine is 1.000 atm: RT gh h e P P / M- = where g is gravity. Use the barometric formula to determine the pressure at the bottom of the mine: P h = -3580 m P = 1.000 atm h = -3580 m J = kg m 2 /s 2 K molK J m s m mol kg RT gh 15 . 293 )( / 31451 . 8 ( ) 3580 )( / 80665 . 9 )( / 352 . ( 2-- =- M =+5.07 17 . 159 ) 000 . 1 ( 07 . 5 3580 = = +- e atm P atm = 160 atm (2 Sig Fig) b) Archimedes discovered that a body will float when the displaced mass is equal to (3) the mass of the object. Another way to express this: a body will float when the density of the body and the density of the fluid are equal. Assumptions: 1) UF 6(g) is an ideal gas 2) The average density of the human body is 965 kg/m 3 What is the minimum pressure of UF 6(g) in which a human will float at 20.ºC? V nRT P = min V n V m M = = ρ 1 m 3 = 10 3 L M M M RT RT V m V RT m P ρ = = = min ( ) atm mol g K molK Latm L g P 9 . 65 ) / 352 15 . 293 082058 . 965 min = = c) Would you be able to float at the bottom of the TauTona mine filled with UF 6(g) ? (1) Pressure is proportional to density P mine > P min , therefore ρ mine > ρ human Yes, you would float. d) The density of the atmosphere also affects the speed of sound. The speed of (3) sound is equal to the root-mean-square speed of the gas particles in the atmosphere. The speed of sound of our atmosphere at sea level is 340.29 m/s. What is the speed of sound in an atmosphere of UF 6(g) at 20.ºC? ( ) kg J mol kg K molK J RT u rms 144 352 . 15 . 293 31451 . 8 3 3 = = = M J = kg m 2 /s 2 u rms = 144 m/s e) What is the mean free path of the UF 6(g) at the bottom of the mine at 20.ºC? The (3) diameter of UF 6(g) = 3.01 Å. (Check your units!) Mean free path = λ ( ) V N d / 2 1 2 π λ...
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MidtermKey - Chemistry 20B Midterm Summer Session Answer...

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