122ch10a_002 - 1 Chapter 10 Homework Solutions 10.3 a F = m...

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1 Chapter 10 - Homework Solutions 10.3) a) F = m × a where, F = force m = mass a = acceleration For a downward force (force exerted by a person on the floor) the accelerations is the acceleration due to gravity, g. F = m × g Since both people have same mass the forces they exert on the floor are equal. b) P = F/A F = force A = area The forces exerted by each person are equal. However, the person standing on one foot is applying this force over smaller area & thus exerts a greater pressure. The Pressure and Area are inversely proportional: A inc., P dec. A dec., P inc. 10.4) The height of the mercury column in a barometer is a direct measure of atmospheric pressure. Atomospheric pressure decreases with increasing altitude (Section 10.1 and 18.1), so atmospheric pressure in Denver, at an altitude of 5000 feet, is lower than atmospheric pressure in Los Angeles, at 132 feet (nearly sea level). Since Denver has a lower atmospheric pressure the height of a mercury column in a barometer in Denver should be lower than that in a barometer in Los Angeles. The fact that atmospheric pressure and the height of a liquid are related is shown: P = F/A F = m C g m = d C V (A=area; m=mass; g=accel due to gravity; d=density; V = volume) m = d C V = d C (A C h) ˆ F = (d C A C h) C g & P = d C h C g This shows that P depends only on the density of the liquid used in the barometer and is independent of the cross-sectional area of the tube (diameter of column) Also, P and the height of the column are directly proportional (i.e. as the atmospheric pressure decreases the height of Hg decreases and vice versa.)
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2 10.6) We need to relate the height of columns of two different liquids. From problem 4, P = d C h C g The pressure is the same (752 torr or 752 mm Hg) for any barometer regardless of the liquid used in the barometer. P liq A = P liq B (d C h) A C g = (d C h) B C g (d C h) A = (d C h) B d B ˆ h A = (-----) h B d A (d C h) 1-id = (d C h) Hg (1-id = 1-iodododecane) d Hg h 1-id = (------) h Hg d 1-id 13.6 g/mL h 1-id = (-------------) C 752 mm 1.20 g/mL = 8.5 2 2 x 10 3 mm = 8.52 m (Pressure of 8.52 m 1-id.) This is 28.0 feet - NOT a very useful barometer. Thus you want to use a liquid with a fairly large density (such as Hg) so the size of the barometer can be reasonable. P % d C h smaller d ==> greater h larger d ==> smaller h
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3 10.9) Note: 1 torr = 1 mm Hg (i.e. can use torr and mm Hg interchangeably) 1 atm (a) 265 torr × ----------- = 0.34 8 68 atm 0.349 atm 760 torr 1 mm Hg (b) 265 torr × -------------- = 265 mm Hg 1 torr 1.01325 x 10 5 Pa (c) 265 torr × ----------------------- = 3.5 3 30 x 10 4 Pa = 3.53 x 10 4 Pa 760 torr 1.01325 x 10 5 Pa 1 bar (d) 265 torr × ----------------------- × -------------- = 0.35 3 3 bar = 0.353 bar 760 torr 1 x 10 5 Pa 10.13) P = F/A = (m C g)/A What are the units of pascals in terms of metric units? 1 Pa = 1 N/m 2 = (1kg C m)/s 2 × 1/m 2 = 1 kg/(m C s 2 ); change mass to kg & area to m 2 (m C g) 125 lb 9.81 m 0.454 kg (39.4) 2 in 2 1 kPa P = --------- = ---------- × ---------- × ------------ × ------------ × ---------- = 1. 7 28 x 10 3 kPa A 0.50 in 2 1 s 2 1 lb 1 m 2 10 3 Pa = 1.7 x 10 3 kPa (The first 3 multiplications give force in the numerator and the next one converts area from in 2 to m 2 giving pascals.) or (m C g) 250 lb 1 atm 101.325 kPa P = --------- = ---------- × -------------- × ----------------- = 1. 7 2 x 10 3 kPa A in 2 14.7 lb/in 2 1 atm ^ 125 lb/0.50 in 2
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4 10.16) a) The Hg in the arm attached to the gas is 13.6 cm (136 mm) higher than in the one open to the atmosphere w. P atm = 1.05 atm P gas = P atm - h 760 mm Hg = (1.05 atm x ------------------) - 136 mm Hg
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