122ch10a_002

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

This preview shows pages 1–5. Sign up to view the full content.

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.)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern