1. Aim # 14: How do we measure the pressure of a confined sample of a gas?
H.W. # 14
Study pp (Sec. 5.1) Study class notes Ans. ques. p. 235 # 39, p. 185 #94 (review problem) (Hint: write the equation for the formation of the precipitate)
Do Now: Zumdahl (8th ed.) p. 176 # 100
p. 178 # 121 (challenge prob.)

2. WHAT IS THE DIFFERENCE BETWEEN A GAS AND A VAPOR?
I Gaseous Elements- H2, N2, O2, F2, Cl2, He, Ne, Ar, Xe, Kr, Rn All other gases are compounds.
WHAT IS THE DIFFERENCE BETWEEN A GAS AND A VAPOR?
II Properties of a gas-
A. Expands to fill the volume of its container
B. Takes the shape of its container
C. Molecules (or atoms) move about freely and randomly at high speeds.
D. Gases consist mostly of empty space. Molecules far apart → weak attractive forces for one another
E. Gases form homogeneous mixtures with one another.

3. III Pressure- force per unit of area
P = F/A measured in N/m2 = Pa (Pascal)
The air has weight, and therefore exerts pressure on the surface of the earth Standard Pressure = 1 atm = kPa = 760 mm Hg = 760 torr = 14.7 lb/in = in Hg

4. IV Measuring Pressure A. the mercury barometer
A column of mercury will stand 760 mm high when the atmospheric pressure is 1 atm (standard pressure).
As atmospheric P↑, the column rises.
As atmospheric P↓, the column falls.

5. B. Unit conversion
Problem: convert atm to mm Hg
Ans: atm x 760 mm Hg = 494 mm Hg atm
Problem: convert 770 mm Hg to kPa
Ans: 770 mm Hg x kPa = 103 kPa mm Hg

6. C. the manometer- used to measure the pressure of a confined sample of a gas.
NOTE: THE LEVEL OF THE LIQUID Hg IS EQUAL ON BOTH SIDES OF THE U-TUBE. WHY?

7. 1. The closed-end manometer
Pgas = h
Used for measurements of P under 1 atm.

8. When Pgas > Patm Pgas = Patm + h
2. The open-ended manometer pressure exerted = pressure exerted + ( ht. of Hg – ht. of Hg ) by the gas by the atm in open arm in closed arm
When Pgas > Patm Pgas = Patm + h
When Pgas < Patm Pgas = Patm - h

9. Ans: Pgas = Patm + h Pgas = 762 mm Hg + 50 mm Hg = 812 mm Hg
Problem: If the atmospheric pressure is 762 mm Hg. What is the pressure exerted by a confined gas if the height of a manometer’s open-ended column is 50 mm greater than the height of its closed-ended column?
Ans: Pgas = Patm + h
Pgas = 762 mm Hg + 50 mm Hg = 812 mm Hg
Problem: A vessel connected to an open-ended mercury manometer is filled with a gas. It exerts a pressure of atm. If the atmospheric pressure is 750 torr, what is the difference in the height between the two columns of mercury? In which arm will the mercury be higher?
Ans: Pgas = Patm + h atm = 750 torr + h
0.963 atm x 760 mm Hg = 750 mm Hg + h atm mm Hg = 750 mm Hg + h h = - 18 mm
The mercury will be higher in th closed arm.

10. Problem: A container of a gas is confined to an open-ended manometer
Problem: A container of a gas is confined to an open-ended manometer as shown below What pressure is the gas in the container exerting, in kPa? In atm?
Patm = 102 kPa h = 30.0 mm Hg
Pgas = 102 kPa mm Hg = 102kPa + (30.0 mm Hg)( kPa) mm Hg
Pgas = 102 kPa + 4 kPa = 106 kPa
106 kPa x atm = 1.05 atm kPa

11. If the mercury in the manometer (d = 13
If the mercury in the manometer (d = 13.6 g/mL) was replaced by water (d = 1.0 g/mL), what would be the height of the column?
30 mm x 13.6 g/mL = 408 mm = 40.8 cm g/mL
Why might we wish to use a less dense liquid such as water or mineral oil in a manometer rather than mercury?
The less dense liquids have a much larger column height, and move through greater distances with the same changes in pressure. Small changes in pressure can be measured more accurately using these liquids rather than mercury.