Transcript Document

Chapter 9 Worked Example 1
The density of water at 20 o C is 0.998 g.cm-3. What height
would the column of liquid be in a water barometer at 20 oC
when the atmospheric pressure corresponds to 760. mm of
mercury?
Solution
P = dw hwg = dHg hHgg
Hence
hw =
dHghHg
dw
(13.595 g cm-3)(760. mm)
=
(0.998 g cm-3)
= 1.04 x 104 mm (10.4 m)
Chapter 9 Worked Example 2
A student attaches a glass bulb containing neon gas to an open-tube
manometer and calculates the pressure of the gas to be 0.890 atm.
(a) If the atmospheric pressure is 762 Torr, what height difference
between the two sides of the mercury in the manometer did the
student find?
(b) Which side is higher, the side of the manometer
attached to the bulb or the side open to the
atmosphere?
(c) If the student mistakenly switches the numbers
for the sides of the manometer when recording
the data in the laboratory notebook, what
would be the reported pressure in the gas bulb?
Solution
Chapter 9 Worked Example 3
In a petroleum refinery a 750.-L container containing ethylene
gas at 1.00 bar was compressed isothermally to 5.00 bar. What
was the final volume of the container?
Solution
Isothermally means at constant temperature, hence Boyle's
law can be used.
P1V1 = P2V2,
V2 = 150. L
(1.00 bar)(750. L) = (5.00 bar)V2
Chapter 9 Worked Example 4
A parcel of air (the technical term in metereology for a
small region of the atmosphere) of volume 1.00 x 103 L
at 20. oC and 1.00 atm rises up the side of a mountain
range. At the summit, where the pressure is 0.750 atm,
the parcel of air has cooled to -10. o C. What is the volume
of the parcel of air at that point?
Solution
P1V1
n1T1
=
P2V2
n2T2
where n1 = n2
(1.00 atm)(1.00 x 103 L)
(293.15 K)
V2 = 1.20 x 103 L
=
(0.750 atm)V2(L)
(263.15 K)
Chapter 9 Worked Example 5
Calculate the volume occupied by 1.0 kg of hydrogen at 25 oC
and 1.0 atm.
Solution
We can use the ideal gas equation PV = nRT, after first finding
the number of moles of H2 in 1.0 kg.
n=
Mass
=
Molar mass
1.0 x 103 g
= 496 mol
2.016 g/mol
(1.0 atm)V(L) = (496 mol)(8.206 x 10-2 L atm K-1mol -1)(298.15 K)
V = 1.2 x 104 L
Chapter 9 Worked Example 6
A sample of methane gas, CH4, was slowly heated at a constant pressure
of 0.90 bar. The volume of the gas was measured at a series of different
temperatures and a plot of volume vs. temperature was constructed.
The slope of the line was 2.88×10-4 L K-1. What was the mass of the
sample of methane?
Solution
Chapter 9 Worked Example 7
A baby with a severe bronchial infection is in respiratory distress.
The anesthetist administers heliox, a mixture of helium and oxygen
with 92.3% by mass O2. What is the partial pressure of oxygen being
administered to the baby if the atmospheric pressure is 730 Torr?
Solution
n(He) =
0.077 g
= 0.0193 mol
n(O2) =
4.00 g mol -1
x(He) =
0.0193 mol
0.0481 mol
0.923 g
= 0.0288 mol
32.0 g mol -1
= 0.401
x(O2) =
0.0288 mol
0.0481 mol
= 0.599
[= 1.00 - 0.401 (only two components)]
Chapter 9 Worked Example 8
Estimate the root mean square speed of water molecules
in the vapor above boiling water at 100. oC.
Solution
Molar mass of water is 18.01 g mol -1 or 0.01801 kg mol-1.
From vrms = (3RT/M) 1/2,
3 x (8.3145 J K-1 mol -1) x (373 K)
vrms
=
=
0.01801 kg mol-1
719 m s-1