Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Chapter 10 P31 Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol
I(ISBN #) 978-049511374-3
THE PROBLEM STATEMENT
Ch 10 P31 . (a) An ideal gas occupies a volume of 1.0 cm3 at 20°C and atmospheric pressure.
Determine the number of molecules of gas in the container.
(b) If the pressure of the 1.0-cm3 volume is reduced to 1.0 x 10-11 Pa (an extremely good
vacuum) while the temperature remains constant, how many moles of gas remain in the
container?
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Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 10 P31 . (a) An ideal gas occupies a volume of 1.0 cm3 at 20°C and atmospheric pressure.
Determine the number of molecules of gas in the container.
(b) If the pressure of the 1.0-cm3 volume is reduced to 1.0 x 10-11 Pa (an extremely good
vacuum) while the temperature remains constant, how many moles of gas remain in the
container?
Basic Solution (including BRAINSTORMING-Definitions, concepts , principles and Discussion)
(a)
Given: P1= 1 atm = 1.013x105 Pa, T1 = 20oC =T, V1 =1.0 cm3 = V2 =V.
Find: (a) N1 = n1/NA NA = 6.02 molecules/mole.
(b)
Given: P2 = 1.0 x 10-11 Pa , T2 = T1 = 20oC =T = T V2 =V1 = V= 1.0 cm3 .
Find: n2
Conversions
V2 =V1 = V= 1.0 cm3 1.0 [cm]3 = 1.0 [cm( )m/cm]3
= 1.0 [cm(10-2)m/cm]3 = 1.0 [(10-2)m]3 = 1.0 x10-6 m .
T2 = T1 = T = 20oC = 273.15 K + 20 = 293.15 K.
Physical Principle
Ideal Gas Law:
PV = nRT
(1)
R = the Universal Gas Constant = 8.31 J/mole K.
T1 = T = 293.15 K T and V1 = V = 1.0 x10-6 m
(a)
n1 = P1V1/(RT1) = P1V1/(RT)
=
(2)
1.013x10-5 Pa *1.0 x10-6 m3/(8.31 J/mole K*293.15 K) = 4.2x10-5 moles
N1 = n1/NA
-5
= 4.2x10
moles/6.02 molecules/mole = 2.5x1019 molecules.
(b) Use Eq. (2) now for n2 with
T2 = T = 293.15 K T and V2 = V = 1.0 x10-6 m
gives
n2
= P2V2/(RT2 ) = P2V/(RT)
= 1.0 x 10-11 Pa 1.0 x10-6 m3/(8.31 J/mole K*293.15 K) =4.1 x10-21 mol.
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