GROUP (20)
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SHEET (1)
INTRODUCTION and
DEFINITION OF
TERMS.
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Question No. (1)
How many H2O molecules are there in a snowflake
weighing 1 mg?
The atomic weights of hydrogen and oxygen are 1 and 16,
respectively. The Avogadro's number (NA) =6.023x 1023
atoms/mole.
The Answer of Question No. (1)
Mass of one mole of H2O = (2 x 1+1 x 16)
= 18 g.
One mole of H2O has Avogadro’s number of molecules
then, 1 mg has = (1 x 10-3 x 6.023 1023)/18
= 3.3461 x1019 molecules.
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Question No. (2)
Determine the number of moles of CO2 in 454 grams of
CO2, The atomic weights of carbon and oxygen are 12
and 16, respectively.
The Answer of Question No. (2)
Mass of one mole of CO2 = (1 x 12+2 x 16)
= 44 g.
Number of moles of CO2 in 454 grams = 10.318 moles.
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Question No. (3)
Calculate the mass in grams of a single carbon atom, the
atomic weight of carbon is 12 and NA=6.023 x1023
atoms/mole.
The Answer of question No. (3)
Mass of a single carbon atom = 12/ NA
=12/ 6.023 x1023
= 1.99236 x10-23 grams.
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Question No. (4)
Calculate the mass in grams of 2.5x109 H2O molecules,
the atomic weights of hydrogen and oxygen are 1 and 16,
respectively. NA = 6.023 x 1023 atoms/mole.
The Answer of Question No. (4)
Mass of 6.023 x 1023 H2O molecules = 18 g.
Mass of 2.5x109 H2O molecules = (2.5x109 x18)/ NA
= 7.471 x10-14 grams.
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Question No. (5)
A sample of Helium gas at 25 ºC is compressed
isothermally from 200 cm3 to 0.2 cm3. Its pressure is now
3 cm-Hg. What was the original pressure of helium?
The Answer of Question No. (5)
V1 = 200 cm3, V2 =0.2 cm3, P2 =3 cm-Hg, P1 =??
Applying Boyle’s law
P1
P2
=
V2
V1
P1 = (0.2 x3)/200
= 3 x10-3 cm-Hg.
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Question No. (6)
A sample of air occupies 1 lit at 25 ºC and 1 atm, what
pressure (in atm) is needed to compress it to 100 cm3 at
this temperature?
The Answer of Question No. (6)
V1 = 1 lit, P1 = 1 atm, V2 =100 cm3 =0.1 lit, P2 =??
Applying Boyle’s law
P1
P2
=
V2
V1
P2 = (1 x1)/0.1
= 10 atm.
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Question No. (7)
One hundred cubic cm of O2 is collected over water at 23
ºC and 800 Torr. Compute the volume of O2 in m3 at P=23
Torr. and 298K?
The Answer of Question No. (7)
V1= 100 cm3, T1=296°K, P1= 800 Torr, P2= 23 Torr
T2= 298°K, V2=??
Number of moles is constant, by applying General
Law of Gases
P1 V1
T1
=
P2 V2
T2
V2= (800 x100 x298)/ (23 x296)
= 3501.763 cm3
= 3.501763 x10-3 m3.
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Question No. (8)
A given mass of gas occupies a volume of 240 mlit at a
pressure of 1.25 atm., by how much would the volume
change if the pressure were altered to 0.75 atm. at the
same temperature?
The Answer of Question No. (8)
V1=240 mlit, P1= 1.25 atm, P2= 0.75 atm, V2=??
Applying Boyle’s law
P1
P2
=
V2
V1
V2= (1.25 x240)/0.75
= 400 mlit.
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Question No. (9)
A hydrogen gas thermometer is found to have a volume of
100 cm3 when placed in an ice-water bath at 0 ºC. What is
the hydrogen temperature if hydrogen volume becomes
87.2 cm3 and its pressure is kept constant?
The Answer of Question No. (9)
V1=100 cm3, T1=273°K, V2=87.2cm3, T2=??
Applying Charles’ law
T2
T1
=
V2
V1
T2= (273 x87.2)/100
= 238.056°K
=-35°C.
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Question No. (10)
To what temperature (in K) must a 1 lit sample of a perfect
gas be cooled from 25 ºC in order to reduce its volume to
100 cm3 at the same pressure?
The Answer of Question No. (10)
V1=1lit =1000cm3, V2=100cm3, T1=298°K, T2=??
Applying Charles’ law
T2
T1
=
V2
V1
T2= (298 x100)/1000
= 29.8 °K.
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Question No. (11)
The volume of a given mass of gas is 360 mlit at 15 ºC, at
what temperature; will the volume be 480 mlit assuming
constant pressure?
The Answer of Question No. (11)
V1=360 mlit, T1=288°K, V2=480 mlit, T2=??
Applying Charles’ law
T2
T1
=
V2
V1
T2= (288 x480)/360
= 384°K
= 111 ºC.
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Question No. (12)
2.50 g of XeF4 gas is placed into an evacuated 3 lit
container at 80 ºC. What will be the pressure value in the
container? The atomic weights of Xenon and Fluorine are
131.3 and 19, respectively.
The Answer of Question No. (12)
V=3 lit, T=353 ºK, P=??
Mass of one mole of XeF4 = (1 x131.3+4 x19)
= 207.3 grams.
Number of moles of 2.50 g of XeF4 = 2.5/207.3
= 0.0121moles.
By applying General Law of Gases PV = nRT
P= (0.0121 x0.082057 x353)/3
= 0.1168 atm.
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Question No. (13)
Could 131 g of Xe gas in a vessel of 1 lit exert a pressure
of 20 atm at 25 ºC if it behaves as a perfect gas? The
atomic weight of Xenon is 131.3.
The Answer of Question No. (13)
Xe gas is an inert gas, so it is monatomic molecule.
n=131/131.3=0.99772moles, v=1 lit, p= 20atm, T=298 ºK,
R= 0.082057 lit.atm/ k.mole
To check the behavior of the gas, we apply the General
Law of Gases
(PV=nRT)
PV/nRT=1 (for ideal or perfect gas)
By substitution = (20 x1)/ (0.99772 x0.082057 x298)
= 0.82 (not equal 1)
So, it is not a perfect gas.
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Question No. (14)
In an industrial process, N2 is heated to 500K in steel tank, if it
enters the system at 300 K and 100 atm. What pressure does it
exert at the final working temperature?
The Answer of Question No. (14)
T2 = 500 ºK, T1 =300 ºK, P1 =100 atm, P2 =??
Volume is constant, then
T2
T1
=
P2= (500 x100)/300
= 166.67atm.
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P2
P1
Question No. (15)
How many moles of O2 are contained in 10 lit of the gas at
a pressure of 75 cm-Hg and a temperature of 27 ºC
assuming ideal behavior?
The Value of R is 0.082 lit.atm/ (deg.mole).
The Answer of Question No. (15)
V=10 lit, P=75 cm-Hg= 0.987atm, T=300 ºK,
R=0.082 lit.atm/ ºK.mole, n =??
General Law of Gases PV=nRT
n= (0.987 x10)/ (0.082 x300)
= 0.40122 moles.
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Question No. (16)
A given mass of gas occupies a volume of 250 mlit at 21
ºC and a pressure of 1.4 atm, at what pressure wills the
volume be 300 mlit, when the temperature is raised to 49
ºC?
The Answer of Question No. (16)
V1=250 mlit, P1=1.4 atm, T1=294 ºK, V2 =300 mlit
T2 =322 ºk, P2 =??
Mass is constant, then,
P1 V1
T1
=
P2 V2
T2
P2 = (1.4 x250 x322)/ (294 x300)
= 1.278 atm.
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Question No. (17-a)
Calculate the pressure exerted by one mole of ethane if it
behaves as an ideal gas, under the conditions (a) 273K in
22.414 dm3?
The Answer of Question No. (17-a)
n = 1, T = 273 ºK, V=22.414 dm3 =22.414 lit, P =??
Applying General Law of Gases
PV =nRT
P= (1 x0.082 x273)/ (22.414)
= 0.9988 atm.
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Question No. (17-b)
Calculate the pressure exerted by one mole of ethane if it
behaves as an ideal gas, under the conditions
(B) 1000K in 100 cm3.
The Answer of Question No. (17-b)
n = 1, T = 1000 ºK, V=100 cm3, P =??
Applying General Law of Gases
PV = nRT
P= (1 x0.082 x1000)/ (0.1) = 820 atm.
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