chapter no.3 gases · pdf file... equal masses of methane and oxygen are mixed in an empty...
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Chapter No.3
GASES
TEXTBOOK EXERCISE
Q1: Select the correct answer out of the following alternative
suggestions: (i) Pressure remaining constant, at which temperature the
volume of a gas will become twice of what it is at 0oC .
a. 546oC b.200
oC c. 546 k d. 273k
Hind: V T.
(ii) Number of molecules in one dm3 of water is close to
a. b. c. d.
Hint: 1 dm3
of H2O =1000cm3
of H2O; 1000cm3
of H2O
=1000g of H2O
No of moles of H2O = moles ( 1 ml of H2O =1 f
of H2O)
(iii) Which of the following will have the same number of
molecules at STP?
a. 280cm3 of CO2 and 280 cm
3 of N2 O
b. 11.2dm3 of O2 and 32 g of O2
c. 44g of CO2 and 11.2 dm3 of CO
d. 28 g of N2 and 5.6 dm3 of oxygen
(iv) If absolute temperature of a gas is doubled and the
pressure is reduced to one half , the volume of the gas will
a. remain unchanged b. Increase four times
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c. reduce to d. be doubled
Hint: PV = RT, V= : V
(v) How should the conditions be changed to prevent the
volume of a given gas from expanding when its mass is
increased?
a. Temperature is lowered and pressure is
increased.
b. Temperature is increased and pressure is lowers.
c. Temperature and pressure both are lowered
d. Temperature and pressure both are increased.
Hint: PV = RT, V= : V ( M and R being
constant)
(vi) The molar volume of CO2 is maximum at
a. STP b. 127oC and 1 atm
c. 0oC and 2 atm d. 273
oC 1 atm
Hint:For1 mole of CO2 PV = RT, V= : V or Vm
(vii) The order of the rate of diffusion of gases NH3, SO2
C12 and CO2 si :
a. NH3 > SO2 > C12 > CO2
b. NH3 > CO2 > SO2 > C12
c. C12 >SO2 > CO2 >NH3
Hint: r
(viii) Equal masses of methane and oxygen are mixed in an empty
container at 25 oC . The fraction of total pressure exerted by
oxygen is
a. b.
c. d.
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Hint: In an empty container, the partial pressure of gas
is directly proportional to the mole-fraction of the gas. Partial
pressure mole fraction, at STP.
(viii) Gases deviate from ideal behavior at high pressure.
Which of the following is correct for non-ideality?
a. At high pressure, the ga molecules move in one
direction only.
b. At high pressure, the collision between the gas
molecules are increased manifold.
c. At high pressure, the intermolecular attractions
become significant.
d. At high pressure, the intermolecular attractions
become significant.
(ix) The deviation of a gas from ideal behavior is maximum
at
a. -10oC and 5.0atm b. -10
oC and 2.0atm
c. 100 oC and 2.0 atm d. 0.oC and 2.0atm
(xi) A real gas obeying van der Waals equation will resemble
ideal gas if
a. both ‘a’ and ‘b’ are larger b. both ‘a’ and
‘b’ are small
c. ‘a’ is small and ‘b’ is larger d. ‘a’ is larger
and ‘b’ is small
Ans: (i)c (ii)d (iii)a (iv)b (v)a (iv)b (vii)b (viii)a
(ix)d (x)a (xi)b
Q2. Fill in the blanks. (i) The product of PV has the S.I. unit of _______.
(ii) Eight grams each of O2 and H2 at 27oC will have total
K.E in the ratio of ________.
(iii) Smell of the cooking gas during leakage from a gas
cylinder is due to of the property of _______of gases.
(iv) Equal _________of ideal gases at the same temperature
and pressure contain ________number of molecules.
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(v) The temperature above which a substance exists only
as gas is called _____.
Ans. (i) atm dm3 (ii) :16 (iii) diffusion
(iv) Volumes: equal (v) Critical temperature
Q3. label the following sentences as true or false. (i) Kinetic energy of molecules is zero at 0
oC.
(ii) A gas in a closed container will exert much higher
pressure at the bottom due to gravity than at the top.
(iii) Real gases show ideal gas behavior at low pressure and
high temperature.
(iv) Liquefaction of gases involves decrease in
intermolecular spaces.
(v) An ideal gas on expansion will show Joule-Thomson
effect.
Ans. (i) False (ii) False (iii) True (iv) False
(v) False
Q4: (a) What is Boyle’s law of gases? Give its experimental
verification.
(b) What are isotherms? What happen to the positions of
isotherms when they plotted at high temperature for a particular
gas.
(c) Why do we get a straight line when pressures exerted on
a gas are plotted against inverse of volumes. This straight line
changes its position in the graph by varying the temperature.
Justify it.
(d) How will you explain that the value of the constant k in
equation PV=k depends upon
(i) The temperature of the gas (ii) the quantity of the
gas.
Ans: (a) Isotherms: The P-V curves obtained at constant
temperature are called isotherms. These curves are obtained by
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plotting a graph between pressure on the x-axis and volume on
the y-axis. Similar curves are obtained at fixed temperatures.
When the isotherms (P-V curves are plotted at high
temperature, they go away from the axes. The reason is that, at
higher temperature, the volume of the gas increase.
Ans.(c) A plot of P versus gives a straight line at constant
temperature. This shows that p is directly proportional to . This
straight line will meet at the origin where both P and are zero.
The P goes down as the gas expands, falling away to zero as the
volume approaches infinity (= =0)
This straight line changes its position because both pressure
and volume varies on varying the temperature. When
temperature is increased both pressure and volume will increase.
Keeping T constant and plotting P versus another straight line
is obtained. This straight line goes away from x-axis. However ,
when temperature is decreased both the values of P and V will
decrease. Again a straight line is obtained. This straight line will
be closer to the x-axis
Ans.(d) General equation: PV=nRT _______(1)
The general gas equation contains the Boyle’s law, for which
nRT are constant at fixed T and n.
Bolyle’s law: PV=k _______(2)
Form Eq (1) and Eq (2), we get
K=nRT
K=
K=constant x mT (at fixed R and M )
K mT _______(3)
This relation indicates that
(i) k T; it means k depends upon the temperature of the
gas
(ii) k m; it means k depends upon the quantity of the gas.
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Q5. (a) What is the Charles’s law? Which scale of temperature is
used to verify that = k (pressure and number of moles are
constant)
(b) A sample of carbon monoxide gas occupies 150.0mL at
25.0o C. it is then cooled at constant pressure until it occupies
100.0mL. What is the new temperature?
(c) Do you think that the volume of any quantity of a gas
becomes zero at -273oC. Is it not against the law of
conservation of mass? How do you deduce the idea of absolute
zero from this information?
Ans.(a) The relation, =k can be verified only when T is taken on the
Kelvin scale.
Ans.(b) V1 =150 mL T1 =273+25=298K
V2 =100 mL T2 =?
Formula Used:
=
or
T2 =
T2 =
T2 =198.67k
T2 =273+toC
toC=T2 – 273
toC=198.67-273
=-74.33oC Answer
Ans.(c) We know that : V1 =Vo (1+)
At -273oC, V-273 = Vo (1-)= Vo (1-1) =Vo x0=0
The volume of the gas become zero at- 273oC. But it is impossible
to imagine that a
Gas which is matter occupies no space. It goes against the law of
conservation of mass.
Hence, it follows that -273oC is the lowest temperature which a
body can ever have. So -273o
C is called the absolute zero of
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temperature. The scale of temperature on which -273oC is taken as
zero is called absolute scale of temperature.
Q6. (a) What is Kelvin scale of temperature ? Plot a graph for
one mole of an ideal gas to prove that a gas become liquid, carlier than
-273o C .
(b) Throw some light on the factor in Charles’s law.
Ans.(b) In Charles’s law, the factor (0.00366 ) is the coefficient of
expansion of given mass of gas at constant pressure. It shows
that a gas expands by parts of its volume at 0o C for a rise of
temperature of 1oC.
Statement of Charles’s Law: “At constant pressure, the volume
of a given mass of gas increases or decreases by of its volume
at0oC for every 1
oC rise or fall in temperature .”
Mathematically. Vt -=Vo +
It means that if we have 273 cm3 of gas at 0
oC, its volume
will increase by 1 cm3 for every 1
oC rise in temperature if it is
heated at constant pressure Thus,
At 1oC , the volume will become 274 cm
3.
At 2oC , the volume will become 271 cm
3.
At 273oC , the volume will become 0 cm
3.
Similarly, if the gas is cooled by 1oC at constant pressure, its
volume will decrease by 1 cm3 . Thus,
At -1oC , the volume of the gas becomes 272 cm
3.
At -2oC , the volume of the gas becomes 275 cm
3.
At -273oC , the volume of the gas becomes 546 cm
3.
Q7. (a) What is the general gas equation? Derive it in various
forms.
(b) Can we determine the molecular mass of an unknown gas
if we know the pressure, temperature and volume along with
the mass of that gas?
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(c) How do you justify from general gas equation that increase
in temperature or decrease of pressure decreases the density
of gas?
(d) Why do we feel comfortable in expressing the densities of
gases in the units of g dm-3
rather than g cm-3
, a which is used
to express the densities of liquids and solids.
Ans.(b) Yes, we can determine the molecular mass of an
unknown gas if we know its, P,T,V and m by applying the
following formula:
M=
Ans.(c) We know form general gas equation,
d=
d=constant x (M and R being constant)
d=
Density is directly proportional to pressure and
inversely proportional to temperature or decrease of pressure,
decreases the density of the gas.
Ans.(d) The densities of gases are very low. They are about 1000
times smaller than the densities of liquids and solids. So, if gas
densities are expressed in g cm-3
, then the values will be very
small. They may go to fourth place of decimal for some gases.
When we express the densities in g dm-3
, then the values of the
densities become reasonable to be expressed. For example, the
density of CH4 gas is 0.7168 g dm-3
at STP, but if it is expressed
in g cm-3
, then it is 0.0007168 g cm-3
at STP. Therefore, we feel
comfortable in expressing the densities of gases in the units of g
dm-3
rather than in g cm-3
.
Q8. Derive the units for gas constant R in general gas equation:
(a) When the pressure is in atmosphere and volume in dm3.
(b) when the pressure is in Nm-2
and volume in m3 .
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( c) when energy Is expressed in ergs.
Q9. (a) what is Avogadro’s law of gases?
(b) Do you think that 1 mole of H2 and 1 mole NH3 at 0oC and 1
atm pressure will have equal Avogadro’s number of
particles? If not, why?
(c) Justify that 1 cm3
of H2 and 1 cm3
of CH4 at STP will have
the same number of molecules, when one molecules of CH4
is 8 times heavier than that of hydrogen.
Ans (b) 1 mole of H2 and 1 mole of NH3 at 0oC and 1 atm pressure
will have equal number of molecules under the same conditions
of temperature and pressure. Hence, 1 cm3 of H2 and 1 cm3 of
CH4 at STP will have the same number of molecules.
Q10. (a) Dalton’s law of partial pressure is only obeyed by those
gases which do not have attractive forces among their molecules.
Explain it.
(c) Derive an equation to find out the partial pressure of a
gas knowing the individual moles of component gases and
the total pressure of the mixture.
(d) Explain that the process of respiration obeys the
Dalton’s law of partial pressure.
(e) How do you differentiate between diffusion and
effusion? Explain Graham’s law of diffusion.
Ans. (a) For Dalton’s law of partial pressure to hold, there will be
no attractive forces among the molecules on the walls of the gases.
The pressure of a gas is due to the collisions of the molecules on
the walls of the container. In the absence of attractive forces each
molecules of gas mixture will hit the walls of the container with
the same number of times and with the same force. Thus the partial
pressure of a given gas is unaffected by the presence of other
gases. In this case, the total pressure. Hence the law will not hold
in the presence of attractive forces among the molecules.
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Q11. (a) What is critical temperature of a gas? What is its
importance for
Liquefaction of gases? Discuss Linde’s method of
liquefaction of gases.
(b) What is Joule-Thomson Effect? Explain its importance in
linde’s method of liquefaction of gases.
Ans. (a) Importance of Critical temperature for liquefaction of
gases.
The critical temperature of the gases provides us the
information about the
Condition under which gases liquefy. For example,O2 ,has a
critical temperature 154.4k(-118,75 oC). It must be cooled below
this temperature before it can be liquefied by applying high
pressure.
Q12. (a) What is Kinetic molecular theory of gases? Give its
postulates.
(b) How dose Kinetic molecular theory of gases explain the
following gas laws:
(i) Boyle’s law (ii) Charles’s law
(iii) Avogadro’s law (iv) Graham’s law of
diffusion.
Q13. (a) Gases show non-ideal behavior at low temperature and
high pressure.
Explain this with the help of a graph.
(b) Do you think that some of the postulates of Kinetic
molecular theory of gases are faulty? Point out these
postulates.
(c) Hydrogen and helium are ideal at room temperature, but
SO2 and C12 are non ideal. How will you explain this?
Ans. (c) Hydrogen (b.p-253oC) and helium b.p-269
oC)have a very
low boiling points .They are far away from their boiling points at
room temperature. Also, they have smaller number of electrons in
their molecules and smaller molecular sizes, i.e., molecular weight.
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So, intermolecular forces are negligible at room temperature.
Hence, they behave as an ideal gases at room temperature.
On the other hand, SO2 (b.p-10oC) and C12 (b.p-34
oC)
have boiling points near to room temperature. They are not far
away from their boiling points at room temperature. Also, they
have larger number of electrons in their molecules and larger
molecular sizes. So, sufficient intermolecular attractive forces are
present at room temperature. Hence, they behave as non-ideal at
room temperature.
Q14. (a) Derive van der Waals equation for real gases.
(b) What is the physical significance of van der Waals
constant, ‘a’ and ‘b’.
Give their units.
Ans. (b) Physical Significance of van der Waals constant ‘a’ and
‘b’
(i) Significance of’ a’: The value of constant ‘a’ is a
measure of the intermolecular attractive forces and greater will be
the ease of its liquefaction.
Units of ‘a’ :
The units of ‘a’ are related to the units of pressure,
volume and number of moles.
P= or
a= =
a= atm dm6
mol-2
In SI units: a= ==Nm4
mole -2
(iii) Significance of ‘b’: The value of constant ‘b’ us related
to the size of the molecule. Larger the size of the molecule,
lager is the value of ‘b’. It is effective volume of the gas
molecules.
Units of ‘b’:
‘b’ is the compressible volume per mole of gas. So the units
of ‘b’ are related to the units of volume and moles.
V=nb or b =
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b==dm3
mol-1
In SI units: b===dm3
mol-1
Q15. Explain the following facts:
(a) The plot of PV versus P is a straight line at constant
temperature and with a fixed number of moles of an ideal
gas.
(b) The straight line in (a) is parallel to pressure-axis and goes
away from the pressure axis at higher pressure for many
gases.
(c) The van der walls constant ‘b’ of a gas is four times the
molar volume of that gas
(d) Pressure of NH3 gas at given conditions (say 1 atm pressure
and room temperature) is less as calculated by van der Waals
equation than that calculated by general gas equation.
(e) Water vapors do not behave ideally at 273 k.
(f) SO2 is comparatively non-ideal at 273 k but behaves ideally
at 327 K.
Ans. (a) At constant temperature and with a fixed number of
moles of an ideal gas, when the pressure of the gas is varied, its
volume changes, but the product PV remains constant. Thus,
P1 V1 = P2 V2 = P3 V3 =
Hence, for any fixed temperature, the product PV when plotted
against P.a straight line parallel to P-axis is obtained. This straight
line indicates that PV remains constant quantity.
Ans. (b) Now, increase the temperature of the same from T1 to T2
.At constant temperature T2 and with the same fixed number of
moles of an ideal gas, when the constant. However, the value of
PV increase with increase in temperature. On plotting graph
between P on x-axis is obtained. This straight line at T2 will be
away from the x-axis. This straight line also shows that PV is a
constant quantity.
Ans. (c) Excluded volume, ‘b’ is four times the molar volume
fo gas. The excluding with each other as shown in Fig. The spheres
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are considered to be non-compressible. So the molecules cannot
approach each other more closely than the distance, 2r . Therefore,
the space indicated by the dotted sphere having radius, 2r will not
be available to all other molecules of the gas. In other words, the
dotted spherical space is excluded volume per pair of molecules.
Let each molecules be a sphere with radius =r
Volume of one molecules (volume of sphere)=
The distance of the closest approach of 2 molecules =2 r
The excluded volume for 2 molecules=
The excluded volume for 1 molecule=
=
=4Vm =b
The excluded volume for ‘n’ molecules=n b
Where Vm is the actual volume of a molecule.
Hence, the excluded volume or co-volume or non-compressible
volume is equal to 4 times the actual volume of the molecules of
the gas.
Ans. (d) The pressure of NH3 calculated by general gas equation is
high because it is considered as an ideal gas. In an ideal gas, the
molecules do not exert any force of attraction on one another. On
the other hand, when the pressure of NH3 is considered as a real
gas. Actually, NH3 is a real gas. It consists of polar NH3 molecules
approaches the walls of the container, it experiences an inward
pull. Clearly, the molecule strikes the wall with a lesser force than
it would have done it these are no attractive forces. As a result of
this , the real gas pressure is less than the ideal pressure.
Ans. (e) Water vapors present at 273K do not behave ideally
because polar water molecules exert force of attraction on one
another.
Ans. (f) At low temperature, the molecules of SO2 possess low
kinetic energy. They come close to each other. The e
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intermolecular attractive forces become very high. So, it behave
non-ideally at 273K. At high temperature, the molecules of SO2
have high kinetic energy. The molecules are at larger distances
from one other another. The intermolecular attractive forces
become very weak. So, it behaves ideally at 327K.
Q16. Helium gas in a 100 cm3
container at a pressure of 500 torr is
transferred to a container with a volume of 250cm3.What will be
the new pressure
(a) if no change in temperature occurs
(b) if its temperature changes from 20 oC to 15
oC?
Solution: (a) Given: P1=500 torr P2 =?
V1 =100cm3 V2 =250cm3
Formula used: P2 V2 = P1V1
P2 =
P2 =
=250torr Answer
(b) Given: P1 =500torr ; P2 =?
V1 =100cm3 ; V2 =250 cm3
T1 =273 +20=293K : T2=273+15=288K
Formula Used: =
P2 = x
=
=196.58 torr Answer
Q17. (a) What are the densities in kg/m3
of the following gases at
STP
(i) Methance, (ii) oxygen (iii) hydrogen
(P=101325Nm-2
, T=273k, molecular masses are in kg
mol-1
)
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(b) Compare the values of densities in proportion to their
mole masses.
(c) How do you justify that increase of volume upto 100 dm3
at 27 oC of 2 moles of NH3 will allow the gas behave ideally.
Solution: (a) (i)Given P=101325Nm
-2
T=273K
Molar mass of CH4 =12+4=16g mol-1
=16x10-3
kg mol-1
R=8.3143NmK-1
mol-1
d=?
Formula Used: d=
d=
d= 0.714kgm-3
P=101325Nm-2
T=273k
Molar mass of O2 =32g mol-1
d= ?
Formula used: d=
d=
d=1.428kgm-3
(iii) P=101325Nm-3
; T=273k
Molar mass of H2 =2x10-3
kg mol-1
:R=8.3143 NmK-1
mol-1
d=?
Formula Used: d=
d=
d=0.089kg m-3
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Q18. A sample of krypton with a volume of 6.25 dm3
, a pressure of
765 torr and a temperature of 20oC is expanded to a volume of 9.55 dm
3
and a pressure of 375 torr. What will be its final temperature in oC?
Solution: P1 = 765torr ; P2 =375torr
V1 =6.25dm3 ; V2 =9.55 dm
3
T1 =273+20 =273k ;T2 =?
Formula used: x
T2 =
T2 =
T2 =
T1 =273+o C
oC =T-273=219.46-273
=-53.54 oC Answer
Q19. Working at a vacuum line, a chemist isolated a gas in a weighing
bulb with a volume of 255 cm3
, at a temperature of 25 oC and
under a pressure in the bulb of 10.0torr. The gas weight 12.1 mg.
what is the molecular mass of this gas?
Solution: V=255 cm
3 =0.255 dm
3
P=10.0torr =
T1 =273+25=298K
m=12.1mg=0.0121g
R=0.0821dm3
atom K-1
mol-1
Formula Used: PV= RT
M=
M=
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M=87.93 g Mol-1
Answer
Q20. What pressure is exerted by a mixture of 2.00g of H2 and 8.00g
of N2 at 273 K in a dm3 vessel?
Solution: Given: V=10dm
3 ; T=273k
R=0.0821 dm3 atm K
-1 mol
-1
P H2 =? PN2 =?
Mass of H2 =2.00g Mass of N2 =8.00g
Molar mass of H2 =2g Molar mass of N2 =28g
mol-1
nH2 = =1 mole ;n N2 = =0.286
mole
n=mH2 +nN2 =1+0.286=1.286 moles
PV=nRT
P x10 dm3 =1.286 molx 0.0821dm
3 atm k
-1 mol
-1 x 273K
P=
P=2.88 atm Answer
Q21. (a) The relative densities of two gases A and B are 1:1.5 find
out the volume of B which will diffused in the same time in which
150 dm3 of A will diffuse?
(b) Hydrogen (H2 ) diffuses through a porpous plate at a rate of
500cm3 per minute at 0oC. What is the rate of diffusion of
oxygen through the same porpous plate 0oC?
(c) The rate of effusion of an unknown gas A through a
pinhole is found to be 0.279 times the rate of effusion of H2
gas through the same pinhole.
Calculate the molecular mass of the unknown gas at STP.
Solution: Given:
dA =1 rA=150dm3
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rB=
(b) rH 2=500 cm3 per minute MH2 = 2 g mol
-1
ro2=? Mo2 =32 g mol-1
=
=4
ro2==125cm3 Answer
(c) Given: rH2=1 rA=0.279
MA =? MH2=g mol-1
=
=
=
0.078=
MA =
MA =25.64 g mol-1
Answer
Q22. Calculate the number of molecules and the number of atoms in
the given amounts of each gas
(a) 20cm3
of cH4 at 0oC and pressure of 700 mm of
mercury
(b) 1 cm3
of NH4 at 100oC and pressure of 1.5 atm
Solution: (a) Given: P=700 mm =0921atm
V=20cm3
=0.02dm3
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R=0.0821dm3 atm K
-1 mol
-1
T=273+0=273K
N=?
Formula Used: PV=nRT
n=
n=
No of molecules =No of moles x NA
=8.22x1020
molecules Answer
Since 1 molecule of CH4 contains =5 atoms
Therefore, No of atoms =5x4.948x1020
=24.74x1020
atoms Answer
(b) Given P=105 atm ; R=0.0821dm3
atm
K-1
mol-1
V=1 cm3
=0.001dm3 ;
T=273+100=373K
n=?
Formula Used: n=
n=
No of molecules =4.89x10-5
x6.02x1023
=29.44x1018
=2.944x1019
molecules
Answer
No of atoms =2.944x1019
x4=1.18x1020
atoms Answer
Q23. Calculate the masses of 1020
molecules of each of H2 ,O2 and CO2
at STP. What will happen to the masses of these gases, when the
temperature of these gases are increased by 100oC and pressure is
decreased by 100 torr.
Solution: (a) Given: Molecules of H2 =10
20
Now , 6.02x1023
molecules of H2 at STP =1mole =2g
1020
molecules of H2 at STP =
=0.332x10-5
g
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=3.32x10-4
g
Answer (b) Given: Molecules of O2 =10
20
Now, 6.02x1023
molecules of O2 at STP =32g
1020
molecules of CO2 STP =
=5.32x 10-3
g
Answer
(c) Given: Molecules of CO2 =1020
Now, 6.02x1023
molecules of CO2 at STP =44g
1020
molecules of CO2 at STP =
=7.30x10-3
g Answer
Q24. (a) Two moles of NH3 are enclosed in a 5dm3
flask at 27oC.
calculate the pressure exerted by the gas assuming that
(i) it behaves like an ideal gas
(ii) it behaves like a real gas
a=1.17 atm dm6
mol-2
b=0.0371 dm3
mol-1
(b) Also calculate the amount of pressure lessened due to
forces of attractions at these conditions of volume and
temperature.
(c) Do you expect the same decreased in the pressure of 2
moles of NH3 having a volume of 40dm3 and at temperature
of 27oC.
Solution: (a.i) Given: V=5dm
3 ; T=273+27=300K
n=2mole ; R=0.821dm3
atm K-
1 mol
-1
P=?
Formula Used: PV=nRT
P=
P==
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P=9.852 atm Answer
(a.ii) Given: n=2mol
a=4.17atm dm3
mol-1
b=0.0371dm3
mol-1
V=5dm3
R=0.0821atm dm3 K
-1 mol
-1
T=273+27=300K
P=?
Formula Used: (V-nb)=nRT
On rearranging the equation
P=-
On substituting the values
P=-
P=-
P=-
P=9.99-0.667
P=9.32atm Answer
(b) Difference of pressure , P=9.852-9.32=053atm Answer
(c) Given: n = 2 mol ;
V=40dm3;T=273+27=300K
p =?
Formula Used: PV=nRT
P=
P=
P=1.232 atm
The decrease in pressure is not the same
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