Chapter No.3 GASES ...(d) Why do we feel comfortable in expressing the densities of gases in the...

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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.


3.


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.


3.


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=?


P=

P==

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21


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 =?


P=

P=

P=1.232 atm

The decrease in pressure is not the same

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Chapter No.3 GASES ...(d) Why do we feel comfortable in expressing the densities of gases in the...

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