Kumar Vaibhav - 18. Electrochemistry

Chapter 13

Kinetic Theory Solutions

SECTION - A

Objective Type Questions 1.

Selec Sel ectt the the appr approp opria riate te pro prope pert rty y of an an idea ideall gas gas (1)) Its molecu (1 molecules les are infinit infinitesim esimally ally small small (2)) There are no (2 no forces of of interaction interaction between between its its molecules molecules (3)) It strictl (3 strictly y obeys obeys the the ideal ideal gas laws laws (4) Al Alll of of th thes ese e

Sol. Answer (4) All of the statements are true for an ideal gas. 2.

A rea reall gas gas be beha have ves s as as an an ide ideal al ga gas s at at (1) Very low pressure and high temperature

(2) High pressure and low temperature

(3) High pressure and high temperature

(4) Low pressure and low temperature

Sol. Answer (1) At very low pressure the force of interaction between particles may be considered negligible. Also at high temperature the force of inter molecular interaction decreases 3.

A gas at a pressure P 0 is contained in a vessel. If the masses of all the molecules are halved and their velocities doubled, then the resulting pressure P will will be (1) 4P 0

(2) 2P 0

(3)) P 0 (3

(4)) (4

P 0 2

Sol. Answer (4) P 0 =

P  =

1 3



mn

v

2

v

P 1 mn  (2v )2  0 3 2v 2

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2

4.

Kinetic Theory

Solution of Assignment

If E is is the energy density in an ideal gas, then the pressure of the ideal gas is (1)) (1

P



2

E

3

(2)) (2

P 

3 2

E

(3)) (3

P 

5 2

(4)) (4

E

P



2 5

E

Sol. Answer (1) 1

E =

2

2

3 2

P =

3



n v

1 mN

P =

5.

mv

v

v

2

E

A sam sampl ple e of of gas gas in a box box is at pr pres essu sure re P 0 and temperature T 0. If number of molecules is doubled and total kinetic energy of the gas is kept constant, then final temperature and pressure will be (1)) T 0 , P 0 (1

(2)) T 0 , 2P 0 (2

T 0

(3)) (3

2

, 2P 0

(4)) (4

T 0

2

, P 0

Sol. Answer (4) 1 mN

P 0 =

3 v

2  v rms

If E 0 is initial KE of one molecule E   2n

nE 0 =

⇒ E 

E 2

1  ⇒  mv 2  2 

1 2



1 2

2

mv 

 v 

v 2

Thus KE of every molecule becomes half. Hence temperature becomes 1 m2N  v

P  

Thus

6.

T

3



v

T 0

2

T 0 2

.

2

    P 0  2

, P   P 0

By increasi increasing ng tempera temperature ture of of a gas by 6°C its press pressure ure increas increases es by 0.4 %, at cons constant tant volume volume.. Then initial initial temperature of gas is (1) 1000 K

(2) 1500 K

(3) 2000 K

(4) 750 K

Sol. Answer (2) P2 P1



T 2

T 1 = T

T 1

T 2 = T T + +6 P2 P1 P2

1  P 1 P 1

0.4 =

T 2 T 1

1

 T  6 100  1  T 

 100  

600 T

 T = = 1500 K



7.

Kinetic Theory

3

Boy oylle’ e’s s la law is is ob obey eyed ed by (1)) Real gas (1 gas of const constant ant mass mass and and temperatur temperature e (2)) Ideal gas (2 gas of consta constant nt mass mass and tempe temperature rature (3)) Both ideal and (3 and real gases gases at constant constant temperatur temperature e and variable variable mass (4)) Both ideal and (4 and real gases of constan constantt mass and variable variable temperat temperature ure

Sol. Answer (2) Boyle's law states that if m and T are are constant. V



1 P

and gas laws are only valid for ideal gases. 8.

For an ideal ideal gas the the fractiona fractionall change change in its volume volume per per degree degree rise in tempe temperatu rature re at consta constant nt pressure pressure is equal to [T [ T is is absolute temperature of gas] (1)) T 0 (1

(2)) T (2

(3)) T –1 (3

(4)) T 2 (4

Sol. Answer (3) PV = PV = n RT

... (1)

P dv = = n R dT

... (2)

Dividing (2) by (1) dV V

dT



T

Fractional change in volume per degree rise in temperature dV V dT

9.



1

T

The raise raise in the the temperat temperature ure of a given given mass mass of an ideal ideal gas at at constant constant press pressure ure and and at tempera temperature ture 27° 27° to double its volume is (1) 327°C

(2) 54°C

(3) 300°C

(4) 600°C

Sol. Answer (1) PV = PV = n RT Initial temperature T 0 = 300 K V 0  T 0 2V 0  2 2T T 0 2T 0 = 600 K

 T = = 600 – 300 = 300 K 10.. The aver 10 average age velo velocit city y of gas gas molecu molecules les is is (1) Pr Prop opor ortio tiona nall to (3) Ze Zero

T

(2)) Pro (2 Propo port rtio iona nall to T (4) No Not possible to determine

Sol. Answer (3) v



v

=



3RT M T


4

11.

Kinetic Theory


Which of the following following methods methods will will enable the the volume of of ideal gas to be be increased increased four times? times? (1)) Doubl (1 Double e the temperatu temperature re and reduce reduce the pressu pressure re to half (2)) Doubl (2 Double e the temperatu temperature re and also double double the the pressure pressure (3)) Reduc (3 Reduce e the temperatu temperature re to half and double double the the pressure pressure (4)) Reduc (4 Reduce e the temperatur temperature e to half and reduce reduce the pressure pressure to half half

Sol. Answer (1) T 0

V 0 = k

P 0

T 0 4k P 0

4V 0 =

k 

=

2T 0 P 0 / 2

 Temperature is doubled and pressure halved. 12.. A co 12 cont ntai aine nerr has has N molecules molecules at absolute temperature T . If the number of molecules is doubled but kinetic energy in the box remain the same as before, the absolute temperature of the gas is

(1)) T (1

(2)) (2

T

(3) 3T

2

(4) 4T

Sol. Answer (2) Initial energy of gas = Final energy Let K.E. of each molecule initially be E 0.

 Total kinetic energy = E 0 × n Let final kinetic energy of each molecule be E f . E 0 × n = E f × 2n E f =

E 0 2

Since temperature is the average kinetic energy of molecules.

T 0 = KE 0

T f =

 Temperature becomes T becomes T f =

KE 0 2

T 0 2

13.. During 13 During an experi experimen mentt an ideal ideal gas is found found to obey obey an additio additional nal law VP 2 = constant. The gas is initially at temperature T and and volume V , when it expands to volume 2V 2 V , the resulting temperature is

(1)) (1

T 2

(2) 2T

(3)) (3

2 T

(4)) (4

T 2



Kinetic Theory

5

Sol. Answer (3) VP 2 = constant As PV = = RT  P = = 2

VR T V

2V

RT V

. Thus from (i)

2

= constant 

2

2

T 2

... (i)

T

2

= constant

V

2

T



⇒

V

T2

 T

2

14. When pressure pressure remaining remaining constant constant,, at what temperature temperature will the the r.m.s. r.m.s. speed speed of a gas molecules molecules increase increase by 10% of the r.m.s. speed at NTP? (1) 57 57.3 K

(2) 57 57.3ºC

(3) 55 557.3 K

(4) –5 –57.3ºC

Sol. Answer (2) 3 RT

V = V =

Let

V  110

K

100

V 1.1

or V V = =

M

V

K T

T 2

T

=

T 2

T 2 = 1.21 T Putting T = = 273 K T 2 = 57.33°C 15. Two Two thermally thermally insulated insulated vessels vessels 1 and 2 are are filled with with air at temperat temperatures ures (T (T 1, T 2), volume (V (V 1, V 2) and pressure (P 1, P 2) respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be (1)) T 1 + T 2 (1

(2)) (2

(T1  T 2 ) 2

(3)) (3

T1T2 ( PV P1V1 P1V1T2

 P2V2 )

 P2V2T1

(4)) (4

T1T2 ( PV P1V1 P1V1T1

 P2V2 )

 P2V2T2

Sol. Answer (3) Total number of moles remain constant.

  n2

n1 + n2 = P1V1 T1



n1

P2V2 T2



P1V1 T



P2V2 T

Solving, we get T = T =

P1V1  P2V2 P1V1 T1



P2V2 T 2


6

Kinetic Theory


16.. The ave 16 average rage spee speed d of gas mole molecule cules s is v at at pressure P . If by keeping temperature constant the pressure of gas is doubled, then average speed will become (1)) (1

2

(2)) v (2

v

(3) 2v

(4)) (4

v

2

Sol. Answer (2) v av 

T

Since temperature is constant v av is constant. 17. Four molecules molecules of a gas gas have speeds speeds 1, 2, 3 and 4 km/s. km/s. The value value of the r.m.s. r.m.s. speed speed of the gas gas molecules molecules is

(1)) (1

1

15

2

km km/s

(2)

1

10

2

km km/s

(3) 2. 2.5 km/s

(4)

15 2

km/s

Sol. Answer (4)

R.M.S. speed =

2

v

=

v

=

v

=

1

v

2 1

 v 22  v 32 ...v 2 n

n

 22  32  42 4

30 4

15 2

18.. The r.m.s 18 r.m.s.. speed speed of the molecu molecules les of enclos enclosed ed gas is is V . What will be the r.m.s. speed if pressure is doubled keeping the temperature same?

(1)) (1

V

(2)) V (2

2

(3) 2V

(4) 4V

Sol. Answer (2) Temperature is a quantities which denotes a value which gives the average kinetic energy of molecules in a gas. This depends on velocities of gas molecules and vice-versa. If temperature does not change V r.m.s. will also not change. 19. The effect effect of tempera temperature ture on Maxwel Maxwell’s l’s speed speed distribution distribution is correctly correctly shown shown by

N v

(1)) (1

T 2 T1

>

N v

T 2 T 1 v

(2)) (2

T1

>

T 2

T 2

N v T 1

(3)) (3 v

T1 T 1

>

T 2

N v

T 2

(4)) (4 v

T 1

T1

>

T 2

T 2 v



Kinetic Theory

7

Sol. Answer (1) The Maxwell’s distribution curve heats at the most probable speed, which depends on temperature. v probable 

T

Hence T 1 > T 2 is correctly shown in option (1) as it shows with peaks of the curve at higher temperature, furthers along the x the x -axis. -axis. 20. Select the incorrect stateme statement nt about about Maxwell’s speed distributio distribution n (1)) The distribution (1 distribution function function depends depends only on the absolute absolute temperatu temperature re (2)) (2

 V av  V mp

V rms

(3)) The area under (3 under the distribution curve gives total number of molecules of the gas gas (4)) The distribution (4 distribution curve curve is symmetric symmetric about about the most probable probable speed speed Sol. Answer (4) The Maxwell’s speed distribution is asymmetric due to the fact that the lowest speed possible is zero. While the highest speed possible is infinity. 21. The ratio of number number of collisions per second second at the walls of containers containers by by He and O 2 gas molecules kept at same volume and temperature, is (assume normal incidence on walls) (1) 2 : 1

(2) 1 : 2

(3)

2 2 :1

(4)) (4

1: 2 2

Sol. Answer (3) 2

 nO2     n  He 

M He M O

2

M O

M He = 4 n O2



n He



nHe

2

= 32

1 2 2

: nO

2

2

2 :1

22. An ant is moving moving on a plane plane horizontal horizontal surface. surface. The number number of degrees degrees of freedom freedom of the ant ant will be (1) 1

(2) 2

(3) 3

(4) 6

Sol. Answer (2) The number of degrees of freedom of movement of ant is 2 as it can move only in two independent directions in the plane surface.

23. If a gas ha has ‘f ‘f ’’ degree of freedom, the ratio of the specific heats of the gases

(1)) (1

1  f 2

(2)) (2

1

f 2

(3)) (3

1

1

f

C p C v

is

(4)) (4

1

2

f


8

Kinetic Theory


Sol. Answer (4) C = C v + R p and an d C v =

f 2

R

f

C p

=

C v

2

R

 R

f 2

C p

=

C v C p

=

C v C p

=

C v

f  2 f 2  f

f

1

2

f

24. Molar specific heat at constant volume, for a non-linear non-linear triatomic triatomic gas is (vibration (vibration mode neglect neglected) ed) (1) 3R

(2) 4R

(3) 2R

(4)) R (4

Sol. Answer (1) Molar heat capacities for a gas is given by C v =

f 2

RT

Where f = = 6 in triatomic molecules C v = 3 RT 25. A mixture mixture of ideal ideal gases has 2 moles moles of He, 4 moles moles of oxygen oxygen and 1 mole of ozone ozone at absolute absolute temperatur temperature e T . The internal energy of mixture is (1) 13 RT

(2) 11 RT

(3) 16 RT

(4) 14 RT

Sol. Answer (3) Degrees of freedom of He (f ( f He) = 3 Degrees of freedom of O 2 ( (f f O ) = 5 2

Degrees of freedom of O 3 ( (f f O ) = 6 3

nHe = 2,

nO 2

=4

nO 3

=1

Energy of mixture = Sum of individual energies =

n

HefHe

 nO2 fO2  nO3 fO3 

= 2  3  4  5  1  6 

RT 2

RT

2

= (3 + 10 + 3) RT = 16 RT Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


Kinetic Theory

9

26. E 0 and E H respectively represents the average kinetic energy of a molecule of oxygen and hydrogen. If the two gases are at the same temperature, which of the following statement is true? (1)) E 0 > E H (1

(2)) E 0 = E H (2

(3)) E 0 < E H (3

(4)) Da (4 Data ta insuf insuffi fici cien entt

Sol. Answer (2) Temperature is an approximate value which refers to average kinetic energy per molecule. If temperature of both is same, average energy will be same according to definition. 27. 14 g of of CO at at 27°C 27°C is is mixe mixed d with with 16 g of O2 at 47°C. The temperature of mixture is (vibration mode neglected) (1) –5°C

(2) 32°C

(3) 37°C

(4) 27°C

Sol. Answer (3) 1 mole of CO and 1 mole of O 2 are mixed. Net internal energy =

=

=

f1 2

5 2

5 2



RTCO

R 300 

f 2 2

5 2

RTO

2

R 350

R (650)

= 5 R (325) (325) = 1625 R 1625 =

5 2

162 5  2 5

RTf in inal

 nf in inal

= T final × nfinal

325 × 2 = T final × 2 T final = 325 K T final = 37°C 28. When one mole mole of monatomic monatomic gas is mixed mixed with one one mole of a diatomic diatomic gas, then then the equivalent equivalent value value of  for the mixture will be (vibration mode neglected) (1) 1.33

(2) 1.40

(3) 1.50

(4) 1.6

Sol. Answer (3)

Y for monatomic gas =

Y for diatomic gas =

1

1

2 5

2 3





5 3

=  1

  1

2

7 5

n

=  2

  1





n1

1  1 1 5 3

1





n2

2  1 1 7 5

1

Solving, we get  = = 3/2 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

10

Kinetic Theory


29. A box of negligi negligible ble mass mass contain containing ing 2 moles moles of an ideal ideal gas of molar molar mass mass M and and adiabatic exponent  moves moves with constant speed v on on a smooth horizontal surface. If the box suddenly stops, then change in temperature of gas will be (   1)Mv 2 (1) (1) 4R

(2)) (2

2

 Mv 2

(3)) (3

2R

Mv

2(   1)R

(   1)Mv 2 (4) (4) 2R

Sol. Answer (4) Mass of gas in the box = 2 M Initial kinetic energy = 1

Mv 2 =

2

1 2

 2M  v 2  Mv 2

nfRT 2

 T =

2Mv

nfR

Substitution f =

T =

2 1  r

and n = 2

(   1) Mv 2 2R

30. On increasing increasing number densit density y for a gas in a vessel, vessel, mean mean free path of of a gas (1) Decreases

(2) Increases

(3) Remains same

(4) Becomes double

Sol. Answer (1) Mean free path of a substance is the average distance a molecule may travel without collision. If the number of molecules per unit volume increases it increases the frequency of collision and decreases the mean free path. SECTION - B

Objective Type Questions 1.

At room room temperat temperature ure the the rms speed speed of the the molecu molecules les of a certa certain in diatomic diatomic gas gas is found found to be be 1920 m/s. m/s. The The gas is (1) H2

(2) F2

(3) Cl2

(4) O2

Sol. Answer (1) V r.m.s. =

1920 =

M = =

3RT M

3  8.314  300 M

3  8 .3 1 4  3 0 0 1920

M = = 0.00202 kg On molar weight = 2.02 g. Hence it is hydrogen. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


2.

Kinetic Theory

11

An idea ideall gas gas is en enclo close sed d in a cont contain ainer er of of volum volume e V at at a pressure P . It is being pumped out of the container by using a pump with stroke volume v . What is final pressure in container after n-stroke of the pump? (assume temperature remains same)

 V  (1) P  (1)   V  v 

PV

n

(2)) (2

(3)) (3

(V  v )n

P

V

 V  (4) P  (4)   V  v 

n

n

v

n

Sol. Answer (1) After stroke PV = constant PV = PV = P 1(V V + + v ) P 1 =

PV

V

(V  v ) v

Similarly after 2nd stroke

 V  P 2 = P    V  v 

2

After nth stroke

P n =

3.

 V    V  v 

n

P 

Three per Three perfec fectt gase gases s at ab abso solut lute e temp tempera eratu tures res T 1, T 2 and T 3 are mixed. If number of molecules of the gases are n1, n2 and n3 respectively then temperature of mixture will be (assume no loss of energy)

(1)) (1

T1

2

 T2  T 3

(2)) (2

3

n1 T1

 n22T2  n32T3

n1

 n2  n3

(3)) (3

n1T1

 n2T2  n3T3

n1

 n2  n3

(4)) (4

T1

 T2  T 3

n1  n2

 n3

Sol. Answer (3) Absolute temperature = T 1, T 2, T 3. Internal energy of gases =

Average temperature =

n1RT1 2



n2 RT2 2

n3 RT 3 2

Internal energy of mixture (n1  n2  n3 )

4.



R

2

Varia ariation tion of atmo atmosph spheric eric pre pressu ssure, re, wit with h heigh heightt from from earth earth is (1) Linear

(2) Parabolic

(3) Exponential

(4) Hyperbolic

Sol. Answer (3) Variation of atmospheric pressure due to height is given by the barometric formula mgh//RT P h = P 0 e –mgh

Hence the decrease will be exponential. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

12

5.

Kinetic Theory


A narrow glass tube, tube, 80 cm long long and opens at both ends, is half immerse immersed d in mercury mercury,, now the top of the tube is closed and is taken out of mercury. A column of mercury 20 cm long remains in the tube. Find atmospheric pressure (1) 20 cm of air column

(2) 60 cm of Hg column

(3) 60 cm of air column

(4) 20 cm of Hg column

Sol. Answer (2) PV = constant P 1V 1 = P 2V 2

[P 1 = P 0 atmospheric pressure]

P 0 × 40 = P 1 × 60

... (i)

P 1 + 20 = P 0

... (ii)

From (i) 2P 0

P 1 =

3

From (ii) 2P 0 3

6.

+ 20 = P 0  P 0 = 60 cm of Hg.

One mole mole of monatomi monatomic c gas and three three moles moles of diatomic gas are are put togethe togetherr in a conta container iner.. The molar specific specific heat (in JK–1 mol –1) at constant volume is (Let R = = 8 JK –1 mol –1) (1) 18

(2) 19

(3) 20

(4) 21

Sol. Answer (1) n1 = 1 mole

f 1 = 3

n2 = 3 moles

f 2 = 5

R = = 8 JK –1 mol –1 Molar specific heat are given by the weighted means of the gases.

C v

f

C v

f

=

=

n1  Cv

1

2

n1  n2

1

3

3 2

8

R

18

=

f

v

R

3

5 2

R

4

=

C v

 n2  C

8



15 8

R

R

= 18



7.

Kinetic Theory

13

Two close closed d conta container iners s of equ equal al volume volume fille filled d with with air at press pressure ure P 0 and temperature T 0. Both are connected by a narrow tube. If one of the container is maintained at temperature T 0 and other at temperature T , then new pressure in the containers will be

(1)) (1

2P0T T

(2)) (2

 T 0

P0T T

 T 0

(3)) (3

P0T

2(T  T 0 )

(4)) (4

T

 T 0 P 0

Sol. Answer (1) P T

= constant

P0

Initially



T0

P0 T0



2P 0 T 0

For two containers P0 T0



P =

or P =

8.

P T



2P 0 T 0

2P0  T  T 0 T0 (T

 T 0 )

2P0T (T  T 0 )

The tempera temperature ture of a gas is –68°C –68°C.. At what what temperat temperature ure will the the average average kinetic kinetic energy energy of of its molecule molecules s be twice that of –68°C? (1) 137°C

(2) 127°C

(3) 100°C

(4) 105°C

Sol. Answer (1)

Average kinetic energy = or

f 2

RT

K.E.avg  T

For K.E. energy to be doubles that of K.E. at – 68°C or 205 K. The temperature must be 2 T or or 410 K When converted to °C = 410 – 273 = 137°C 9.

An ideal ideal gas is filled filled in a closed closed contai container ner and and containe containerr is moving moving with uniform uniform accel accelerat eration ion in horizo horizontal ntal direction. Neglect gravity. Pressure inside the container is (1) Uniform everywhere

(2) Less in front

(3) Less at back

(4) Less at top

Sol. Answer (2) Each particle closes experience a pseudo force initials, themselves to give low pressure every where. This is because of Pascal's law. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

14

Kinetic Theory


10.. One kg 10 kg of a diat diatomic omic gas is is at press pressure ure of of 8 × 1 10 0 4 N/m2. The density of the gas is 4 kg/m 3. The energy of the gas due to its thermal motion will be (1) 3 × 104 J

(2) 5 × 104 J

(3) 6 × 104 J

(4) 7 × 104 J

Sol. Answer (2) PV = PV = n RT

... (1)

f = 4 kg/m 3

 v = = 0.25 m3 1

8 × 104 ×

4

PV

... (2)

Energy Ene rgy of of gas is given given by =

=

= 11.

f 2

5 2

RT

n

 PV

5 2

[From (1)]

4 4  2  10 10  5  10 10 J

A co cont ntain ainer er cont contain ains s 32 32 g of O2 at a temperature T . The pressure of the gas is P . An identical container containing 4 g of H2 at a temperature 2T 2 T has has a pressure of (1) 8 P

(2) 4 P

(3)) P (3

(4)) P 18 (4

Sol. Answer (2) Given, 1 mole of O 2 at temperature T , pressure P and 2 moles of H 2 at a temperature 2T 2 T P 1 =

P 0 =

P H =

n  RT V RT V

 P

4 RT V



4P

12.. An ideal 12 ideal gas gas is is expan expandin ding g such such that that PT = = constant. The coefficient of volume expansion of the gas is (1)) (1

1

(2)) (2

T

2 T

(3)) (3

3 T

(4)) (4

4 T

Sol. Answer (2) PT = = Constant or

T

2

V

= Constant [PV [ PV = = nRT ]  T 2 = KV

... (i)

Differentiating w.r.t. T , we get 2T

V





dV VdT

K dV

2T

⇒

V dT



VK

2TV

VT

2





dV VdT

2

T



Kinetic Theory

15

13. 50 cal of heat heat is required required to raise the temperature temperature of 1 mole of an ideal gas from from 20°C to to 25°C, while while the pressure of the gas is kept constant. The amount of heat required to raise the temperature of the same gas through same temperature range at constant volume is ( R = = 2 cal/mol/K) (1) 70 cal

(2) 60 cal

(3) 40 cal

(4) 50 cal

Sol. Answer (3) 50

C P =

= 10 cal K –1 mol–1

AT

C P = C V + R C V = C P = R C V = 8 cal K–1 mol–1 14. Pressure versus temperatur temperature e graph of an an ideal gas is as shown shown in figure. figure. Density of the gas at point A A is is 0. Density at point B will be P B

3P 0 P 0

A

T 0

(1)) (1

3 4

0

(2)) (2

3 2

0

T

2T 0 4

(3)) (3

3

0

(4)) (4

20

Sol. Answer (2) PV = PV = nRT

P =

 M

RT

Initially  = 0,

0R

P 0 =

M



3P 0 =

x

T 0

R

M

P = P = P 0,

T = T = T 0

in initially

... (i)

2T 0

fifinal

... (ii)

Dividing (ii) by (i)

 3=

3 2

x

0

2

0  

x


16

Kinetic Theory


15.. The energy 15 energy (in eV) eV) posses possessed sed by by a neon neon atom atom at 27°C 27°C is (1) 1.72  10–3

(2) 4.75  10–4

(3) 3.88  10–2

(4) 3.27  10–5

Sol. Answer (3) Neon is a monoatomic gas. So, at 300 K its internal energy is given by

For one molecule

3 2

1 2

fkT

kT

16. If heat energy energy is given given to an ideal ideal gas at constant constant pressure, pressure, then then select the the graph which which best represents represents the variation of VT with with temperature (T ( T ). ). VT

VT

(1)) (1

VT

(2)) (2

VT

(3)) (3

T

(4)) (4

T

T

T

Sol. Answer (1) PV = PV = nRT P nR

and an d

V T





T V

nR P

= constant (K ( K )

VT = VT = KT 2 Assuming VT VT = = y and x and x = = T y = = Kx 2 Which is equation of a parabola will focus on y -axis -axis > facing upwards. 17. If hydrogen gas is heated to a very high high temperature, then the fraction of energy possessed possessed by gas molecules correspond to rotational motion (1)) (1

3

(2)) (2

5

2 7

(3)) (3

3 7

(4)) (4

2 5

Sol. Answer (2) Hydrogen is a diatomic molecules and if vibrational degrees of freedom are increased the degrees of freedom will be 3 translation 2 rotational and two vibrational.

 So total 7 degree of freedom. Fraction of energy possessed due to rotational motion : Degree of freedom due to rotation total degree of freedom =

2 7



Kinetic Theory

17

18.. The 18 The te temp mper erat atur ure e (T ) of one mole of an ideal gas varies with its volume ( V ) as T = = –V 3 + V 2, where  and  are positive constants. The maximum pressure of gas during this process is

(1)) (1

2R (2) (2) 4

 2R

(3)) (3

(  )R 22

(4)) (4

 2R 2

Sol. Answer (2) T = = –  V 3 + V 2 and an d PV PV = = nRT

... (i) ... (ii)

n=1 So,

RT

P =

V

Multiplying RT V

or

R V

in (i)

= (–  V 2 + V V ))R

P = = (–  V 2 + V V ))R ... (iii) dP

= (– 2 V + )R

dV

2

dP

Maxima is when

dV

d P

= 0 and

dV

2

in negative, so

O = (–2V + )R  2

V =

Put in value of V in in equation (iii)

 2 2  2R R  P    P = P =  P = =  42 2  4 19. Nitroge Nitrogen n gas is filled filled in in an insula insulated ted conta container iner.. If  fraction of moles dissociates without exchange of any energy, then the fractional change in its temperature is (1)) (1

– 5



(2)) (2

 3

(3)) (3

–3 2



(4)) (4

5 2  3

Sol. Answer (1) Degree of freedom of diatomic nitrogen = 5 Degree of freedom of monoatomic nitrogen = 3 Let initial number of moles be n and  fraction dissociated. So fraction dissociated = n fraction remaining = n – n. n break into two so new atoms formed is actually 2 n. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

18

Kinetic Theory


n

Initial energy is given by =

Final energy = (n  n )

5

=

2

5

=

2

f 2

 RT =

n

5 2

 RT

5 3 RT2  2n  RT2 2 2

nRT2



nRT 2



5 2

nRT2

 n3RT2

nRT 2

2

(5  2) nRT 2

=

2

Change in energy is given on zero. 5 nRT 2

5T 5

(5  ) nRT 2

=

2

= T 2



T T = = T 2 – T or T =

5T 5

 T =

 T 5

Fractional change in temperature = temperature =

T T

or 

 5

20.. An ideal 20 ideal gas under undergoes goes a polyt polytropi ropic c given given by equatio equation n PV n = constant. If molar heat capacity of gas during this process is arithmetic mean of its molar heat capacity at constant pressure and constant volume then value of n is (1) Zero

(2) –1

(3) +1

(4) 

Sol. Answer (2) Polytropic process PV n = constant Given heat capacities is average of C P and C V . So C = C =

or C C = =

or C C = =

CP

 C V 2

2CV

 R

2

CV

 R 2

... (i)



Kinetic Theory

19

Now formula for specific heat of polytropic process is given by

C = C =

or

R y 1

R y 1 R 2



R





R



2

... (ii)

1 n

R y 1



R

as

1 n

C V 

R

y 1

R 1 n

or n = –1 21. If  moles of a monoatomic gas are mixed with  moles of a polyatomic gas and mixture behaves like diatomic gas, then [neglect the vibrational mode of freedom] (1) 2 = 

(2))  = 2 (2

(3))  = –3 (3

(4) 3 = –

Sol. Answer (1) 22. If different different ideal gases gases are at the same same temperature temperature,, pressure pressure and have same same volume, volume, then all gases gases have same same (1) De Density

(2) Nu Number of molecules

(3) Most probable speed

(4) Internal energy per mole

Sol. Answer (2) PV = nRT or

PV RT

n

At the same pressure volume and temperature each molecule will have same number of moles i.e i.e.. same number of molecules of gas. 23. The internal internal energy energy of 10 10 g of nitrogen nitrogen at N.T N.T.P .P.. is about about (1) 2575 J

(2) 2025 J

(3) 3721 J

(4) 4051 J

Sol. Answer (2) 10

Number of moles of N 2 =

U = =

=

f 2

28

nRT

5 2



5 14

 R  273

= 2025 J 24.. The mean 24 mean free free path path of of a molecu molecule le of He He gas is . Its mean free path along any arbitrary coordinate axis will be (1))  (1

(2)) (2

 3

(3)) (3

 3

(4) 3


20

Kinetic Theory


Sol. Answer (3) Mean free path of a molecule is the resultant of path along three separate axis and they will be equal.  2x  2y  22

=

So,

2 where x =  y  2 = (say) a

=

a

 a2  a2



a=

or

2

3

25. Accordin According g to C.E. van der der Waal, the interatomic interatomic potential potential varies varies with the the average interatom interatomic ic distance distance (R ( R ) as (1)) R –1 (1

(2)) R –2 (2

(3)) R –4 (3

(4)) R –6 (4

Sol. Answer (4) Interatomic potential varies with average interatomic distance as R –6 which is a fact. 26. The value value of critical critical tempera temperature ture in terms terms of of van der der Waals’ Waals’ constants constants a and b is given by (1)) (1



T C

8a 27Rb

(2)) (2

T C



27a 8Rb

(3)) (3

T C

(3)) (3

f



a 2Rb

(4)) (4

T C

(4)) (4

f



a 27Rb

Sol. Answer (1) Critical temperature is given as : 8a

T C =

27 R b

27. To find out degree degree of freedom freedom,, the expre expression ssion is (1)) (1

f



2

(2)) (2

  1

f



  1 2



2

  1



1

  1

Sol. Answer (1) C V =

 =

=

 =

f

R

2

f 0

C V + R R = =

f

2

R

R 

(f  2) 2

R

C P C V

(f  2)R  2 2  f R 1

2

 – 1 =

f =

C P =

f 2

f

2

  1



Kinetic Theory

21

28. Nit itro rog gen ga gas s N2 of mass 28 g is kept in a vessel at pressure of 10 atm and temperature 57ºC. Due to leakage of N2 gas its pressure falls to 5 atm and temperature to 27ºC. The amount of N 2 gas leaked out is (1)) (1

5 63

g

(2)) (2

63 5

g

(3)) (3

28 63

g

(4)) (4

63 28

g

Sol. Answer (2) Mass = 28 g P i = 10 atm

T i = 57°C = 330 K

P f = 5 atm

T f = 27°C = 300 K

Volume is kept constant. P i = K K × × ni T i

... (i)

P f = K × nf T f

... (ii)

Dividing (i) by (ii) Pi Pf

nf T f 10

ni

=

nf

or

ni T i



n i

=

nf

ni

=

nf

5

2



300 330

10 11

20 11

Now ni = 1 mole of N 2 11

nf =

20

moles 11

or

Mass of N2 left =

20

× 28 11

 Quantity released = 28 –

=

9 20

20

× 28

 28 =

63 5

g

29. A diatomic diatomic gas of molecular molecular mass mass 40 g/mol g/mol is filled in a rigid container container at temperatu temperature re 30ºC. It It is moving with velocity 200 m/s. If it is suddenly stopped, the rise in the temperature of the gas is (1)) (1

32

ºC

R

(2)) (2

320

ºC

R

(3)) (3

3200 R

ºC

(4)) (4

3.2

ºC

R

Sol. Answer (2) Let there by n moles of gas. Mass of gas = 40ng or

40n 1000

or 0.04n 0.04n kg


22

Kinetic Theory


K.E. of gas in container =

1 2

× 0.04 n × (200)2

= 0.02 × n × 4 × 10 4 = 8 × 102 × n J Now heat capacity of gas (C ( C ) =

or C C = =

5 2

f 2

nR

R × R ×n

or C T = 8 × 102 × n or

5 2

× R R × × n T = = 8 × 10 2 × n

T =

8  10

2

R 16

T T = =

5



2 5

 10 2 =

320 R

°C

30. The ratio ratio of average average translato translatory ry kinetic kinetic energy energy of He He gas molecule molecules s to O 2 gas molecules is (1)) (1

21

25

(2)) (2

21

25

(3)) (3

3 2

(4) 1

Sol. Answer (4) Both He and O2 have 3 translatory degrees of freedom. At the same temperature, energy is divided equally in all degrees of freedom. Hence ratio of the translatory kinetic energy is one. SECTION - C

Previous Years Questions 1.

Two ves vesse sels ls sep separa arate tely ly con conta tain in two two ide ideal al gase gases s A A and B at the same temperature, the pressure of A A being twice that of B. Under such conditions, the density of A A is found to be 1.5 times the density of B. The ratio [Re-AIPMT-2015] of molecular weight of A of A and and B is (1)) (1

2

1

(2)) (2

2

3

(3)) (3

3 4

(4) 2

Sol. Answer (3) A

B

2P

P

1.5 d

d

We known, PV = PV = nRT

PV =

m M

RT



So,

23

m 1  M RT V  

P



P



P1

d

d1



P2

RT

M

2P

M 1

M 2 d 2

1.5 d M 2



M1

15



M 2



M1



P

2.

Kinetic Theory

20

d

3



4

The Th e rat ratio io of th the e spe speci cifi fic c hea heats ts

 

  2

C v

  in terms of degrees of freedom (n ( n) is given by

 

n

(1)) 1  (1

C P

(2))  1  (2

1

 n

 

(3)) 1  (3

  3

n

 

(4))  1  (4

[AIPMT-2015]

2 n

 

Sol. Answer (4) We known

So,

n

Cv



CP

 R  Cv  R 

R

...(i)

2

=

 

n

R

2

n

R  1 

...(ii)



2

On dividing equation (ii) by (i)

 

n

R  1 

C P



C v

R



2

n

 Y

2

 

So,  1 

Y

3.

 1

n

n

  Y 2 2

2 n

The me mean an fre free e path path of mole molecu cules les of a gas gas,, (rad (radius ius r ) is inversely proportional to (1)) r 3 (1

(2)) r 2 (2

(3)) r (3

(4)) (4

[AIPMT-2014] r


24

Kinetic Theory


Sol. Answer (2) The mean free path of molecules of gas is given by 1



d = = diameter of molecule

where

2

2d n

d = = 2r So,   4.

1 r

2

The amoun amountt of heat heat energy energy require required d to raise raise the the tempera temperature ture of of 1 g of Helium Helium at at NTP, NTP, from T 1K to T 2K is [NEET-2013]

(1)) (1

3 2

N ak B(T 2 – T 1)

(2)

3

N ak B(T 2 – T 1)

4

(3)

3 4

 T 2    T 1 

N ak B 

(4)) (4

3 8

N ak B(T 2 – T 1)

Sol. Answer (4) Mole of helium is 4 g So, number of moles of helium =

1 4

moles

NTP there is constant pressure C P of gas =

f 2

R = =

3R 2

Q = C P × n T =

= 5.

3 2

R

1 4

 (T2  T 1 )

3R (T2  T 1 ) 8

Two container A A and and B are partly filled with water and closed. The volume of A A is is twice that of B and it contains half the amount of water in B. If both are at same temperature, the water vapour in the container will have pressure in the ratio of (1) 1 : 2

(2) 1 : 1

(3) 2 : 1

(4) 4 : 1

Sol. Answer (2) Vapour pressure for the same liquid is always the same. So the ratio will be P : : P or or 1 : 1. 6.

At cons consta tant nt volu volume me,, tempe tempera ratu ture re is inc incre reas ased ed then then (1)) Colli (1 Collision sion on walls walls will be be less less (2)) Number of collisions (2 collisions per unit unit time time will increase (3)) Colli (3 Collision sions s will be be in straig straight ht lines lines (4)) Coll (4 Collisio isions ns will will not not chan change ge

Sol. Answer (2) At constant volume if temperature is increased pressure will increase. Since pressure is increased due to collisions of particles will the wall of the container. So collisions per unit time will increase. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


7.

Kinetic Theory

25

A po polyatomic ga gas wit with h n degree of freedom has a mean energy per molecule given by (1)) (1

nKT

(2)) (2

N

nKT

(3)) (3

2N

nKT

(4)) (4

2

nKT 4

Sol. Answer (3) Numbers of degrees of freedom = n. Internal energy of gas = R

K = =

n 2

RT

where N is is the Avogadro's number.

N

or NK NK = = R Internal energy =

n

2

N K T

Internal energy per molecule =

8.

n NKT 2

N

or

nKT 2

For a certa certain in gas gas the the ratio ratio of spec specific ific hea heats ts is is given given to be be  = = 1.5. For this gas (1)) (1

C v



3R

(2)) (2

J

C p



3R

J

(3)) (3

C p



5R

J

(4)) (4

C v



5R

J

Sol. Answer (2) For a certain gas

C p C v

 1.5

C = C v + R p 1

R C v

R = R =

 1.5

C v 2

or C v = 2R C = 3R p 9.

Accordi Acc ording ng to to kinet kinetic ic theor theory y of of gases gases,, at abso absolute lute zero tem tempera perature ture (1) Water freezes

(2) Liquid helium freezes

(3) Molecular motion stops

(4) Liquid hydrogen freezes

Sol. Answer (3) According to kinetic energy of gases at absolute temperature molecular motion stops as for ideal gases only in kinetic energy of gases is considered which is given by K.E. =

f 2

nKT

So, T = = 0 K and motion will be zero. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

26

Kinetic Theory


10. One mole of an an ideal monoatomic monoatomic gas gas requires 207 J heat to to raise the temperatur temperature e by 10 K when heated heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is [Given the gas constant R = 8.3 J/mol-K] (1) 198.7 J

(2) 29 J

(3) 215.3 J

(4) 124 J

Sol. Answer (4) 207

C P =

= 20.7 J/mol-K

10

R = = 8.3 and an d C V = C P – R = 12.4 J/mol-K

Q = C V T = 124 J 11. Re Rela lati tion on betw betwee een n press pressur ure e (P ( P ) and average kinetic energy per unit volume volume of gas (E (E ) is (1)) (1

P



2 3

E

(2)) (2

P



1 3

E

(3)) (3

P



1 2

E

(4)) P (4 P = = 3E

Sol. Answer (1) Relation between pressure (P ( P ) and average kinetic energy is given by 1

P =

E =

3

1 2

2

mNv

mv

... (i)

2

... (ii)

Using (i) and (ii) 2

P =

3

E

12. If C s be the velocity of sound in air and C be be the rms velocity, then 1/2

(1)) C s < C (1

(2)) C s = (2 =C C

(3)) (3

Cs

    C   3

(4) No None ne of th thes ese e

Sol. Answer (3) Velocity of sound in air is given by C s



C = C =

C s =

 P

or

 3RT M

C

 RT M

 P

 3



Kinetic Theory

27

13. The tempera temperature ture of gas gas is raised raised from from 27°C 27°C to 927°C. 927°C. The The rms speed speed is

(1)) (1

927 27

times the earlier value

(2) Remain the same

(3) Ge Gets halved

(4) Ge Get doubled

Sol. Answer (4) 3RT

V =

V =

M

K T

V 1 =

K 300

V 2 =

K 1200

V 1 V 2



300 1200

2 300 300

V1

 V 2

or V 2 = 2V 1 14.. The equa 14 equation tion of state state,, corres correspon ponding ding to 8 g of O2 is (1)) PV (1 PV = = 8RT

(2)) (2

PV



RT 4

(3)) PV (3 PV = = RT

(4)) (4

PV



RT 2

Sol. Answer (2)

8 g of O 2 =

1 4

moles

PV = PV = nRT

PV =

1 4

RT

1  as n  4 

15.. At 0 K, which 15 which of the follow following ing prope propertie rties s of a gas will will be zero? zero? (1) Kinetic energy

(2) Potential energy

(3) Density

(4) Mass

Sol. Answer (1) By definition at absolute zero the kinetic energy of a gas is zero. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

28

Kinetic Theory


16. At 10oC the value of the density of a fixed mass of an ideal gas divided by its pressure is x . At 110oC, this ratio is 283

(1)) (1

383

383

(2)) x (2

x

(3)) (3

283

10

(4)) (4

x

110

x

Sol. Answer (1) PV = PV = nRT P



RT

=

x = x =

M

RT M

T 1 = 283 K R

x 1 = x 1 x 2

M

T 2 = 383 K

 283

x 2 =

R M

 383

283

=

383

383

x 2 =

283

17. The degrees degrees of freedom freedom of a triatomic triatomic gas gas is (consider (consider moderate moderate tempera temperature) ture) (1) 6

(2) 4

(3) 2

(4) 8

Sol. Answer (1) A non-linear triatomic gas has 3 translators and 3 rotatory degrees of freedom. 18.. The equat 18 equation ion of stat state e for 5 g of of oxygen oxygen at at a pressu pressure re P and and temperature T , when occupying a volume V , will be (where R is is the gas constant) 5

(1)) PV = (1

32

RT

(2)) PV (2 PV = = 5RT 5 RT

(3)) PV (3 PV = =

5

5

RT

2

(4)) PV (4 PV = =

16

RT

Sol. Answer (1) 5

Number of moles (n ( n) =

32

–PV = = nRT 5

PV = PV =

32

RT

SECTION - D

Assertion-Reason Type Questions 1.

A : For a rea reall gas gas intern internal al ener energy gy depe depends nds on its its temp tempera erature ture as well well as as volum volume e also. also. R : For a real gas interatomic potential potential energy depends depends on volume and kinetic energy energy depends on temperature. temperature.

Sol. Answer (1) The assertion is correct and the reason is the correct explanation of assertion. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


2.

Kinetic Theory

29

A : The gravita gravitational tional force betwe between en the the gas gas molecul molecules es is ineffe ineffective ctive due to extreme extremely ly small small size and very high speed. R : No force of interaction interaction acts acts between between molecules molecules of an ideal gas. gas.

Sol. Answer (2) Both the statements are true but reason is the incorrect explanation of assertion. The assertion happens to be self explanatory. 3.

A : Av Aver erag age e vel veloc ocit ity y of of gas gas mo mole lecu cule les s is is zer zero. o. R : Due to random motion of gas molecules, velocities velocities of different molecules cancel cancel each other. other.

Sol. Answer (1) The assertion is true but reason is correct explanation of assertion. 4.

A : At cons constan tantt volum volume e on on incre increasin asing g tempe temperatu rature re the coll collisio ision n frequ frequency ency incr increase eases. s. R : Col Collis lision ion frequ frequenc ency y  temperature of gas.

Sol. Answer (3) The assertion is true but collision frequency does not increase linearly with temperature. 5.

A : Two gases gases with with the the same same average average tran translat slationa ionall kinetic kinetic energ energy y have have same same tempera temperature ture even if one one has grea greater ter rotational energy as compared to other. R : Only average average translational translational kinetic energy energy of a gas contributes contributes to its temperature. temperature.

Sol. Answer (1) The assertion is true and reason is correct explanation of assertion. 6.

A : Al Alll mol molec ecul ular ar mo moti tion on ce ceas ases es at –2 –273 73.1 .15° 5°C. C. R : Temperature below –273.15°C –273.15°C cannot cannot be attained. attained.

Sol. Answer (2) Both the statements are correct but reason is not the correct explanation for assertion. 7.

A : Mag Magnitu nitude de of of mean mean velo velocity city of the the gas gas mole molecule cules s is sam same e as the their ir mean mean spe speed. ed. R : The only difference difference between mean mean velocity and mean mean speed is that mean velocity velocity is a vector and mean speed is a scalar.

Sol. Answer (4) Mean velocity of a gas is not the same as mean speed. Hence both the statements are correct. 8.

A : Mea Mean n free free pat path h of of gas gas mole molecul cules es var varies ies inv invers ersely ely as den density sity of the gas gas.. R : Mean free path varies inversely as pressure of the gas.

Sol. Answer (2) Mean free path is given by =

V 2

2 N  d

Where N is is total number of molecules. V is is volume. d is is the diameter of molecule. N V

is the number velocity of gas hence assertion is true. But the mean free path does not depend on pressure.


30

9.

Kinetic Theory


A : Number Number of of air mole molecule cules s in a room in winter winter is more more than than the the numbe numberr of molec molecules ules in the the same same room room in summer. R : At a given pressure and volume, the number of molecules molecules of a given mass mass of a gas is directly directly proportional to the absolute temperature.

Sol. Answer (3) The assertion is true as at a lower temperature there is a higher density. According to PV PV = = nRT



1 n

Hence reason is false. 10. A : Evapo Evaporation ration occur occurs s at any any tempera temperature ture whereas whereas the boiling boiling point point depend depends s on the external external press pressure. ure. R : Evaporation Evaporation of a liquid occurs from the surface surface of a liquid at all temperature whereas boiling boiling takes place at a temperature determined by the external pressure. Sol. Answer (1) The assertion is true and the reason is the correct explanation of the assertion.

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Kumar Vaibhav - 18. Electrochemistry

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