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CHAPTER 24

KINETIC THEORY OF GASES

1.

Volume of 1 mole of gas

RT

0.082 273

–3

–2

3

PV = nRT V =

=

= 22.38 ≈ 22.4 L = 22.4 × 10 = 2.24 × 10 m

P

1

2.

n=

3.

click

here

1 1 10 3

10 3

PV

1

=

=

=

RT

0.082 273

22.4

22400

1

23

19

No of molecules = 6.023 × 10 ×

= 2.688 × 10

22400

3

–5

V = 1 cm ,

T = 0°C,

P = 10 mm of Hg

1.36 980 10 6 1

PV

ƒgh V

–13

=

=

= 5.874 × 10

RT

RT

8.31 273

23

–13

11

No. of moluclues = No × n = 6.023 × 10 × 5.874 × 10 = 3.538 × 10

n=

4.

n=

1 1 10 3

10 3

PV

=

=

RT

0.082 273

22.4

5.

10

3

32

–3

g = 1.428 × 10 g = 1.428 mg

22.4

Since mass is same

n1 = n2 = n

nR 300

nR 600

P1 =

,

P2 =

V0

2V0

mass =

2V0

2V0

P1

nR 300

1

=

= =1:1

P2

V0

nR 600 1

6.

600 K

–3

V = 250 cc = 250 × 10

–3

–3

–3

–6

–3

P = 10 mm = 10 × 10 m = 10 × 13600 × 10 pascal = 136 × 10 pascal

T = 27°C = 300 K

136 10 3 250

PV

136 250

=

10 3 =

10 6

8.3 300

RT

8.3 300

136 250

No. of molecules =

10 6 6 10 23 = 81 × 1017 ≈ 0.8 × 1015

8.3 300

5

6

P1 = 8.0 × 10 Pa,

P2 = 1 × 10 Pa,

T1 = 300 K,

Since, V1 = V2 = V

n=

7.

8.

9.

T2 = ?

P1V1

PV

8 10 5 V

1 10 6 300

1 10 6 V

= 2 2

=

T2 =

= 375° K

T1

T2

300

T2

8 10 5

3

6

3

T = 300 K,

P=?

m = 2 g, V = 0.02 m = 0.02 × 10 cc = 0.02 × 10 L,

M = 2 g,

m

2

PV = nRT PV =

RT P × 20 = 0.082 300

M

2

0.082 300

5

5

P=

= 1.23 atm = 1.23 × 10 pa ≈ 1.23 × 10 pa

20

nRT

m RT

ƒRT

P=

=

=

V

M V

M

–3

3

ƒ 1.25 × 10 g/cm

7

R 8.31 × 10 ert/deg/mole

T 273 K

M=

ƒRT

1.25 10 3 8.31 10 7 273

4

=

= 0.002796 × 10 ≈ 28 g/mol

P

13.6 980 76

24.1

V0

300 K

Kinetic Theory of Gases

10. T at Simla = 15°C = 15 + 273 = 288 K

–2

P at Simla = 72 cm = 72 × 10 × 13600 × 9.8

T at Kalka = 35°C = 35 + 273 = 308 K

–2

P at Kalka = 76 cm = 76 × 10 × 13600 × 9.8

PV = nRT

PM

m

m

RT PM =

RT ƒ =

PV =

RT

M

V

PSimla M RTKalka

ƒSimla

=

ƒKalka

RTSimla

PKalka M

=

72 10 2 13600 9.8 308

2

288 76 10 13600 9.8

1

ƒKalka

=

= 0.987

ƒSimla

1.013

11. n1 = n2 = n

nRT

nRT

,

P2 =

P1 =

V

3V

P1

nRT 3 V

=

=3:1

P2

V

nRT

=

72 308

= 1.013

76 288

V

V

3V

V

PT

PT

P2T

P1 -

12. r.m.s velocity of hydrogen molecules = ?

–3

T = 300 K,

R = 8.3,

M = 2 g = 2 × 10 Kg

3RT

C=

M

C=

3 8.3 300

2 10 3

= 1932. 6 m/s ≈1930 m/s

Let the temp. at which the C = 2 × 1932.6 is T

2 × 1932.6 =

3 8 .3 T

2 10

3

(2 × 1932.6) =

2

3 8 .3 T

2 10 3

(2 1932.6)2 2 10 3

= T

3 8 .3

T = 1199.98 ≈ 1200 K.

13. Vrms =

3P

ƒ

5

P = 10 Pa = 1 atm,

ƒ=

1.77 10 4

10 3

3 10 5 10 3

= 1301.8 ≈ 1302 m/s.

1.77 10 4

14. Agv. K.E. = 3/2 KT

–19

3/2 KT = 0.04 × 1.6 × 10

–23

–19

(3/2) × 1.38 × 10 × T = 0.04 × 1.6 × 10

=

T=

2 0.04 1.6 10 19

3 1.38 10 23

4

= 0.0309178 × 10 = 309.178 ≈ 310 K

8RT

8 8.3 300

=

M

3.14 0.032

Dis tan ce

6400000 2

T=

=

= 445.25 m/s

Speed

445 .25

15. Vavg =

28747 .83

km = 7.985 ≈ 8 hrs.

3600

–3

16. M = 4 × 10 Kg

=

8 8.3 273

8RT

=

= 1201.35

M

3.14 4 10 3

–27

–24

–24

Momentum = M × Vavg = 6.64 × 10 × 1201.35 = 7.97 × 10 ≈ 8 × 10 Kg-m/s.

Vavg =

24.2

Kinetic Theory of Gases

8RT

8 8.3 300

=

M

3.14 0.032

8RT1

8RT2

Now,

=

2

4

17. Vavg =

T1

1

=

T2

2

8RT

M

18. Mean speed of the molecule =

Escape velocity =

8RT

=

M

T=

VavgN2

2gr

8RT

= 2gr

M

2grM

2 9.8 6400000 3.14 2 10 3

=

= 11863.9 ≈ 11800 m/s.

8R

8 8 .3

8RT

M

19. Vavg =

VavgH2

2gr

=

8RT

28

=

2

8RT

28

=

2

14 = 3.74

20. The left side of the container has a gas, let having molecular wt. M1

Right part has Mol. wt = M2

Temperature of both left and right chambers are equal as the separating wall is diathermic

3RT

=

M1

21. Vmean =

3RT

8RT

8RT

=

M1

M2

M2

8RT

=

M

M1

M

3

3

=

1 =

= 1.1775 ≈ 1.18

M2

M2

8

8

8 8.3 273

= 1698.96

3.14 2 10 3

Total Dist = 1698.96 m

1698.96

10

= 1.23 × 10

No. of Collisions =

1.38 10 7

5

22. P = 1 atm = 10 Pascal

–3

T = 300 K,

M = 2 g = 2 × 10 Kg

8RT

8 8.3 300

=

= 1781.004 ≈ 1780 m/s

M

3.14 2 10 3

(b) When the molecules strike at an angle 45°,

(a) Vavg =

Force exerted = mV Cos 45° – (–mV Cos 45°) = 2 mV Cos 45° = 2 m V

No. of molecules striking per unit area =

=

10

5

2 2 10

3

1780

=

3

2 1780

Force

2mv Area

=

1

2

=

2 mV

Pr essure

2mV

10 31 = 1.19 × 10–3 × 1031 = 1.19 × 1028 ≈ 1.2 × 1028

6 10 23

PV

PV

23. 1 1 = 2 2

T1

T2

P1 200 KPa = 2 × 10 pa

T1 = 20°C = 293 K

102 V1

V2 = V1 + 2% V1 =

100

5

P2 = ?

T2 = 40°C = 313 K

P 102 V1

2 10 5 V1

2 10 7 313

= 2

P2 =

= 209462 Pa = 209.462 KPa

293

100 313

102 293

24.3

Kinetic Theory of Gases

–3

3

5

24. V1 = 1 × 10 m ,

P1V1 = n1R1T1

n=

P1 = 1.5 × 10 Pa,

P1V1

1.5 10 5 1 10 3

=

R1T1

8.3 400

T1 = 400 K

n=

1 .5

8 .3 4

1 .5

1 .5

M =

32 = 1.4457 ≈ 1.446

8 .3 4

8 .3 4

5

–3

3

V2 = 1 × 10 m ,

P2 = 1 × 10 Pa,

P2V2 = n2R2T2

m1 =

n2 =

T2 = 300 K

P2 V2

10 5 10 3

1

=

=

= 0.040

R 2 T2

8.3 300

3 8 .3

m2 = 0.04 × 32 = 1.285

m = m1 – m2 =1.446 – 1.285 = 0.1608 g ≈ 0.16 g

5

5

5

25. P1 = 10 + ƒgh = 10 + 1000 × 10 × 3.3 = 1.33 × 10 pa

4

5

–3 3

T1 = T2 = T,

V1 = (2 × 10 )

P2 = 10 ,

3

4 3

V2 = r ,

r=?

3

P1V1

PV

= 2 2

T1

T2

1.33 10 5

4

4

(2 10 3 )3

10 5 r 2

3

3

=

T1

T2

1.33 × 8 × 10 × 10 = 10 × r

5

26. P1 = 2 atm = 2 × 10 pa

3

T1 = 300 K

V1 = 0.002 m ,

P1V1 = n1RT1

5

n=

–9

5

r=

3

3

–3

10.64 10 3 = 2.19 × 10 ≈ 2.2 mm

P1V1 2 10 5 0.002

4

=

= 0.1606

=

RT1

8.3 300

8 .3 3

5

P2 = 1 atm = 10 pa

3

V2 = 0.0005 m ,

P2V2 = n2RT2

n2 =

T2 = 300 K

P2 V2

5

1

10 5 0.0005

=

=

= 0.02

RT2

8.3 300

3 8.3 10

n = moles leaked out = 0.16 – 0.02 = 0.14

27. m = 0.040 g,

T = 100°C,

MHe = 4 g

3

3 m

T = ?

U = nRt = RT

2

2 M

3 m

3 m

Given RT 12 = RT

2 M

2 M

1.5 × 0.01 × 8.3 × 373 + 12 = 1.5 × 0.01 × 8.3 × T

58.4385

= 469.3855 K = 196.3°C ≈ 196°C

T =

0.1245

2

28. PV = constant

2

2

P1V1 = P2V2

nRT1

nRT2

V12 =

V2 2

V1

V2

T1 V1 = T2 V2 = TV = T1 × 2V T2 =

T

2

24.4

Kinetic Theory of Gases

29. PO2 =

nO2

no2 RT

,

PH2 =

nH2 RT

V

V

1.60

m

=

=

= 0.05

32

MO2

nO nH2

RT

Now, Pmix = 2

V

2.80

m

nH2 =

=

= 0.1

MH2

28

(0.05 0.1) 8.3 300

2

= 2250 N/m

0.166

30. P1 = Atmospheric pressure = 75 × ƒg

V1 = 100 × A

P2 = Atmospheric pressure + Mercury pessue = 75ƒg + hgƒg (if h = height of mercury)

V2 = (100 – h) A

P1V1 = P2V2

75ƒg(100A) = (75 + h)ƒg(100 – h)A

2

75 × 100 = (74 + h) (100 – h) 7500 = 7500 – 75 h + 100 h – h

2

2

h – 25 h = 0 h = 25 h h = 25 cm

Height of mercury that can be poured = 25 cm

31. Now, Let the final pressure; Volume & Temp be

After connection = PA Partial pressure of A

PB Partial pressure of B

P 2V

P V

Now, A

= A

TA

T

Pmix =

P

P

Or A = A

T

2TA

A

…(1)

PA

Similarly,

P

PB

= B

2TB

T

: TA

V

…(2)

Adding (1) & (2)

PA PB

P

P

P

1P

= A B = A B

2TA 2TB

T

T

2 TA TB

P

P

1P

= A B

T

2 TA TB

32. V = 50 cc = 50 × 10–6 cm3

5

P = 100 KPa = 10 Pa

(a) PV = nrT1

PV =

[ PA + PB = P]

M = 28.8 g

m

PMV 10 5 28.8 50 10 6

50 28.8 10 1

RT1 m =

=

=

= 0.0635 g.

RT1

8.3 273

8.3 273

M

(b) When the vessel is kept on boiling water

PV =

PVM

10 5 28.8 50 10 6

50 28.8 10 1

m

RT2 m =

=

=

= 0.0465

RT2

8.3 373

8.3 373

M

(c) When the vessel is closed

0.0465

–6

P × 50 × 10 =

8.3 273

28.8

0.0465 8.3 273

6

P=

= 0.07316 × 10 Pa ≈ 73 KPa

28.8 50 10 6

24.5

B

PB

: TB

V

Kinetic Theory of Gases

33. Case I Net pressure on air in volume V

= Patm – hƒg = 75 × ƒHg – 10 ƒHg = 65 × ƒHg × g

Case II Net pressure on air in volume ‘V’ = Patm + ƒHg × g × h

P1V1 = P2V2

ƒHg × g × 65 × A × 20 = ƒHg × g × 75 + ƒHg × g × 10 × A × h

65 20

= 15.2 cm ≈ 15 cm

62 × 20 = 85 h h =

85

34. 2L + 10 = 100 2L = 90 L = 45 cm

Applying combined gas eqn to part 1 of the tube

( 45 A )P0

( 45 x )P1

=

300

273

273 45 P0

P1 =

300( 45 x )

Applying combined gas eqn to part 2 of the tube

45 AP0

( 45 x )AP2

=

300

400

400 45 P0

P2 =

300( 45 x )

P1 = P2

273 45 P0

400 45 P0

=

300( 45 x )

300( 45 x )

V

20 cm

P 20 A

P A

=

…(1)

400

T

P 10 A

P(30 x )

=

…(2)

100

T

Equating (1) and (2)

1

x

=

30 – x = 2x 3x = 30 x = 10 cm

2

30 x

The separator will be at a distance 10 cm from left end.

24.6

h

V

27°C

L

l

1

2

P0

10

P0

L-x

L+x

P1

P2

0°C

(45 – x) 400 = (45 + x) 273

18000 – 400 x = 12285 + 273 x

(400 + 273)x = 18000 – 12285 x = 8.49

273 46 76

P1 =

= 85 % 25 cm of Hg

300 36.51

Length of air column on the cooler side = L – x = 45 – 8.49 = 36.51

35. Case I Atmospheric pressure + pressure due to mercury column

Case II Atmospheric pressure + Component of the pressure due

to mercury column

20cm

P1V1 = P2V2

43cm

(76 × ƒHg × g + ƒHg × g × 20) × A × 43

= (76 × ƒHg × g + ƒHg × g × 20 × Cos 60°) A × ℓ

96 × 43 = 86 × ℓ

96 43

= 48 cm

ℓ=

86

36. The middle wall is weakly conducting. Thus after a long

10 cm

20 cm

time the temperature of both the parts will equalise.

The final position of the separating wall be at distance x

400 K

100 K

P

from the left end. So it is at a distance 30 – x from the right

P

end

Putting combined gas equation of one side of the separating wall,

P1 V1

P V2

= 2

T1

T2

10 cm

10 cm

27°C

10

0°C

60°

ℓ

x

T P

30 – x

T P

Kinetic Theory of Gases

37.

dV

= r dV = r dt

dt

Let the pumped out gas pressure dp

Volume of container = V0 At a pump dv amount of gas has been pumped out.

Pdv = –V0df PV df = –V0 dp

P

P

dp

=

p

t

dtr

V

0

P = P e rt / V0

0

Half of the gas has been pump out, Pressure will be half =

ln 2 =

38. P =

rt

V0

t = ln2

1 vt / V0

e

2

0

r

P0

V

1

V0

nRT

=

V

RT

=

V

RT

=

V0

2

P0

V

1

V0

P0

V

1

V0

P0

2

V

1

V0

2

2

[PV = nRT according to ideal gas equation]

[Since n = 1 mole]

[At V = V0]

P0V0 = RT(1 +1) P0V0 = 2 RT T =

P0 V0

2R

39. Internal energy = nRT

Now, PV = nRT

PV

Here P & V constant

nT =

R

nT is constant

Internal energy = R × Constant = Constant

40. Frictional force = N

Let the cork moves to a distance = dl

Work done by frictional force = Nde

Before that the work will not start that means volume remains constant

P

P

P

1

= 2 P2 = 2 atm

1 = 2

T1

T2

300

600

Extra Pressure = 2 atm – 1 atm = 1 atm

Work done by cork = 1 atm (Adl)

Ndl = [1atm][Adl]

1 10 5 (5 10 2 )2

1 10 5 25 10 5

=

2

2

dN

N

Total circumference of work = 2r

=

dl

2r

N=

=

1 10 5 25 10 5

1 10 5 25 10 5

4

=

= 1.25 × 10 N/M

0.2 2r

0.2 2 5 10 5

24.7

Kinetic Theory of Gases

41.

P1V1

PV

= 2 2

T1

T2

2P0

P0

P0 V

PV

=

P = 2 P0

T0

2T0

Net pressure = P0 outwards

Tension in wire = P0 A

Where A is area of tube.

[ Since liquid at the same level have same pressure]

42. (a) 2P0x = (h2 + h0)ƒg

2P0 = h2 ƒg + h0 ƒg

h2

h2 ƒg = 2P0 – h0 ƒg

2P0

2P0

2P0 h 0 ƒg

=

h0

h2 =

ƒg

ƒg

ƒg

(b) K.E. of the water = Pressure energy of the water at that layer

P

1

2

mV = m

2

ƒ

V =

2

2P

2

=

ƒ

ƒP0 ƒg(h1 h 0

1/ 2

2

V=

ƒ

P

ƒ

g

(

h

h

0

1

0

(c) (x + P0)ƒh = 2P0

2P0 + ƒg (h –h0)= P0 + ƒgx

P0

= h2 + h1

X=

ƒg h1 h0

i.e. x is h1 meter below the top x is –h1 above the top

43. A = 100 cm2 = 10–3 m

m = 1 kg,

P = 100 K Pa = 105 Pa

ℓ = 20 cm

Case I = External pressure exists

Case II = Internal Pressure does not exist

P1V1 = P2V2

1 9 .8

1 9 .8

10 5

× V

V =

3

10 3

10

(10 + 9.8 × 10 )A × ℓ = 9.8 × 10 × A × ℓ

5

–1

2

3

10 × 2 × 10 + 2 × 9.8 × 10 = 9.8 × 10 × ℓ

5

ℓ =

3

2 10 4 19.6 10 2

9.8 10 3

3

= 2.24081 m

44. P1V1 = P2V2

mg

P0 A P0 Aℓ

A

1 9. 8

10 5 0.2 = 105 ℓ

10 10 4

(9.8 × 10 + 10 )× 0.2 = 10 ℓ

3

5

109.8 × 10 × 0.2 = 10 ℓ

109.8 0.2

= 0.2196 ≈ 0.22 m ≈ 22 cm

ℓ =

10 2

3

5

5

24.8

h0

h1

Kinetic Theory of Gases

45. When the bulbs are maintained at two different temperatures.

The total heat gained by ‘B’ is the heat lost by ‘A’

Let the final temp be x

So, m1 St = m2 St

n1 x = 62n2 – n2 x

n1 M × s(x – 0) = n2 M × S × (62 – x)

x=

V

V

A

B

62n 2

62n 2

=

= 31°C = 304 K

n1 n 2

2n 2

For a single ball

P1V1

PV

= 2 2

T1

T2

Initial Temp = 0°C

P = 76 cm of Hg

V1 = V2

Hence n1 = n2

P V

76 V

403 76

= 2

P2 =

= 84.630 ≈ 84°C

273

304

273

46. Temp is 20°

Relative humidity = 100%

So the air is saturated at 20°C

Dew point is the temperature at which SVP is equal to present vapour pressure

So 20°C is the dew point.

47. T = 25°C

P = 104 KPa

VP

[SVP = 3.2 KPa,

RH = 0.6]

SVP

3

3

3

VP = 0.6 × 3.2 × 10 = 1.92 × 10 ≈ 2 × 10

When vapours are removed VP reduces to zero

Net pressure inside the room now = 104 × 103 – 2 × 103 = 102 × 103 = 102 KPa

48. Temp = 20°C

Dew point = 10°C

The place is saturated at 10°C

Even if the temp drop dew point remains unaffected.

The air has V.P. which is the saturation VP at 10°C. It (SVP) does not change on temp.

RH =

VP

SVP

The point where the vapour starts condensing, VP = SVP

We know P1V1 = P2V2

3

RH SVP × 10 = SVP × V2

V2 = 10RH 10 × 0.4 = 4 cm

50. Atm–Pressure = 76 cm of Hg

When water is introduced the water vapour exerts some pressure which counter acts the atm pressure.

The pressure drops to 75.4 cm

Pressure of Vapour = (76 – 75.4) cm = 0.6 cm

49. RH =

VP

0 .6

=

= 0.6 = 60%

SVP

1

51. From fig. 24.6, we draw r, from Y axis to meet the graphs.

Hence we find the temp. to be approximately 65°C & 45°C

52. The temp. of body is 98°F = 37°C

At 37°C from the graph SVP = Just less than 50 mm

B.P. is the temp. when atmospheric pressure equals the atmospheric pressure.

Thus min. pressure to prevent boiling is 50 mm of Hg.

53. Given

SVP at the dew point = 8.9 mm

SVP at room temp = 17.5 mm

Dew point = 10°C as at this temp. the condensation starts

Room temp = 20°C

R. Humidity =

RH =

8 .9

SVP at dew po int

=

= 0.508 ≈ 51%

SVP at room temp

17.5

24.9

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