NCERT Textual Exercise (Solved)

179

Kinetic Theory

1. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.

Sol. Given,Diameter,d=3Å, r = d/2 = 1.5Å = 1.5 × 10–8 cm

V = molecular volume = 43

3πr N (WhereN–Avogadro’snumber)

= 43

227

1 5 10 6 023 10 8 528 3 23×

× × × =−( . ) ( . ) . cc

V′ = actual volume occupied by 1 mole of O2 at STP = 22400 cc

∴ VV '

. .= = × −8 5222400

3 8 10 4

2. Molarvolumeisthevolumeoccupiedby1moleofanyidealgasatstandardtemperatureandpressure(STP)whichare1atmosphericpressureand0°C.Show that it is 22.4 litres.

Sol. Foronemoleofanidealgas,PV = RT

∴ V RTP

= [Where,R = 8.31 J mol–1 K–1,T=273K,P = 1 atmosphere

= 1.013 × 105 Nm–2]

= 8 31 2731 013 105

..

××

= 0.0224 m3 = 22.4 litres

3. An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atmandatemperatureof27°C.Aftersomeoxygeniswithdrawnfromthecylinder,thegaugepressuredropsto11atm.anditstemperaturedropsto17°C.Estimatethemassofoxygentakenoutofthecylinder.

(R = 8.1 J mole–1 K–1,molecularmassofO2=32cc) Sol. Initially,thevolumeoftheoxygenO2cylinder,V1 = 30 litre = 3 × 10–3 m3

P1 = 15 atm = 15 × 1.01 × 105 Pa,T1 = 27 + 273 = 300 K If n1 be the moles of O2gasinthecylinder,P1V1 = n1RT1

∴ n1

5 315 1 01 10 30 108 3 300

18 253=× × × ×

×=

−( . ) ( ).

.

M = molecular weight of O2 = 32g; Initial cylinder mass = m1 = n1M = 18.253 × 32 = 584.1 g

Let,n2 – moles of O2 left in the cylinder

∴ nPVRT2

2 2

2

=

[where,V2 = 30 × 10–3m3,P2 = 11 × 1.01 × 105Pa,T2 = 17 + 273=290°K]

NCERT Textual Exercise (Solved)

180

Kinetic Theory

= ( . )( ).

11 1 01 10 30 108 3 290

5 3× × ××

−

= 13.847

∴ m2–finalmassofO2 in cylinder = 13.847 × 32 = 443.1 g ∴ Massofgastakenout m1 – m2 = 584.1 – 443.1 = 141 g 4. An air bubble of volume 1 cm3 rises from the bottom of a lake 40 m deep

atatemperatureof12°C.Towhatvolumedoesitgrowwhenitreachesthesurface,whichisatatemperatureof35°C?Given1atm=1.01×105 Pa.

Sol. Given,V1 = 1.0 cm3 = 1 × 10–6 m3,T1=12°C=285K, P1 = 1 atm + h1ρg = 1.01 × 105 + 40 × 103 × 9.8 = 493000 Pa Let,V2 = volume of bubble at the surface of the lake

PVT

PVT

1 1

1

2 2

2

=

∴ V PVTT P21 1 2

1 2

= [Where T2=35°C=308K,P2 = 1atm = 1.01 × 105 Pa]

= ( ) ( . )

..493000 1 0 10 308

285 1 01 105 275 10

6

56 3× × ×

× ×= ×

−− m

5. Estimate the total number of air molecules (inclusive of O2,N2,watervapour,andotherconstituent)inaroomofcapacity25m3atatemperatureof27°Cand 1 atm pressure.

[Boltzmann constant = 1.38 × 10–23 JK–1] Sol. Given,V = 25 m3,T=27+273=300K,k = 1.38 × 10–23 JK–1. Now,PV = nRT = n(Nk)T = N'kT where,nN = N' = total no. of air molecules in the given gas

N PV

kT' ( . )

( . ).= =

× ×× ×

= ×−

1 01 10 251 38 10 300

6 10 105

2326

6. Estimatetheaveragethermalenergyofaheliumatomat(i)roomtemperature(27°C)(ii)The temperatureof thesurfaceof theSun(6000K)(iii)Thetemperature of 10 million kelvin (the typical core temperature in the case ofastar).

Sol. Given,T=27°C=300°K

Averagethermalenergy,=3/2kT = 32

1 38 10 300 6 2 1023 21× × × = ×− −. . J

NCERT Textual Exercise (Solved)

181

Kinetic Theory

(ii) At,T = 6000 K; Avg. thermal energy

= 32kT =

32

1 38 10 6000 1 24 1023 19× × × = ×− −( . ) . J

(iii) At,T=10millionkelvin,T = 107 K

Avg. thermal energy = 32kT =

32

1 38 10 1023 7× × ×−( . ) = 2.1 × 10–16 J.

7. Three vessels of equal capacity have gases at the same temperature and pressure.Thefirstvesselcontainsneon(monoatomic).Thesecondcontainsoxygen(diatomic),andthethirdcontainsuraniumhexafluoride(polyatomic).Dothevesselcontainequalnumberofrespectivemolecules?Is therootmeansquarespeedofmoleculesthesameinthecases?Ifno,inwhichcaseis vrmsthelargest?

Sol. Atsametemperatureandpressure,allthevesselshavethesamevolume. ByAvogadro’slaw,threevesselscontainequalnumberofmolecules Avogadro’s no. N = 6.023 × 1023

As v kTmrms =

3

∴ vmrms ∝1

Therefore,atagiventemperature,rmsspeedofmoleculeswill not be the sameinallthethreecases.Asneonhasthesmallestmass,therefore,rmsspeed will be the largest in case of neon.

8. At what temperature is the root mean square speed of an atom in an argon gascylinderequaltothermsspeedofaheliumgasatomat–20°C?(AtomicmassofAr=39.9uandHe=4u).

Sol. Let C,C'bethermsvelocityofArandHeattemperaturesT and T ′K,respectively.

Given,M=39.9,M′=4,T' = – 20 + 273 = 253 K

Now C RTM

RT and C RTM

R= = = =

×3 339 9

3 3 2534.

' ''

Since C = C',therefore,339 9

3 2534

RT R.=

×

∴ T = 2523.7 K

NCERT Textual Exercise (Solved)

182

Kinetic Theory

9. Estimate the mean free path and collision frequency of a nitrogen molecule inacylindercontainingnitrogenat2atm.andtemperature17°C.Taketheradiusofanitrogenmoleculetoberoughly1Å.Comparethecollisiontimewiththetime,themoleculemovesfreelybetweentwosuccessivelycollision.(Moleculemassofnitrogen=28u)

Sol. Given P = 2 atm = 2 × 1.03 × 105 Nm–2; T=17°C=290K,d = 2 × 1 = 2Å = 2 × 10–10m

k = 1.38 × 10–23 J molecule–1K–1; M = 28 × 10–23 kg

λ

π ρ= =

× ×× × × × ×

−

−

kTd2

1 38 10 2901 414 3 14 2 10 2 026 102

23

10 2 5

.. . ( ) ( . )

⇒ λ = 1.11 × 10–7 m

v RT

Mrms = =× ×

×=−

3 3 8 31 29028 10

508 243

. . m/s

Collisionfrequency=No.ofcollisionspersecond=508 24

1 11 104 58 107

9..

.×

= ×−

10. Fromacertainapparatus,thediffusionrateofH2 has an average value of 28.7 cm3s–1. The diffusion of another gas under the same condition is measured to have an average rate of 7.2 cm3s–1. Identify the gas.

Sol. AccordingtoGraham’slawofdiffusion,rr

MM

1

2

2

1= ,

where,r1=diffusionrateofH2 = 28.7 cm3s–1

r2 = diffusion rate of unknown gas = 7.2 cm3s–1 M1=molecularwt.ofH2 = 2u

∴28 77 2 2

2..

=M

∴ M2=32,whichismolecularmassofO2 gas.