Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen: Overview, Questions, Preparation

This is a long answer type question as classified in NCERT Exemplar

(a) The moon has small gravitational force and hence, the escape velocity is small .

As the moon in the proximity of the earth as seen from the sun, the moonhas the same amount of heat per unit area as that of the earth.

The air molecules have large range of speeds . even though the rms speed of the air molcules is smaller than the escape velocity on the moon, a significant number of molecules have speed grater than the escape velocity.

(b) As the molecules move higher their potential energy increases and hence kinetic energy decreases and hence temperature reduces. At greater height more volume is available and gas expands and hence some cooling takes place.

This is a long answer type question as classified in NCERT Exemplar

This is a long answer type question as classified in NCERT Exemplar

Time require to avoid the collision T= l/v where l = mean free path =1/ $\sqrt[]{2}\pi d^{2}n$

Where n = N/V

n=number of aeroplanes/volume

= $\frac{10}{20\times 20\times 1.5}=0.0167km$ ^-3

T= $\frac{1V}{\sqrt[]{2}\pi d^{2}N}\times \frac{1}{v}$

T= $\frac{1}{\sqrt[]{2}\times 3.14\times {20}^2\times 0.0167\times {10}^{-6}\times 150}$ = $\frac{{10}^6}{1776.25\times 2.505}$

$=\frac{{10}^6}{4449.5}=224.74h\approx 225h$

This is a long answer type question as classified in NCERT Exemplar

Given volume V = 1m³

area = 0.01mm²

= 8.01 $\times {10}^{-6}$ m²= ${10}^{-8}$ m²

Temperature both inside and outside are equal

So initial pressure inside the box = 1.50atm

Final pressure inside the box= 0.1atm

Assuming V_x= speed of nitrogen molecule in x direction

n_i = number of molecules per unit volume in a time interval of $?T$

Let area of the wall, number of particles colliding in time

$?t$ = $\frac{1}{2}n$ _i (v_{x $?t$} )A , here we use ½ because particle moves both in positive and negative direction.

V_x²+ V_y²+ V_z²= V_rms²

V_x²= V_rms²/3 if all three velocities are equal.

½ mv_rms²= 3/2K_BT/m

This is a long answer type question as classified in NCERT Exemplar

Consider the diagram

let n =number of molecules per unit volume

V_rms= rms speed of gas molecule

When block is moving with speed v_o, relative speed of molecules w.r.t front face =v+v_o

Coming head on, momentum transferred to block per collision =2m (v+v_o)

Number of collisions in time $?t$ = $\frac{1}{2}$ (v+v_o)n $?t$ A  where A is the area of cross section.

So momentum transferred in time $?t$ =m (v+v_o)²nA $?t$   this is from front surface

Similarly momentum transferred in time $?t$ = m (v-v_o)²nA $?t$ ) this is from back surface

Drag force = mnA (v+v_o)²- (v-v_o)²)

= mnA (4w_o)=4mnAvv_o

= 4 $\rho A$ vv_o

So $\rho$ =mn/V=M/V

If v = velocity along x axis

Then we can write KE= ½ mv²=1/2 K_BT

This is a short answer type question as classified in NCERT Exemplar

molar mass = mass of avogadro’s number of atoms= 6.023 $\times {10}^{23}$ atoms.

197 g of gold contains =6.023 $\times {10}^{23}atoms$

1g of gold contain= $\frac{6.023\times {10}^{23}}{197}$ atoms

39.4 g of gold atoms = $\frac{6.023\times {10}^{23}\times 39.4}{197}=1.2\times {10}^{23}$ atoms

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V $\propto T$

V/T = constant

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

T₁=273+27=300K

T₂= 273+327= 600K

V₁= 100cc

V₂=V₁ (600/300)

V₂=2V₁

V₂= 2 (100)=200cc

This is a short answer type question as classified in NCERT Exemplar

V_rms= $\sqrt[]{\frac{3RT}{M}}$

V_rms= $\sqrt[]{T}$

$\frac{V_{rms1}}{V_{rms2}}=\sqrt[]{\frac{T_1}{T_2}}$

T₁=27^oC= 27+273=300K

T₂=127^oC= 127+273= 400K

$\frac{100}{V_{rms2}}=\sqrt[]{\frac{300}{400}}=\frac{\sqrt[]{3}}{2}$

V_rms²= $\frac{2\times 100}{\sqrt[]{3}}=\frac{200}{\sqrt[]{3}}m/s$

This is a short answer type question as classified in NCERT Exemplar

This is a short answer type question as classified in NCERT Exemplar

Oxygen gas having 5 degrees of freedom

Energy per mole of the gas =5/2RT

For 2 moles of the gas total internal energy =2 $*$ 5/2RT=5RT

Neon is a monoatomic gas having 3 degrees of freedom

Energy per mole =3/2RT

We have 4 moles of Ne

Energy = 4 $*$ 3/2RT=6RT

Total energy =5RT+6RT=11RT

This is a short answer type question as classified in NCERT Exemplar

Mean free path l=1/ $\sqrt[]{2}{d}^{2}n$

So n= number of molecules /volume

d = diameter of the molecule

l $\propto \frac{1}{d^2}$

d₁=1A^o, d₂=2A^o

l $\propto \frac{1}{d^2}$

$\frac{l_1}{l_2}={\left(\frac{d_2}{d_1}\right)}^2={\left(\frac{2}{1}\right)}^2=\frac{4}{1}$

l₁:l₂=4:1

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V₁=2L ,V₂=3L

µ₁ = 4.0and µ₂ =5.2

p₁= 1.00 atm and p₂ = 2.00 atm

p₁V₁= µ₁RT₁

p₂V₂= µ₂RT₂

when the partition is removed the gases get mixed without any loss of energy . the mixture now attains a common equilibrium pressure and total volume of the system is sum of the volume of individual chambers V₁ and V₂

$\mu ={\mu }_1+{\mu }_2$ , V =V₁+V₂

From the kinetic theory of gases pV=2/3 E

For mole 1  ,P₁V₁= 2/3 ${\mu }_1{E}_1$

For mole 2  , P₂V₂= 2/3 ${\mu }_2{E}_2$

Total energy is ( ${\mu }_1{E}_1+{\mu }_2{E}_2$ )= 3/2 ( $P_1{V}_1+P_2{V}_2$ )

PV==2/3E_total = 2/3 $\mu E_{permole}$

P( $V_1+V_2$ )= $\frac{2}{3}\times \frac{3}{2}(p_1{V}_1+p_2{V}_2)$

P= $\frac{P_1{V}_1+P_2{V}_2}{V_1+V_2}$ = $\frac{1\times 2+2\times 3}{2+3}atm$

P=8/5 =1.6atm

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(a) The average KE will be the same

M= molar mass of the gas

m=mass of each molecular of the gas

R= gas constant

v_{rms $\propto \frac{1}{\sqrt[]{m}}$}

(b) k = Boltzmann constant

T= absolute temperature

m_A>m_B>m_C

V_rms.Arms.Brms.C

This is a short answer type question as classified in NCERT Exemplar

Radius = 1A^o

Volume of hydrogen molecules = 4/3 $\pi$ r³

= 4/3 (3.14) (10^-10)³ $\approx 4\times {10}^{-30}$ m³

Number of moles of H₂ = mass/molecular mass=0.5/2=0.25

Molecules of H₂ present = number of moles of H₂ present $\times 6.023\times {10}^{23}$

= 0.25 $\times 6.023\times {10}^{23}$

So volume of molecules present = molecule number $\times$ volume of each molecules

= 0.25 $\times 6.0233\times {10}^{23}\times 4\times {10}^{23}$

$\approx$ 6 $\times {10}^{-7}m$ ³

P_iV_i= P_fV_f

V_{f $\left(\frac{P_i}{p_f}\right)V$} = _i= $\frac{1}{100}(3\times {10}^{-2})$ ³

V_f= 2.7 $\times {10}^{-7}m$ ³

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When air is pumped, more molecules are pumped and boyle's law is staed for situation where number of molecules remains constant . in this case as the number of air molecules keep increases, hence mass change. Boyle's law is only applicable in situations, where number of gas molecule of remains fixed.

This is a short answer type question as classified in NCERT Exemplar

Number of helium = 5

T=7^oC=7+273=280K

(a) hence number of atoms = number of moles $*$ Avogadro's number

= $5*6.023*{10}^{23}=30.015*{10}^{23}=3*{10}^{24}$ atoms

(b) now average kinetic energy per molecule = 3/2 K_BT

= total internal energy

= $\frac{3}{2}{K}_{b}T*$ number of atoms

= $\frac{3}{2}*1.38*{10}^{-23}*280*3*{10}^{24}=1.74*{10}^{4}J$

This is a short answer type question as classified in NCERT Exemplar

Volume occupied by 1 mole = 1mole of the gas at NTP= 22400mL=22400cc

So number of molecules in 1cc of hydrogen= $\frac{6.023\times {10}^{23}}{22400}=2.688\times {10}^{19}$

H₂ is a diatomic gas, having a total of 5 degrees of freedom

So total degrees possesed by all the molecules

= 5 $\times 2.688$ $\times {10}^{19}=1.344\times {10}^{20}$

This is a short answer type question as classified in NCERT Exemplar

According to kinetic interpretation of temperature, absolute temperature of a given sample of a gas is proportional to total translational kinetic energy of its molecule.

Hence any change in absolute temperature of a gas will contribute to corresponding change in translational KE and vice versa.

N= number of moles

m=molar mass of the gas

when the container stops its total kinetic energy transferred to the gas molecules in the form of translational KE, thereby increasing the absolute temperature.

KE of molecules due to velocity KE= $\frac{1}{2}\left(mn\right)v_o^2$

Increase in translational KE =n $\frac{3}{2}R?$ T

According to kinetic theory  of gases

$\frac{1}{2}\left(mn\right)v_o^2$ = n $\frac{3}{2}R?$ T

$?$ T= $\frac{m{v}_o^2}{3R}$

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(b) As the motion of the vessel as a whole does not affect the relative motion of the gas molecules with respect to the walls of the vessel, hence pressure of the gas inside the vessel, as conserved by us, on the ground remains the same.

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(d) In an ideal gas when a molecules collides elastically with a wall, the momentum transferred to each molecule will be twice the magnitude of its normal momentum. For the face EFGH, it transfer only half of that.

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(b) Boyle's law is applicable when temperature is constant

PV=nRT=constant

PV= constant

So pressure is inversely proportional to volume.

So process is called isothermal process.

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(c) The pressure inside the gas will be 

P= p_a+Mg/A

A= area of piston

P_a= atmospheric pressure

Mg = weight of piston

When temperature is increases

pV=nRT so volume increases at constant pressure.

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(a) Pressure is inversely proportional to slope

So $\frac{p_1}{}$$\frac{p_{}}{}$$\frac{2_{}}{}=\frac{m_2}{m_1}\mathrm{<1}$

So p₁ > p₂

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(d) When number of molecules breaks then number of moles would become double.

P $\propto nT$ where n is the number of moles and T is temperature of ideal gas

As gases breaks number of moles becomes twcice of initial so n₂ =2n₁

$\frac{P_2}{P_1}=\frac{n_2{T}_2}{n_1{T}_1}=\frac{2n_1\left(3000\right)}{n_1\left(300\right)}$ = 20

p₂=20p₁

so 20 times

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(b) For a function f (v), the number of molecules n=f (v) which are having speeds between v and v+dv

So f₁ (v) and f₂ (v) will obey’s maxwell distribution law

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(d) pV=nRT

So n= PV/RT

As number moles are fixed

$\frac{P_1{V}_1}{{RT}_1}=\frac{P_1{V}_1}{{RT}_1}$

P₂=p₁V₁ (T₂/V₂T₁)

= $\frac{p\mathrm{V\; (}\left(1.1\mathrm{T)}\right)}{1.05\mathrm{V\; (}\left(\mathrm{T)}\right)}$

= P $\times \frac{1.1}{1.05}\approx 1.05P$

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(b), (d) due to the presence of external positive charge on face ABCD. The usual expression for pressure on the basis of kinetic energy will not be valid as ions would also experience electrostatic force. So presence of positive charge the isotropy is also lost.

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(c) According to kinetic theory, we know walls only exert perpendicular forces on molecule. They do not exert parallel force . so there is only change in translational motion.

So pV=2/3E

Where E is representing only translational part of energy.

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(a), (d) The total energy associated with the molecule is

E= $\frac{1}{2}m{v}_x^2+\frac{1}{2}m{v}_y^2+\frac{1}{2}m{v}_z^2+\frac{1}{2}{I}_x{w}_x^2+\frac{1}{2}{I}_y{w}_y^2$

The number of independent terms in the above expressions is 5. So we can predict velocities of molecule by maxwell’s distribution, hence the above expression obeys maxwell ‘s distribution .

So 2 rotational and 3 translational energies are associated with each molecule

So rotational energy at a given temperature is 2/3 of the translational KE of each molecule.

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(a), (c) pV=nRT

(a) when pressure p =constant

volume is directly proportional to temperature

(b) when T= constant

from PV =constant

so the graph is rectangular hyperbola

(c) when V =constant

from pressure is directly proportional to temperature.

So the graph is straight the passes through the origin.

PV $\propto T$

PV/T=constant

So the graph hence through origin. So d is correct.

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When the gas is compressed adiabatically, the total work done on the gas increases its internal energy which in turn increases the KE of the gas molecule and hence, the collisions between molecules also increases.