Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

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

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

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

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

molar mass = mass of avogadro's number of atoms= 6.023 $*{10}^{23}$ atoms.

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

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

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

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

Given volume V = 1m³

area = 0.01mm²

= 8.01 $*{10}^{-6}$ 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

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

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

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

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

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