Kinetic Theory

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

(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

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

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

P=8/5 =1.6atm

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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