Physics

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 $\approx 4*{10}^{-30}$ m³

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

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

= 0.25 $*6.023*{10}^{23}$

So volume of molecules present = molecule number $*$ volume of each molecules

= 0.25 $*6.0233*{10}^{23}*4*{10}^{23}$

$\approx$ 6 $*{10}^{-7}m$ ³

P_iV_i= P_fV_f

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

V_f= 2.7 $*{10}^{-7}m$ ³

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

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

Answer – (b, c)

Explanation – According to the relation that p/q=r/s if galvanometer shows no deflection. In the above equation $\frac{R1}{R2}=\frac{R3}{R4}$ the value of R4 depends upon R1 and R2 . if the value of R1 and R2 will be feeble the value of R4 would be affected.

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 Multiple Choice Questions as classified in NCERT Exemplar

Explanation- according to the relation $\rho$ = $\frac{m}{ne2t}$ also T $\propto \frac{1}{t}$