Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

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

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

So total degrees possesed by all the molecules

= 5 $*2.688$ $*{10}^{19}=1.344*{10}^{20}$

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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

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

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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.

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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

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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

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

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