Tag: law of equipartition of energy and mean free path

Questions Related to law of equipartition of energy and mean free path

The total kinetic energy of $1$ mole of $N _2$ at $27$C will be approximately

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. 3739.662 J 

  2. 1500 calorie

  3. 1500 kilo calorie

  4. 1500 erg.

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Correct Option: A
Explanation:

The kinetic enrgy of one mole is given by:

KE=$\dfrac{3}{2} K _BT$
The kinetic enrgy of 1 mole of $N _2$ atoms is:
KE=$\dfrac{3}{2}K _B T$ where $N$ is Avogadro's number,$K _B$ is Boltzmann's constant and $T$ is temperature
KE=$\dfrac{3}{2} \times (6.022 \times 10^{23})\times (1.38 \times 10^{-23}) \times 300$
$=3739.662 J$

The de-Broglie wavelength of a particle accelerated with $150\ volt$ potential is $10^{-10}\ m$. If it accelerated by $600\ volts$ p.d. its wavelength will be

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. $0.25\ A^{o}$

  2. $0. 5\ A^{o}$

  3. $1.5\ A^{o}$

  4. $2\ A^{o}$

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Correct Option: A
Explanation:

Given,

$\lambda =\dfrac{hc}{eV}\ \ \ \ where,\ V=potential$

$\lambda \ \alpha \ \dfrac{1}{V}$

${{10}^{-10}}\ \alpha \ \dfrac{1}{150}\ ......\ (1)$

$\lambda \ \alpha \ \dfrac{1}{600}\ ......\ (2)$

Divide (2) by (1)

$ \dfrac{\lambda }{{{10}^{-10}}}=\dfrac{150}{600}=\dfrac{1}{4} $

$ \Rightarrow \lambda =0.25\times {{10}^{-10}}m\ =0.25\ {{A}^{o}} $ 

Three particles are situated on a light and rigid rod along Y-axis as shown in the figure. If the system is rotating with angular velocity of $2 rad/sec$ about X axis, then the total kinetic energy of the system is :

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. $92 J $

  2. $184 J $

  3. $ 276 J $

  4. $46 J $

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Correct Option: A

A gas has molar heat capacity $C = 4.5\ R$ in the process $PT = constant$. Find the number of degrees of freedom (n) of molecules in the gas.

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. $n = 7$

  2. $n = 3$

  3. $n = 5$

  4. $n = 2$

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Correct Option: C

A gas undergoes a process such that $P \alpha \dfrac{1}{T}$. If the molar heat capacity for this process is $24.93 \,J/mol \,K$, then what is the degree of freedom of the molecules of the gas?

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. $8$

  2. $4$

  3. $2$

  4. $6$

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Correct Option: A

The degrees of freedom of a triatomic gas is? (consider moderate temperature)

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. $6$

  2. $4$

  3. $2$

  4. $8$

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Correct Option: A
Explanation:

The general epression for degree of freedom is $DOF=3N-n$

here, DOF means degree of freedom, N is number of particle, and n is the number of holonomic constraints.
for a triatomic molecule, the number of particle is 3 and since the separation between three atoms are fixed so, the number of constraints is 3.
hence, $DOF=(3\times 3)-3$
$DOF=9-3$
$DOF=6$

A vessel contains a non-linear triatomic gas. If $50$% of gas dissociate into individual atom, then find new value of degree of freedom by ignoring the vibrational mode and any further dissociation 

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. $2.15$

  2. $3.75$

  3. $5.25$

  4. $6.35$

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Correct Option: B
Explanation:

Let's assume we have $1$ mole of triatomic gas

$\therefore 3Na$ is present
So, $0.5$ moles= $1.5 Na$ atoms
$1$ part of $0.5$ moles remains untouched
Degree of dissociation= $0.5 \times 6=3$
Degree of freedom for $0.5Na= 1.5 \times 0.5=0.75$
Total=$3+0.75=3.75$

For gas, if the ratio of specific heats at constants pressure $P$ and constant volume $V$ is $\gamma $, then the value of degree of freedom is:

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. $\dfrac{\gamma +1}{\gamma -1}$

  2. $\dfrac{\gamma -1}{\gamma +1}$

  3. $\dfrac{1}{2}(\gamma-1)$

  4. $\dfrac{2}{\gamma-1}$

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Correct Option: A

The speed of a longitudinal wave in a mixture containing 4 moles of He and 1 mole of Ne at 300 K will be

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. 930 m/s

  2. 541 m/s

  3. 498 m/s

  4. None of these

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Correct Option: A
Explanation:
The mixture contains $n _1=4\ mol$ of $He$ and $n _2=4\ mole$ of $Ne$ act. $T=300\ K$.
Forth $He$ and $Ne$ are monoatomic, so we take $r$ mix $=5/3\quad \left [r=\dfrac {cp}{cv}\right]$ 
$M'=\dfrac {M _1n _1+M _2n _2}{n _1+n _2}$
$=\dfrac {4\times 4+20\times 1}{4+1}=\dfrac {24}{5}=4.8\ g/mol$
$=4.8\times 10^{-3}\ kg/mol$
$\therefore \ $ Speed of sound through mixture
$v=\sqrt {\dfrac {vRT}{M'}}=\sqrt {\dfrac {5}{3}\times \dfrac {8.3\times 300}{4.8\times 10^{-3}}}\sim 929.83\ m/s$


$2$ grams of mono atomic gas occupies a volume of $2$ litres at a pressure of $8.3 \times 10^5$ Pa and $127^0C$. Find the molecular weight of the gas.

physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

  1. $2$ grams/mole

  2. $16$ grams/mole

  3. $4$ grams/mole

  4. $32$ grams/mole

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Correct Option: C