Tag: behaviour of perfect gas and kinetic theory of gases

Questions Related to behaviour of perfect gas and kinetic theory of gases

The total Kinetic energy of $1\ mole$ of ${N}^{} _{2}$ at $27^{o} _{}{C}$ 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. $1500\ J$

  2. $15633\ cal$

  3. $1500\ kcal$

  4. $1500\ erg$

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Correct Option: B
Explanation:

For $n$ mole of any gas the total  kinetic energy is given as $E=\dfrac{3}{2}nRT$

Where $R$ is gas constant having value $8.31J/mole-K$ or $8.31\times 4.18 cal /mole-K=34.74\text{Cal per mole per Kelvin}$
$T$ is temperature in Kelvin which is $T=27+273=300K$
So putting all values we get $E=1.5\times 1 \times 34.74\times 300=15633Calorie$

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes $5.66$V while its temperature falls to $T/2$. How many degrees of freedom do the gas molecules have?

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. 7

  2. $5$.

  3. 6

  4. 8

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Correct Option: B
Explanation:
Adiabatic equation of perfect gas is given as $TV^{r-1}=$ constant
$m=T _{1}V _{1}^{(r-1)}=T _{2}V _{2}^{(r-1)}$
$T _{1}=T _{1}V _{1}=V _{1}V _{2}=5.66\ V$
and $T _{2}=\dfrac{T}{2}$
$TV^{r-1}=\dfrac{T}{2}(5.66\ V)^{r-1}$
$2=5.66^{r-1}$
Taking $\log$ on both sides
$(r-1)\log 5.66=\log 2(r-1)$
$r=\dfrac{\log 2}{\log 5.66}=1+0.3010/0.75$
$r=1+0.4$
$r=1.4$ for $r=1.4$ Agree of freedom $=5$

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. $2\ A^{o}$

Show answer
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. $8$

Show answer
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. $6.35$

Show answer
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