Tag: law of equipartition of energy

Questions Related to law of equipartition of energy

A circular disc of mass $m$ and radius $r$ is rolling about its axis with a constant speed $v$. Its kinetic energy 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. $\cfrac{1}{4}mv^2$

  2. $\cfrac{1}{2}mv^2$

  3. $\cfrac{3}{4}mv^2$

  4. $mv^2$

Show answer
Correct Option: C

The number of degrees of freedom for each atom of a monoatomic gas 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. 3

  2. 5

  3. 6

  4. 1

Show answer
Correct Option: A
Explanation:

A monoatomic gas has 3 degrees of freedom for translation motion.
Hence, option A is correct.

The internal energy of a 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. is the sum total of kinetic and potential energies.

  2. is the total transitional kinetic energies

  3. is the total kinetic energy of randomly moving molecules.

  4. is the total kinetic energy of gas molecules

Show answer
Correct Option: C
Explanation:

At a given temperature, the pressure of a container is determined by the number of times gas molecules strike the container walls. If the gas is compressed to a smaller volume, then the same number of molecules will strike against a smaller surface area; the number of collisions against the container will increase, and, by extension, the pressure will increase as well.
Increasing the kinetic energy of the particles will increase the pressure of the gas.
So, the internal energy of a gas Is the total kinetic energy of randomly moving molecules.
Hence, option C is correct.

State whether true or false:

Linear molecules have $3N-5$ vibrational degrees of freedom, whereas non linear molecules have $3N-6$ vibrational degrees of freedom, where N is no. of atoms present in a molecule.

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. True

  2. False

Show answer
Correct Option: A
Explanation:

Vibrational degree of freedom:
(a) For linear molecule = 3N - 5.
(b) For non-linear molecule = 3N - 6. where N is no. of atoms present in a molecule.

The number of degrees of freedom in an oxygen molecule 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. 3

  2. 4

  3. 5

  4. 6

Show answer
Correct Option: C
Explanation:

Diatomic, like your oxygen molecule, can move in three dimensions and spin in two directions.

Total Degree of Freedom = 5.

Significant motion for the molecules of a monoatomic gas corresponds to :

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. translatory

  2. vibratory

  3. rotatory

  4. none of these

Show answer
Correct Option: A
Explanation:

 Monoatomic gas has only 3 degree of freedom (all translational)

To find out degree of freedom, the correct expression 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. $f=\dfrac { 2 }{ \gamma -1 }$

  2. $f=\dfrac { \gamma +1 }{ 2 }$

  3. $f=\dfrac { 2 }{ \gamma +1 }$

  4. $f=\dfrac { 1 }{ \gamma +1 }$

Show answer
Correct Option: A
Explanation:

$\because \gamma =1+\dfrac { 2 }{ f } $
$\Longrightarrow \dfrac { 2 }{ f } =\gamma -1\Longrightarrow f=\dfrac { 2 }{ \gamma -1 } $

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. $1500\ erg$

Show answer
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

Show answer
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.

Show answer
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=$\dfrac{3}{2} \times (6.022 \times 10^{23})\times (1.38 \times 10^{-23}) \times 300$
$=3739.662 J$