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Multiple choice 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

A circular disc of mass $m$ and radius $r$ is rolling about its axis with a constant speed $v$. Its kinetic energy is 

  1. $\cfrac{1}{4}mv^2$

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

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

  4. $mv^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total kinetic energy = translational KE + rotational KE. For a rolling disc, K = (1/2)mv^2 + (1/2)I(omega^2). With I = (1/2)mr^2 and v = r*omega, K = (1/2)mv^2 + (1/2)(1/2)mr^2(v^2/r^2) = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.

Multiple choice 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

The internal energy of a gas:

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice 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

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.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice 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

To find out degree of freedom, the correct expression is :

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

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

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

  4. $f=\dfrac { 1 }{ \gamma +1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice 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

The total Kinetic energy of $1\ mole$ of ${N}^{} _{2}$ at $27^{o} _{}{C}$ will be approximately :-

  1. $1500\ J$

  2. $15633\ cal$

  3. $1500\ kcal$

  4. $1500\ erg$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $E=1.5\times 1 \times 34.74\times 300=15633Calorie$

Multiple choice 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

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?

  1. 7

  2. $5$.

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $=5$
Multiple choice 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

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

  1. <span><span><span class="mrow"><span class="mn">3739.662 J</span></span></span></span>&nbsp;

  2. 1500 calorie

  3. 1500 kilo calorie

  4. 1500 erg.
Reveal answer Fill a bubble to check yourself
A Correct answer
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=$\dfrac{3}{2} \times (6.022 \times 10^{23})\times (1.38 \times 10^{-23}) \times 300$
$=3739.662 J$