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Degree of freedom: law of equipartition of energy - class-XI

Description: degree of freedom: law of equipartition of energy
Number of Questions: 62
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Tags: kinetic theory of gases behavior of perfect gas and kinetic theory behaviour of perfect gas and kinetic theory of gases kinetic theory of matter physics kinetic theory
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How many degrees of freedom the gas molecules have if under STP the gas density $\rho = 1.3 kg/m^3$ and the velocity of sound propagation in it is $330 ms^{-1}$?

  1. $3$

  2. $5$

  3. $7$

  4. $8$


Correct Option: B

At room temperature (27$^0$ C) the rms speed of the moleculesof certain diatomic gas is found to be 1920 ms$^{-1}$ then the molecule is:

  1. $H _2$

  2. $F _2$

  3. $O _2$

  4. $Cl _2$


Correct Option: A
Explanation:
Let the room temperature is $T = 27^0C=27+273=300K$
Now, $V _{rms}=\sqrt{\dfrac{3RT}{m}}$
$\Rightarrow M=\dfrac{3RT}{V _{rms}^2}$
By putting the value we get,
$M=\dfrac{3\times8.314\times300}{1920^2}=2\times10^{-3}kg=2g$
Thus, it is an Hydrogen.

If temperature of body increases by 10%, then increase in radiated energy of the body is :

  1. 10 %

  2. 40 %

  3. 46 %

  4. 1000 %


Correct Option: C

The law of equipartition of energy was given by :

  1. Claussius

  2. Maxwell

  3. Boltzmann

  4. Carnot


Correct Option: A
Explanation:

The law is given by Claussius which states that for any dynamical system in a thermal equilibrium, the total energy is equally divided among the degrees of freedom.

The value of $\gamma$ for gas X is 1.66, then x is :

  1. Ne

  2. O$ _3$

  3. N$ _2$

  4. H$ _2$


Correct Option: A
Explanation:

Given that value of gamma is 1.66 i.e $\dfrac{5}{3}$ which implies that it is a monoatomic gas, and Neon (Ne) is the only monoatomic gas among the given options.

Gas exerts pressure on the walls of container because the molecules-

  1. Are loosing the kinetic energy

  2. Are getting stuck to the walls

  3. Are transferring their momentum to walls

  4. Are accelerated toward walls.


Correct Option: C
Explanation:

Gas molecules are in random motion having some momentum and while colliding with the walls they transfer their momentum to the walls and this collective transfer of momentum from all the molecules to the walls appears as pressure exerted by gas on the container wall.

A man is climbing up a spiral type staircase. His degrees of freedom are :

  1. 1

  2. 2

  3. 3

  4. more than 3


Correct Option: C
Explanation:

There will be three degrees of freedom. Two are along x-direction and y-directions due to translation and the last degree of freedom due to angular rotation as the  man climbs up.

Hence, Option C is correct.

A system consists of N particles, which have independent K relations among one another. The number of degrees of freedom of the system is given by :

  1. 3 NK

  2. N/3K

  3. 3 N/K

  4. 3N - K


Correct Option: D
Explanation:

For a system of N particles having K independent relations among them, the degrees of freedom of the system is given by 3N-K. 3N  is due to three degrees of freedom associated with each particle if all the particles are independent of each other (i.e K=0) and due to K relation among them, degrees of freedom reduces to 3N-K

Hence, Option D is correct.

The correct relation connecting the universal gas constant (R), Avogadro number N$ _A$ and Boltzmann constant (K) is :

  1. $R = NK^2$

  2. $K = NR$

  3. $N = RK$

  4. $R=NK$


Correct Option: D
Explanation:

Units of R, N and K are $ Joule \times mole^{-1} \times K^{-1} $, $ mole^{-1} $ and $Joule \times  K^{-1}$
So $ R = N \times K $

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. $mv^2$


Correct Option: C

The number of degrees of freedom for each atom of a monoatomic gas is :

  1. 3

  2. 5

  3. 6

  4. 1


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:

  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


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.

  1. True

  2. False


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 

  1. 3

  2. 4

  3. 5

  4. 6


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 :

  1. translatory

  2. vibratory

  3. rotatory

  4. none of these


Correct Option: A
Explanation:

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

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


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 :-

  1. $1500\ J$

  2. $15633\ cal$

  3. $1500\ kcal$

  4. $1500\ erg$


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?

  1. 7

  2. $5$.

  3. 6

  4. 8


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

  1. 3739.662 J 

  2. 1500 calorie

  3. 1500 kilo calorie

  4. 1500 erg.


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

  1. $0.25\ A^{o}$

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

  3. $1.5\ A^{o}$

  4. $2\ A^{o}$


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 :

  1. $92 J $

  2. $184 J $

  3. $ 276 J $

  4. $46 J $


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.

  1. $n = 7$

  2. $n = 3$

  3. $n = 5$

  4. $n = 2$


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?

  1. $8$

  2. $4$

  3. $2$

  4. $6$


Correct Option: A

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

  1. $6$

  2. $4$

  3. $2$

  4. $8$


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 

  1. $2.15$

  2. $3.75$

  3. $5.25$

  4. $6.35$


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:

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

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

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

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


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

  1. 930 m/s

  2. 541 m/s

  3. 498 m/s

  4. None of these


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.

  1. $2$ grams/mole

  2. $16$ grams/mole

  3. $4$ grams/mole

  4. $32$ grams/mole


Correct Option: C

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.

  1. 2.15

  2. 3.75

  3. 5.25

  4. 6.35


Correct Option: B

When an ideal monoatomic  gas is heated zt constant pressure , which of the following may be true

  1. $\dfrac {dU}{dQ} = \frac {3}{5}$

  2. $\dfrac {dW}{dQ} = \frac {2]}{5}$

  3. $\dfrac {dU}{dQ} = \frac {4}{5}$

  4. $dW + dU = dQ $


Correct Option: D

If $\gamma $ be the ration of specific heats of a perfect gas, the number of degree of freedom of a molecule of the gas is:

  1. $\dfrac{{25}}{2}\left( {\gamma - 1} \right)$

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

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

  4. $\dfrac{9}{2}(\gamma - 1)$


Correct Option: C

On increasing temperature of the reacting system by $10$ degrees the rate of reaction almost doubles. The most appropriate reason for this is

  1. collision frequency increases

  2. activation energy decreases by increases in temperatuer

  3. the fraction of molecules having energy equal to threshold energy or more increase

  4. the value of threshold energy decreases


Correct Option: A

n moles of an ideal monoatomic gas undergoes an isothermal expansion at temperature T during which its volume becomes 4 times. The work done on the gas and change in internal energy of the gas respectively is

  1. n RT Ln 4,0

    • n RT Ln 4,0
  2. n RT Ln 4 $\frac { 3 n R T } { 2 }$

    • n RT Ln 4, $\frac { 3 n R T } { 2 }$

Correct Option: A
Explanation:

$\begin{array}{l} w=nRT\, \, \, \ln { \left( { \frac { { 4v } }{ v }  } \right)  }  \ =nRT\, \, \ln { 4 } \, \, \, \, \, \, & \, \, \, \Delta u=0 \end{array}$

$\therefore$ Option $A$ is correct.

A mixture Of $n _ { 2 }$ moles of mono atomic gas and $n _ { 2 }$ moles of diatomic gas has $\frac { C _ { p } } { C _ { V } } = y = 1.5$

  1. $n _ { 1 } = n _ { 2 }$

  2. $2 n _ { 1 } = n _ { 2 }$

  3. $n _ { 1 } = 2 n _ { 2 }$

  4. $2 n _ { 1 } = 3 n _ { 2 }$


Correct Option: A
Explanation:

$\begin{array}{l} As\, { y _{ mix } }=\dfrac { { { C _{ { p _{ mix } } } } } }{ { { C _{ { v _{ mix } } } } } } where\, { C _{ { p _{ mix } } } }=\dfrac { { { n _{ 1 } }{ C _{ { p _{ 1 } } } }+{ n _{ 2 } }{ C _{ { p _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ and\, \, { C _{ { v _{ mix } } } }=\dfrac { { { n _{ 1 } }{ C _{ { v _{ 1 } } } }+{ n _{ 2 } }{ C _{ { v _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ So,\, \, { y _{ mix } }=\dfrac { { { n _{ 1 } }{ C _{ { p _{ 1 } } } }+{ n _{ 2 } }{ C _{ { p _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ Given\, ,\, for\, monoatomic\, { C _{ p } },\dfrac { 5 }{ 2 } R\, and\, { C _{ { v _{ 1 } } } }=\dfrac { 3 }{ 2 } R \ For\, diatomic\, { C _{ { p _{ 2 } } } }=\dfrac { { 7R } }{ 2 } \, and\, { C _{ { v _{ 2 } } } }=\dfrac { 5 }{ 2 } R \ { y _{ mix } }=\dfrac { { { n _{ 1 } }\times \dfrac { 5 }{ 2 } R+{ n _{ 2 } }\times \dfrac { 7 }{ 2 } R } }{ { { n _{ 1 } }\times \dfrac { 3 }{ 2 } R+{ n _{ 2 } }\times \dfrac { 5 }{ 2 } R } } =\dfrac { { 5{ n _{ 1 } }+7{ n _{ 2 } } } }{ { 3{ n _{ 1 } }+5{ n _{ 2 } } } } =\dfrac { 3 }{ 2 }  \ 10{ n _{ 1 } }+14{ n _{ 2 } }=9{ n _{ 1 } }+15{ n _{ 2 } } \ { n _{ 1 } }={ n _{ 2 } } \ Hence, \ option\, \, A\, \, is\, correct\, \, answer. \end{array}$

70 calorie of heat required to rise the temperature of 2 mole of an ideal gas at constant pressure from ${30^o}$C to ${35^o}$C. The degrees of freedom of the gas molecule are,,

  1. 3

  2. 5

  3. 6

  4. 7


Correct Option: C

Three perfect gases at absolute temperatures $T _1, T _2$ and $T _3$ are mixed. The masses of their molecules are $m _1, m _2$ and $m _3$ and the number of molecules are $n _1, n _2$ and $n _3$ respectively. Assuming no loss of energy, the final tempreture of the mixture is 

  1. $\dfrac{T _1 + T _2 + T _3}{3}$

  2. $\dfrac{n _1T _1 + n _2T _2 + T _3 T _3}{n _1 + n _2 + n _3}$

  3. $\dfrac{n _1T _1^2 + n _2T _2^2 + n _3 T _3^2}{n _1 T _1 + n _2 T _2 + n _3 T _3}$

  4. $\dfrac{n _1^2T _1^2 + n _2^2T _2^2 + n _3^2 T _3^2}{n _1 T _1 + n _2 T _2 + n _3 T _3}$


Correct Option: B

The law of equipartition of energy is applicable to the system whose constituents are :

  1. in random motion

  2. in orderly motion

  3. at rest

  4. moving with constant speed


Correct Option: A
Explanation:
The original idea of equipartition  is to assign a average kinetic energy to each degree of freedom if we take transnational case only. So if there would be ordered motion instead of random motion then there is no need of equilibrium theorem  as it would be useless to take about averages as they all are in ordered motion.
Hence, the answer is in random motion.

The heat capacity at constant volume of a sampleof 192 g of gas in a container of volume 80$\mathrm { L }$ at atemperature of $402 ^ { \circ } \mathrm { C }$ and at a pressure of$4.2 \times 10 ^ { 5 } \mathrm { Pa }$ is 124.5$\mathrm { JK }$ . The number of thedegrees of freedom of the gas molecules is 

  1. 3

  2. 5

  3. 7

  4. 6


Correct Option: A

The kinetic energy associated with per degree of freedom of a molecule is

  1. $\dfrac { 1 }{ 2 } M^{ 2 } _{ rms }$

  2. $kT$

  3. $kT/2$

  4. $3 kT/2$


Correct Option: D

Statement -1 : The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume.
and
Statement -2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

  1. Statement - 1 is True, Statement -2 is True, Statement -2 is a correct explanation for Statements-1

  2. Statement - 1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statements-1

  3. Statement - 1 is True, Statement -2 is False

  4. Statement - 1 is False, Statement -2 is True


Correct Option: A

The mass of glucose that should be dissolved in 100 g of water in order to produce same lowering of vapour pressure as is produced by dissolving 1 g of urea (mol. Mass = 60) in 50 g of water is : (Assume dilute solution in both cases) 

  1. 1 g

  2. 2 g

  3. 6 g

  4. 12 g


Correct Option: C

In a process $PT=Constant$, if molar heat capacity of a gas is $C=37.35J/mol=K$, then find the number of degrees of freedom of molecules in the gas.

  1. $n=10$

  2. $n=5$

  3. $n=6$

  4. $n=7$


Correct Option: B

The degree of freedom per molecule of a gas is $3$. The heat absorbed by the gas at constant pressure is $150\,J$. Then increase in internal energy is 

  1. $90\,J$

  2. $50\,J$

  3. $120\,J$

  4. $30\,J$


Correct Option: B

How many degrees of freedom are associated with 2grams of He at NTP?

  1. 3

  2. $3.01\times10^{23}$

  3. $9.03\times10^{23}$

  4. 6


Correct Option: C
Explanation:

Moles of He =$\displaystyle\ \frac{2}{4}$ = $\displaystyle\ \frac{1}{2}$
Molecules = $\displaystyle\ \frac{1}{2}\times6.02\times10^{23}$ = $3.01\times10^{23}$
As there are 3 degrees of freedom corresponding of 1 molecule of a monatomic gas.
$\therefore$ Total degrees of freedom = $3\times3.01\times10^{23}$
$= 9.03\times10^{23}$

At ordinary temperatures, the molecules of a diatomic gas have only translational and rotational kinetic energies. At high temperatures, they may also have vibrational energy. As a result of this compared to lower temperatures, a diatomic gas at higher temperatures will have-

  1. lower molar heat capacity

  2. higher molar heat capacity

  3. lower isothermal compressibility

  4. higher isothermal compressibility


Correct Option: B
Explanation:

Vibrational energy involves additional degrees of freedom. Thus the degrees of freedom for a diatomic gas increases at higher temperatures.

Molar heat capacity is proportional to the number of degrees of freedom of the gas.
Thus the molar heat capacity also increases for a diatomic gas at higher temperatures.

When the temperature is increased from 0$^o$C to 273$^o$C, in what ratio the average kinetic energy of molecules changes?

  1. 1

  2. 5

  3. 4

  4. 2


Correct Option: D
Explanation:

Average K.E. $= \displaystyle \frac{3}{2}$ RT
At 0$^o$C, average K.E. $= \displaystyle \frac{3}{2} \times R \times 273$
                                           $[T = (0 + 273) K]$
At 273$^o$C,    
average K.E. $= \displaystyle \frac{3}{2} \times R \times (273 + 273)$
$= \displaystyle \frac{3}{2} \times R \times 2 \times 273$
$\therefore $ Ratio = 2

When x amount of heat is given to a gas at constant pressure, it performs $\displaystyle \frac{x}{3}$ amount of work. The average number of degrees of freedom per molecule of the gas is-

  1. 3

  2. 4

  3. 5

  4. 6


Correct Option: B
Explanation:

$\displaystyle \frac{W}{Q}=\frac{P\Delta V}{nC _{P}\Delta T}=\frac{nR\Delta T}{nC _{P}\Delta T}=\frac{x/3}{x}$  (standard result)



$\displaystyle \Rightarrow C _{P}=3R=\left ( \frac{f}{2}+1 \right )R\Rightarrow f=4$

The mean kinetic energy of a gas molecule is proportional to 

  1. $\displaystyle \sqrt { T } $

  2. $\displaystyle { T }^{ 3 }$

  3. $\displaystyle T$

  4. None of the above


Correct Option: C
Explanation:

The average kinetic energy of gas molecules is directly proportional to absolute temperature only; this implies that all molecular motion ceases if the temperature is reduced to absolute zero.
Hence, option C is correct.

The degrees of freedom of a diatomic gas at normal temperature is

  1. 3

  2. 4

  3. 5

  4. 6


Correct Option: C
Explanation:

In three-dimensional space, three degrees of freedom are associated with the movement of a particle. A diatomic  gas molecule thus has 6 degrees of freedom. This set may be decomposed in terms of translations, rotations, and vibrations of the molecule. The center of mass motion of the entire molecule accounts for 3 degrees of freedom. In addition, the molecule has two rotational degrees of motion and one vibrational mode The rotations occur around the two axes perpendicular to the line between the two atoms. The rotation around the atom-atom bond is not a physical rotation. At normal temp,  vibration is not possible. Hence, the total no of degrees of freedom is $f= 3+ 2=5$

The number of vibrational degrees of freedom for a $CO _2$ molecule is

  1. 4

  2. 5

  3. 6

  4. 9


Correct Option: A
Explanation:

$Answer:-$ A

There are always $3N$ total independent degrees of freedom for a molecule, where $N$ is the number of atoms. These come about because when each atom moves, it has three independent degrees of freedom: its position in each of the $x, y, z$ directions.

Now, having independent degrees of freedom for each atom isn't all that useful. Instead, we can make combinations of different degrees of freedom. The important thing when doing so is that the number of independent degrees of freedom are preserved: it's just that what a particular degree of freedom does to the atoms changes.

The standard breakdown of degrees of freedom subtracts out global movement in each of the three directions. So you have $3N$ total degrees of freedom, but you can set aside $3$ of them as translation of the whole molecule in each of the $x, y, z$ directions, leaving $(3N-3)$ degrees of freedom.

Likewise, it's standard to subtract out the whole molecule rotation. For most larger molecules, there's three different degrees of rotational freedom: rotation around each of the $x, y, z$ directions. But for linear molecules like $CO _2$, one of those rotations (around the axis of the molecule) doesn't actually change the position of the atoms. Therefore it's not a "degree of freedom" which counts against the $3N$ total. So while for non-linear molecules there are $(3N-3-3) = (3N-6)$ degrees of freedom which are independent from the global rotational and translational ones, for linear molecules there are $(3N-3-2) = (3N-5)$ degrees of freedom which are independent from the global rotational and translational ones. -- A quick clarification. The reason why we ignore this rotation is not because the center of mass doesn't move. The center of mass doesn't move for $any$ of the global rotations: in the typical assignment of degrees of freedom the axis of rotation goes through the center of mass. Instead, the reason the rotation is ignored is that $none$ of the atoms move due to the "rotation".

So since $CO _2$ has three atoms and is linear, it has ($3\times3 - 5 = 4 $)degrees of freedom which are independent of the global rotation and translation. We call these the vibrational modes.

For polyatomic molecules having 'f' vibrational modes, the ratio of two specific heats, $\dfrac{C _p}{C _v}$ is ............

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

  2. $\dfrac{2+f}{3+f}$

  3. $\dfrac{4+f}{3+f}$

  4. $\dfrac{5+f}{4+f}$


Correct Option: C
Explanation:

By the law of equipartition of energy, for one mole of polyatomic gas
$C _p=(4+f)R \ and \ C _v=(3+f)R$
$\therefore \dfrac{C _p}{C _v}=\dfrac{(4+f)R}{(3+f)R}=\dfrac{(4+f)}{(3+f)}$

If for a gas $\dfrac{R}{C _V}=0.67$, this gas is made up of molecules which are.

  1. Monatomic

  2. Diatomic

  3. Polyatomic

  4. Mixture of diatomic and polyatomic molecules


Correct Option: A
Explanation:

For a gas, we know $\dfrac{R}{C _V}=\gamma -1$
or $0.67=\gamma -1$ or, $\gamma =1.67$
Hence the gas is monatomic.

The average degree of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is

  1. 75 J

  2. 100 J

  3. 150 J

  4. 125 J


Correct Option: B
Explanation:

$\Delta u=\dfrac{f}{2}RT=3RT$

$\Delta w=nR\Delta T=25.5$
$\Delta Q=\Delta V+\Delta W$
$=3nR\Delta T+nR\Delta T=4nR\Delta T$
$=4\times 25=100\ J$

A gas performs Q work when it expand at constant pressure. During this process heat absorbed by the gas is 4Q. The average number of degrees of freedom for the gas is:

  1. 5

  2. 6

  3. 4

  4. 3.5


Correct Option: B

N moles of an ideal diatomic gas is contained in a cylinder at temperature $T.$ On supplying some heat to cylinder, $N/3$ moles of gas disassociated into atoms while temperature remains constant. Heat supplied to the gas is

  1. $\dfrac {NRT}{3}$

  2. $\dfrac {5NRT}{2}$

  3. $\dfrac {8NRT}{3}$

  4. $\dfrac {NRT}{6}$


Correct Option: B

The heat capacity at constant volume of a sample of a monoatomic gas is $35\ J/K$. Find the number of moles.

  1. $12.81 \ \ mol  $

  2. $21.81 \ \ mol  $

  3. $4.81 \ \ mol  $

  4. $2.81 \ \ mol  $


Correct Option: D
Explanation:

For monoatomic gas, degrees of freedom is 3. 


Since ${ C } _{ V }=\dfrac { f }{ 2 } nR$

Hence, $35=\dfrac { 3 }{ 2 } n(8.314)$

$n=\dfrac { 70 }{ 3\times 8.314 } =2.81mol$

Answer is $2.81mol.$

Relation between pressure ($P$) and energy density ($E$) of an ideal gas is-

  1. $P=2/3E$

  2. $P=3/2E$

  3. $P=3/5E$

  4. $P=E$


Correct Option: A
Explanation:
Kinetic energy $=\dfrac{1}{2}{ MV } _{ rms }$
$\Rightarrow \dfrac{1}{2}M\left( \dfrac { 3RT }{ M }  \right) $        $[M=$ molar mass,$ { V } _{ rms }=\sqrt { \dfrac { 3KT }{ { m } }  } =\sqrt { \dfrac { 3RT }{ M }  } ]$
$=\dfrac{3}{2}RT$
$\Rightarrow K.E=\dfrac{3}{2}PV$          $[PV=RT]$
$\Rightarrow \dfrac{K.E}{V}=\dfrac{3}{2}P$
$\Rightarrow E=\dfrac{3P}{2}$        $E=$ Energy density.
Hence, the answer is $P=\dfrac{2}{3}E.$

A vessel of volume $0.3 \ { { m }^{ 3 } }$ contains Helium at $20.0$. The average kinetic energy per molecule for the gas is:

  1. $6.07\times { 10 }^{ -21 }J$

  2. $7.3\times { 10 }^{ 3 }J$

  3. $14.6\times { 10 }^{ 3 }J$

  4. $12.14\times { 10 }^{ -21 }J$


Correct Option: A
Explanation:
Given temperature of gas $=20°C$
                                            $=293K$
$\Rightarrow$ Average translational kinetic energy $=\dfrac{3}{2}KT$
                                                                   $=\dfrac{3}{2}\times1.38\times10^{-23}\times293$
                                                                   $=6.07\times10^{-21}J$
Hence, the answer is $6.07\times10^{-21}J.$
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