Tag: law of equipartition of energy and mean free path

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

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

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. $12.81 \ \ mol  $

  2. $21.81 \ \ mol  $

  3. $4.81 \ \ mol  $

  4. $2.81 \ \ mol  $

Show answer
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-

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. $P=2/3E$

  2. $P=3/2E$

  3. $P=3/5E$

  4. $P=E$

Show answer
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:

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

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

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

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

Show answer
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 $6.07\times10^{-21}J.$