Tag: law of equipartition of energy

Questions Related to law of equipartition of energy

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. $P=E$

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

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