Tag: law of equipartition of energy and mean free path

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

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.

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. 2.15

  2. 3.75

  3. 5.25

  4. 6.35

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Correct Option: B

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

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

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

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

  4. $dW + dU = dQ $

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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:

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

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

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

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

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Correct Option: C

The degree of freedom of a diatomic gas at normal temperature 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. $7$

  2. $6$

  3. $5$

  4. $4$

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Correct Option: C

The magnetic monment of a diamagnetic atom 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. Much greater than one

  2. 1

  3. Between zero and one

  4. Equal to zero

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Correct Option: D

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

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Correct Option: A

An ant is walking on the horizontal surface. The number of degree of freedom of ant will be

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. $1$

  2. $2$

  3. $3$

  4. $6$

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Correct Option: C

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

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. n RT Ln 4,0

    • n RT Ln 4,0

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

    • n RT Ln 4, $\frac { 3 n R T } { 2 }$
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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 $\frac { C _ { p } } { C _ { V } } = y = 1.5$

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

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

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

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

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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,,

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. 7

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Correct Option: C