Tag: behaviour of perfect gas and kinetic theory of gases

Questions Related to behaviour of perfect gas and kinetic theory of gases

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

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 

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

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

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Correct Option: B

The law of equipartition of energy is applicable to the system whose constituents 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. in random motion

  2. in orderly motion

  3. at rest

  4. moving with constant speed

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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 

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. 7

  4. 6

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Correct Option: A

The kinetic energy associated with per degree of freedom of a molecule 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. $\dfrac { 1 }{ 2 } M^{ 2 } _{ rms }$

  2. $kT$

  3. $kT/2$

  4. $3 kT/2$

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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.

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. 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

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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) 

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 g

  2. 2 g

  3. 6 g

  4. 12 g

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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.

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. $n=10$

  2. $n=5$

  3. $n=6$

  4. $n=7$

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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 

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. $90\,J$

  2. $50\,J$

  3. $120\,J$

  4. $30\,J$

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Correct Option: B