Tag: degree of freedom: law of equipartition of energy

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Multiple choice 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

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 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From First Law: dQ = dU + dW. This is always true for any process. For monatomic gas at constant pressure, dU = (3/2)nRdT and dW = PdV = nRdT, so dQ = (5/2)nRdT. Then dU/dQ = 3/5 and dW/dQ = 2/5, not 4/5. Option D is the fundamental First Law statement.

Multiple choice 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

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. $\dfrac{9}{2}(\gamma - 1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The relationship between the adiabatic index gamma and degrees of freedom f is gamma = 1 + 2/f. Solving for f gives f = 2 / (gamma - 1).

Multiple choice 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

The magnetic monment of a diamagnetic atom is 

  1. Much greater than one

  2. 1

  3. Between zero and one

  4. Equal to zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Diamagnetic materials have no permanent magnetic moment; the induced magnetic moment opposes the external magnetic field, resulting in a net magnetic moment of zero in the absence of an external field.

Multiple choice 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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice 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

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

  1. $1$

  2. $2$

  3. $3$

  4. $6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An ant walking on a horizontal surface is constrained to move in a 2D plane (x and y coordinates). However, if the ant is considered a rigid body capable of rotation about its center of mass, it possesses 3 degrees of freedom (2 translational and 1 rotational).

Multiple choice 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

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

  3. n RT Ln 4 $\frac { 3 n R T } { 2 }$
    • n RT Ln 4, $\frac { 3 n R T } { 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice 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

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. $2 n _ { 1 } = 3 n _ { 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice 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

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. $\dfrac{n _1^2T _1^2 + n _2^2T _2^2 + n _3^2 T _3^2}{n _1 T _1 + n _2 T _2 + n _3 T _3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The final temperature of a mixture of non-reacting gases is the weighted average of their temperatures based on the number of moles (or molecules) of each gas, assuming equal degrees of freedom. The formula is T_final = (n1*T1 + n2*T2 + n3*T3) / (n1 + n2 + n3).