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 x amount of heat is given to a gas at constant pressure, it performs $\displaystyle \frac{x}{3}$ amount of work. The average number of degrees of freedom per molecule of the gas is-

  1. 3

  2. 4

  3. 5

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

$\displaystyle \frac{W}{Q}=\frac{P\Delta V}{nC _{P}\Delta T}=\frac{nR\Delta T}{nC _{P}\Delta T}=\frac{x/3}{x}$  (standard result)



$\displaystyle \Rightarrow C _{P}=3R=\left ( \frac{f}{2}+1 \right )R\Rightarrow f=4$

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 mean kinetic energy of a gas molecule is proportional to 

  1. $\displaystyle \sqrt { T } $

  2. $\displaystyle { T }^{ 3 }$

  3. $\displaystyle T$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The average kinetic energy of gas molecules is directly proportional to absolute temperature only; this implies that all molecular motion ceases if the temperature is reduced to absolute zero.
Hence, option C 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

The degrees of freedom of a diatomic gas at normal temperature is

  1. 3

  2. 4

  3. 5

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

In three-dimensional space, three degrees of freedom are associated with the movement of a particle. A diatomic  gas molecule thus has 6 degrees of freedom. This set may be decomposed in terms of translations, rotations, and vibrations of the molecule. The center of mass motion of the entire molecule accounts for 3 degrees of freedom. In addition, the molecule has two rotational degrees of motion and one vibrational mode The rotations occur around the two axes perpendicular to the line between the two atoms. The rotation around the atom-atom bond is not a physical rotation. At normal temp,  vibration is not possible. Hence, the total no of degrees of freedom is $f= 3+ 2=5$

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 number of vibrational degrees of freedom for a $CO _2$ molecule is

  1. 4

  2. 5

  3. 6

  4. 9
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Answer:-$ A

There are always $3N$ total independent degrees of freedom for a molecule, where $N$ is the number of atoms. These come about because when each atom moves, it has three independent degrees of freedom: its position in each of the $x, y, z$ directions.

Now, having independent degrees of freedom for each atom isn't all that useful. Instead, we can make combinations of different degrees of freedom. The important thing when doing so is that the number of independent degrees of freedom are preserved: it's just that what a particular degree of freedom does to the atoms changes.

The standard breakdown of degrees of freedom subtracts out global movement in each of the three directions. So you have $3N$ total degrees of freedom, but you can set aside $3$ of them as translation of the whole molecule in each of the $x, y, z$ directions, leaving $(3N-3)$ degrees of freedom.

Likewise, it's standard to subtract out the whole molecule rotation. For most larger molecules, there's three different degrees of rotational freedom: rotation around each of the $x, y, z$ directions. But for linear molecules like $CO _2$, one of those rotations (around the axis of the molecule) doesn't actually change the position of the atoms. Therefore it's not a "degree of freedom" which counts against the $3N$ total. So while for non-linear molecules there are $(3N-3-3) = (3N-6)$ degrees of freedom which are independent from the global rotational and translational ones, for linear molecules there are $(3N-3-2) = (3N-5)$ degrees of freedom which are independent from the global rotational and translational ones. -- A quick clarification. The reason why we ignore this rotation is not because the center of mass doesn't move. The center of mass doesn't move for $any$ of the global rotations: in the typical assignment of degrees of freedom the axis of rotation goes through the center of mass. Instead, the reason the rotation is ignored is that $none$ of the atoms move due to the "rotation".

So since $CO _2$ has three atoms and is linear, it has ($3\times3 - 5 = 4 $)degrees of freedom which are independent of the global rotation and translation. We call these the vibrational modes.

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

For polyatomic molecules having 'f' vibrational modes, the ratio of two specific heats, $\dfrac{C _p}{C _v}$ is ............

  1. $\dfrac{1+f}{2+f}$

  2. $\dfrac{2+f}{3+f}$

  3. $\dfrac{4+f}{3+f}$

  4. $\dfrac{5+f}{4+f}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By the law of equipartition of energy, for one mole of polyatomic gas
$C _p=(4+f)R \ and \ C _v=(3+f)R$
$\therefore \dfrac{C _p}{C _v}=\dfrac{(4+f)R}{(3+f)R}=\dfrac{(4+f)}{(3+f)}$

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 for a gas $\dfrac{R}{C _V}=0.67$, this gas is made up of molecules which are.

  1. Monatomic

  2. Diatomic

  3. Polyatomic

  4. Mixture of diatomic and polyatomic molecules
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a gas, we know $\dfrac{R}{C _V}=\gamma -1$
or $0.67=\gamma -1$ or, $\gamma =1.67$
Hence the gas is monatomic.

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 average degree of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is

  1. 75 J

  2. 100 J

  3. 150 J

  4. 125 J
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta u=\dfrac{f}{2}RT=3RT$

$\Delta w=nR\Delta T=25.5$
$\Delta Q=\Delta V+\Delta W$
$=3nR\Delta T+nR\Delta T=4nR\Delta T$
$=4\times 25=100\ J$

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 gas performs Q work when it expand at constant pressure. During this process heat absorbed by the gas is 4Q. The average number of degrees of freedom for the gas is:

  1. 5

  2. 6

  3. 4

  4. 3.5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At constant pressure, Q = n*Cp*deltaT and W = n*R*deltaT. Given Q = 4Q (this seems to be a typo in the prompt, likely Q_heat = 4*W). If Q_heat = 4*W, then Cp*deltaT = 4*R*deltaT, so Cp = 4R. Since Cp = (f/2 + 1)R, then f/2 + 1 = 4, f/2 = 3, f = 6.

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 diatomic gas is contained in a cylinder at temperature $T.$ On supplying some heat to cylinder, $N/3$ moles of gas disassociated into atoms while temperature remains constant. Heat supplied to the gas is

  1. $\dfrac {NRT}{3}$

  2. $\dfrac {5NRT}{2}$

  3. $\dfrac {8NRT}{3}$

  4. $\dfrac {NRT}{6}$
Reveal answer Fill a bubble to check yourself
B 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

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

  1. $12.81 \ \ mol  $

  2. <span>$21.81 \ \ mol &nbsp;$</span>

  3. <span>$4.81 \ \ mol &nbsp;$</span>

  4. <span>$2.81 \ \ mol &nbsp;$</span>
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $2.81mol.$