Tag: behaviour of perfect gas and kinetic theory of gases

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

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

How many degrees of freedom are associated with 2grams of He at NTP?

  1. 3

  2. $3.01\times10^{23}$

  3. $9.03\times10^{23}$

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

Moles of He =$\displaystyle\ \frac{2}{4}$ = $\displaystyle\ \frac{1}{2}$
Molecules = $\displaystyle\ \frac{1}{2}\times6.02\times10^{23}$ = $3.01\times10^{23}$
As there are 3 degrees of freedom corresponding of 1 molecule of a monatomic gas.
$\therefore$ Total degrees of freedom = $3\times3.01\times10^{23}$
$= 9.03\times10^{23}$

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

At ordinary temperatures, the molecules of a diatomic gas have only translational and rotational kinetic energies. At high temperatures, they may also have vibrational energy. As a result of this compared to lower temperatures, a diatomic gas at higher temperatures will have-

  1. lower molar heat capacity

  2. higher molar heat capacity

  3. lower isothermal compressibility

  4. higher isothermal compressibility
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vibrational energy involves additional degrees of freedom. Thus the degrees of freedom for a diatomic gas increases at higher temperatures.

Molar heat capacity is proportional to the number of degrees of freedom of the gas.
Thus the molar heat capacity also increases for a diatomic gas at higher temperatures.

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 the temperature is increased from 0$^o$C to 273$^o$C, in what ratio the average kinetic energy of molecules changes?

  1. 1

  2. 5

  3. 4

  4. 2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Average K.E. $= \displaystyle \frac{3}{2}$ RT
At 0$^o$C, average K.E. $= \displaystyle \frac{3}{2} \times R \times 273$
                                           $[T = (0 + 273) K]$
At 273$^o$C,    
average K.E. $= \displaystyle \frac{3}{2} \times R \times (273 + 273)$
$= \displaystyle \frac{3}{2} \times R \times 2 \times 273$
$\therefore $ Ratio = 2

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

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

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