Tag: law of equipartition of energy

Questions Related to law of equipartition of energy

The law of equipartition of energy is applicable to the system whose constituents are :

  1. in random motion

  2. in orderly motion

  3. at rest

  4. moving with constant speed


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 

  1. 3

  2. 5

  3. 7

  4. 6


Correct Option: A

The kinetic energy associated with per degree of freedom of a molecule is

  1. $\dfrac { 1 }{ 2 } M^{ 2 } _{ rms }$

  2. $kT$

  3. $kT/2$

  4. $3 kT/2$


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.

  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


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) 

  1. 1 g

  2. 2 g

  3. 6 g

  4. 12 g


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.

  1. $n=10$

  2. $n=5$

  3. $n=6$

  4. $n=7$


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 

  1. $90\,J$

  2. $50\,J$

  3. $120\,J$

  4. $30\,J$


Correct Option: B

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


Correct Option: C
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 = $3\times3.01\times10^{23}$
$= 9.03\times10^{23}$

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


Correct Option: B
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.

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


Correct Option: D
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