Q 1 :

The translational degrees of freedom (ft) and rotational degrees of freedom (fr) of CH4 molecule are              [2024]

  • ft=3 and fr=3

     

  • ft=2 and fr=3

     

  • ft=2 and fr=2

     

  • ft=3 and fr=2

     

(1) 

       CH4 is polyatomic and non-linear,

        The degrees of freedom of CH4 are

  • Translational degrees of freedom ft=3
  • Rotational degrees of freedom fr=3


Q 2 :

Energy of 10 non rigid diatomic molecules at temperature T is         [2024]

  • 35 KB T

     

  • 35 RT

     

  • 72 RT

     

  • 70 KB T

     

(1) 

        Degree of freedom of non-rigid diatomic

         MoleculesfRotation+Translation+Vibration

                                                                                                      

                                                        2                         3                         2

        Degree of freedom, f=2+3+2=7

        Energy of one molecule = f2KBT

        energy of 10 molecules

        =10(f2KBT)=10(72KBT)=35KBT

 



Q 3 :

The average kinetic energy of a monoatomic molecule is 0.414eV at temperature

(Use KB=1.38×10-23J/mol-K)                      [2024]

  • 3000 K

     

  • 3200 K

     

  • 1600 K

     

  • 1500 K

     

(2)   

        For monoatomic molecule degree of freedom = 3.

          K=32KBT

         T=0.414×1.6×10-19×23×1.38×10-23=3200K

 



Q 4 :

The total kinetic energy of 1 mole oxygen at 27°C is :

[Use universal gas constant (R) = 8.31J/ mole K]                [2024]

  • 6845.5 J

     

  • 5942.0 J

     

  • 6232.5 J

     

  • 5670.5 J

     

(3)   

            Kinetic energy = f2nR

                                    =52×1×8.31×300J=6232.5J

 



Q 5 :

N moles of a polyatomic gas (f = 6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of N is                      [2024]

  • 4

     

  • 3

     

  • 6

     

  • 2

     

(1) 

        feq=n1f1+n2f2n1+n2

         For diatomic gas feq=5

        5=(N)(6)+(2)(3)N+2

        5N+10=6N+6N=4

 

 



Q 6 :

The parameter that remains the same for molecules of all gases at a given temperature is                   [2024]

  • kinetic energy

     

  • mass

     

  • speed

     

  • Momentum

     

(1)     

        KE=f2kT

          Conceptual

 



Q 7 :

The kinetic energy of translation of the molecules in 50 g of CO2 gas at 17°C is:          [2025]

  • 3986.3 J

     

  • 4102.8 J

     

  • 4205.5 J

     

  • 3582.7 J

     

(2)

Kinetic energy of translation =32nRT

n=50 g44 g=2522mol

T = 17°C = 290 K

 Kinetic energy of translation

(KE)Translational=32(2522)(8.3)(290) J=4102.8 J

 



Q 8 :

The helium and argon are put in the flask at the same room temperature (300 K). The ratio of average kinetic energies (per molecule) of helium and argon is:

(Give: Molar mass of helium = 4 g/mol, Molar mass of argon = 40 g/mol)          [2025]

  • 1 : 10

     

  • 10 : 1

     

  • 1:10

     

  • 1 : 1

     

(4)

K.E =  f2kT

For He and Ar, f=3

KEHeKEAr=11



Q 9 :

According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is     [2023]

  • 32R

     

  • 72R

     

  • 92R

     

  • 52R

     

(2)

Diatomic gas molecules have three translational degrees of freedom, two rotational degrees of freedom, and it is given that it has one vibrational mode. So there are two additional degrees of freedom corresponding to one vibrational mode. Therefore, total degrees of freedom = 7

CV=fR2=7R2



Q 10 :

A flask contains hydrogen and oxygen in the ratio of 2 : 1 by mass at temperature 27°C. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is              [2023]

  • 2 : 1

     

  • 1 : 1

     

  • 1 : 4

     

  • 4 : 1

     

(2)

Kav=52kT

Ratio = 1 : 1



Q 11 :

The number of air molecules per cm3 increased from 3×1019 to 12×1019. The ratio of collision frequency of air molecules before and after the increase in number respectively is           [2023]

  • 0.75

     

  • 0.50

     

  • 0.25

     

  • 1.25

     

(3)

Collision frequency,

f=Vλ=V(12πd2nv)=2πd2vnv

  fn, nv is number density

f1f2=nv1nv2=3×101912×1019=0.25



Q 12 :

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at 27°C is            [2023]

  • 627°C

     

  • 327°C

     

  • 927°C

     

  • 1227°C

     

(2)

Kinetic energy=f2kT,  where T is the absolute temperature.

If K1 is the kinetic energy at 27°C

and K2 is the kinetic energy at new temperature T.

K1K2=T1T212=300T

T=600KT=327°C



Q 13 :

Match List I with List II:                      [2023]

  List I   List II
A. 3 Translational degrees of freedom I. Monoatomic gases
B. 3 Translational, 2 rotational degrees of freedoms II. Polyatomic gases
C. 3 Translational, 2 rotational and 1 vibrational degrees of freedom III. Rigid diatomic gases
D. 3 Translational, 3 rotational and more than one vibrational degrees of freedom IV. Non rigid diatomic gases

Choose the correct answer from the options given below:

  • (A) – (IV), (B) – (III), (C) – (II), (D) – (I)

     

  • (A) – (IV), (B) – (II), (C) – (I), (D) – (III)

     

  • (A) – (I), (B) – (III), (C) – (IV), (D) – (II)

     

  • (A) – (I), (B) – (IV), (C) – (III), (D) – (II)

     

(3)

Factual

Type of gases No. of degrees of freedom
Monoatomic gas 3T
Diatomic + rigid 3T + 2R
Diatomic + non-rigid 3T + 2R + 1V
Polyatomic gas 3T + 3R + more than 1V

T = Translational degree of freedom

R = Rotational degree of freedom

V = Vibrational degree of freedom



Q 14 :

A flask contains Hydrogen and Argon in the ratio 2 : 1 by mass. The temperature of the mixture is 30°C. The ratio of average kinetic energy per molecule of the two gases (Kargon/Khydrogen) is: (Given: Atomic weight of Ar = 39.9)                [2023]

  • 1

     

  • 2

     

  • 39.92

     

  • 39.9

     

(1)

Average kinetic energy per molecule=32kT

KArKH=11



Q 15 :

Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is 32×108s-1 then collision frequency in gas A is____________/s          [2026]

  • 32×108

     

  • 8×108

     

  • 2×108

     

  • 4×108

     

(4)

Collision frequency (Z)=2πd2N8RTπM

Temp, N are same

Zd2M

dA=dB2

MA=4MB

ZAZB=dA2MA×MBdB2=(MBMA)(dAdB)2

=(14)(12)2

ZAZB=12×14=18

ZA=32×1088=4×108s-1