Q 1 :

If the collision frequency of hydrogen molecules in a closed chamber at 27°C is Z, then the collision frequency of the same system at 127°C is          [2024]

  • 34Z

     

  • 32Z

     

  • 43Z

     

  • 23Z

     

(4)    

         Collision frequency Tf2f1=T2T1

          f2f1=400300f2=23f1=23Z

         f2=43=f1=23Z

 



Q 2 :

If n is the number density and d is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e. mean free path) is represented by                 [2024]

  • 12nπd2

     

  • 2nπd2

     

  • 12nπd2

     

  • 12n2π2d2 

     

(1)   

        The mean free path, λ=12nπd2

         n= number of molecules per unit volume,

         and d= diameter of the molecule

 



Q 3 :

A sample contains mixture of helium and oxygen gas. The ratio of root mean square speed of helium and oxygen in the sample, is                       [2024]

  • 132

     

  • 122

     

  • 14

     

  • 221

     

(4)   

          Vrms=3RTM

          VHeVO2=MO2MHe=324=8=22

 



Q 4 :

Given below are two statements:

 

Statement (I): The mean free path of gas molecules is inversely proportional to square of molecular diameter.

 

Statement (II): Average kinetic energy of gas molecules is directly proportional to absolute temperature of gas.

 

In the light of the above statements, choose the correct answer from the options given below:                   [2024]

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement II are true

     

  • Statement I is true but Statement II is false

     

  • Both Statement I and Statement II are false

     

(2)   

        Statement 1: Mean free path

         λ=12πd2n using d=2r

        λ=142πr2nλ1r2

        Statement 2:

        E=32kTET

 



Q 5 :

The temperature of a gas is – 78°C and the average translational kinetic energy of its molecules is K. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes 2K is              [2024]

  • – 39°C

     

  • 117°C

     

  • 127°C

     

  • – 78°C

     

(2)   

       Average translational kinetic energy

        E=32KBT

       At – 78°C = 273 – 78 = 195 K

       K=32KB×(195K)                                            ...(i)

      Now when kinetic energy becomes 2K, the temperature is T K.

        2K=32KB×(T K)                                    ...(ii)

      From equations (i) and (ii)

      K2K=195T

         T = 390 K

          in °C, 390 – 273 = 117°C

 



Q 6 :

Two vessels A and B are of the same size and are at same temperature. A contains 1 g of hydrogen and B contains 1 g of oxygen. PA and PB are the pressures of the gases in A and B respectively, then PA/PB is                   [2024]

  • 16

     

  • 8

     

  • 4

     

  • 32

     

(1)    

          Ideal gas equation, PAVA=nARTA

          and PBVB=nBRTB

         from (i) and (ii), PAVAPBVB=nARTAnBRTB

        Given VA=VB and TA=TB

         PAPB=nAnB=1/21/32=16

 



Q 7 :

The temperature of a gas having 2.0×1025 molecules per cubic meter at 1.38 atm is

(Given, k=1.38×10-23JK-1)                    [2024]

  • 300 K

     

  • 500 K

     

  • 100 K

     

  • 200 K

     

(2)   

        ldeal gas equation,

         PV=NNART

         N = Total no. of molecules

         P=NVkT

         1.38×1.01×105=2×1025×1.38×10-23×T

         T=1.01×1032500K

 



Q 8 :

At what temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C ?                   [2024]

  • 80 K

     

  • – 73 K

     

  • 4 K

     

  • 20 K

     

(4) 

        Vrms=3RTM(Vrms)H2=(Vrms)O2

         T1M1=T2M2

          T1=M1M2T2=232(273+47)

          =232×320=20K

 



Q 9 :

If the root mean square velocity of hydrogen molecule at a given temperature and pressure is 2 km/s, the root mean square velocity of oxygen at the same condition in km/s is                 [2024]

  • 2.0

     

  • 0.5

     

  • 1.5

     

  • 1.0

     

(2)   

             (Vrms)H2(Vrms)O2=3RTMH23RTMO2

           2(Vrms)O2=MO2MH2=322

           (Vrms)O2=0.5

 



Q 10 :

P-T diagram of an ideal gas having three different densities ρ1,ρ2,ρ3 (in three different cases) is shown in the figure. Which of the following is correct?           [2024]

  • ρ1>ρ2

     

  • ρ2>ρ3

     

  • ρ1<ρ2

     

  • ρ1=ρ2=ρ3

     

(1)

Ideal Gas Equation, PV=nRT=mMRT

P=(mV)RTM=ρ(RTM) (Where m is the mass of gas and M is the molecular mass of gas)

For the same temperature, P1>P2>P3    ρ1>ρ2>ρ3

So, the correct answer is (1)



Q 11 :

The given figure represents two isobaric processes for the same mass of an ideal gas, then            [2024]

  • P1>P2

     

  • P2>P1

     

  • P1=P2

     

  • P2<P1

     

(1)

PV=nRTV=(nRP)T

Slope=nRPSlope1P

(Slope)2>(Slope)1P2<P1

 



Q 12 :

For a particular ideal gas which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?          [2025]

  •  

  •  

  •  

  •  

(1)

Mean squared velocity =(vms)2=3RTM i.e., (mean squared velocity)  (Temperature). Hence, graph is a straight line.



Q 13 :

The ratio of vapour densities of two gases at the same temperature is 425, then the ratio of r.m.s. velocities will be:          [2025]

  • 254

     

  • 25

     

  • 52

     

  • 425

     

(3)

vrms=3RTM

v1v2=M2M1=254=52



Q 14 :

Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4% when heated by 1°C. Its initial temperature must be:          [2025]

  • 25°C

     

  • 2500 K

     

  • 250 K

     

  • 250°C

     

(3)

Isochoric process, P  T

PT=constant  PP=TT

0.4100=1T  T=1000.4=250 K



Q 15 :

The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are 3×107 m and 600 m/s, respectively. Find the frequency of its collisions.         [2025]

  • 2×1010/s

     

  • 9×105/s

     

  • 2×109/s

     

  • 5×102/s

     

(3)

Frequency of collision =average speedmean free path

                                            =6003×107=2×109/s



Q 16 :

There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K, at steady state the pressure in the vessels will be (in kPa).          [2025]

  • 4.4

     

  • 6

     

  • 24

     

  • 18

     

(2)

Number of masses will remain constant

n1+n2=nf

P1V1RT1+P2V2RT2=PfVfRTf

8×2VR×1000+7×VR×500=Pf(3V)R×600

161000+141000=Pf200

301000=Pf200  Pf=6 kPa



Q 17 :

A container of fixed volume contains a gas at 27°C. To double the pressure of the gas, the temperature of gas should be raised to ________ °C.          [2025]



(327)

For V = constant, P1T1=P2T2

P300=2PT2

T2=600 K=327°C



Q 18 :

Given below are two statements:                           [2023]

Statement I: The temperature of a gas is −73°C. When the gas is heated to 527°C, the root mean square speed of the molecules is doubled.

Statement II: The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.

In the light of the above statements, choose the correct answer from the options given below:

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement II are true

     

  • Both Statement I and Statement II are false

     

  • Statement I is true but Statement II is false

     

(4)

Statement-I

T1=-73°C=200 K

T2=527°C=800 K

V1V2=3RT1M3RT2M=T1T2=200800=12

V2=2V1  (True)

Statement-II

KE=32PV



Q 19 :

The root mean square velocity of molecules of gas is          [2023]

  • inversely proportional to square root of temperature 1T

     

  • proportional to square root of temperature T

     

  • proportional to square of temperature T2

     

  • proportional to temperature T

     

(2)

Vrms=3RTM

so VrmsT



Q 20 :

A bicycle tyre is filled with air having pressure of 270 kPa at 27°C. The approximate pressure of the air in the tyre when the temperature increases to 36°C is       [2023]

  • 270 kPa

     

  • 262 kPa

     

  • 278 kPa

     

  • 360 kPa

     

(3)

Taking volume constant: P1T1=P2T2

P2=P1T1×T2=270×309300=278 kPa



Q 21 :

At 300 K, the rms speed of oxygen molecules is α+5α times to that of its average speed in the gas. Then, the value of α will be (use π=227)            [2023]

  • 32

     

  • 28

     

  • 24

     

  • 27

     

(2)

3RTM=α+5α8πRTM

3=α+5α8π  α=28



Q 22 :

Heat energy of 735 J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but does not oscillate. The increase in the internal energy of the gas will be               [2023]

  • 525 J

     

  • 441 J

     

  • 572 J

     

  • 735 J

     

(2)

ΔQ=nCpΔT=735 J

5nRΔT2=735 J

ΔU=nCvΔT=32(nRΔT)=32×25×735=441 J



Q 23 :

The average kinetic energy of a molecule of the gas is                [2023]

  • proportional to absolute temperature

     

  • proportional to volume

     

  • proportional to pressure

     

  • dependent on the nature of the gas

     

(1)

Translational K.E. on average of a molecule is 32kT, which is independent of nature, pressure, and volume.



Q 24 :

The temperature of an ideal gas is increased from 200 K to 800 K. If the r.m.s. speed of the gas at 200 K is v0, then the r.m.s. speed of the gas at 800 K will be       [2023]

  • v0

     

  • 4v0

     

  • v04

     

  • 2v0

     

(4)

Vrms=3RTMVrmsT

Increasing the temperature 4 times, the rms speed gets doubled.



Q 25 :

An air bubble of volume 1 cm3 rises from the bottom of a lake 40 m deep to the surface at a temperature of 12°C. The atmospheric pressure is 1×105 Pa, the density of water is 1000 kg/m3 and g=10 m/s2. There is no difference of the temperature of water at the depth of 40 m and on the surface. The volume of the air bubble when it reaches the surface will be                [2023]

  • cm3

     

  • cm3

     

  • cm3

     

  • cm3

     

(3)

P=P0+ρgh=105 Pa+103×10×40=5×105 Pa

At T is constant, PV=P0V0

5×105 Pa×1 cm3=105 Pa×V0V0=5 cm3



Q 26 :

Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and the third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square speed (vrms) and choose the correct answer from the options given below:                    [2023]

  • vrms(mono)=vrms(dia)=vrms(poly)

     

  • vrms(mono)>vrms(dia)>vrms(poly)

     

  • vrms(dia)<vrms(poly)<vrms(mono)

     

  • vrms(mono)<vrms(dia)<vrms(poly)

     

(2)

vrms(mono)=3RT4×10-3

vrms(dia)=3RT71×10-3

vrms(poly)=3RT146×10-3

So the correct relation is

vrms(mono)>vrms(dia)>vrms(poly)



Q 27 :

The root mean square speed of molecules of nitrogen gas at 27°C is approximately: 

(Given mass of a nitrogen molecule =4.6×10-26 kg and take Boltzmann constant kB=1.4×10-23 JK-1)          [2023]

  • 523 m/s

     

  • 1260 m/s

     

  • 91 m/s

     

  • 27.4 m/s

     

(1)

Vrms=3kBTm=3×1.4×10-23×3004.6×10-26=523 m/s



Q 28 :

If the r.m.s. speed of chlorine molecule is 490 m/s at 27°C, the r.m.s. speed of argon molecules at the same temperature will be (Atomic mass of argon = 39.9 u, molecular mass of chlorine = 70.9 u)             [2023]

  • 751.7 m/s

     

  • 451.7 m/s

     

  • 651.7 m/s

     

  • 551.7 m/s

     

(3)

Vrms=3RTM, vArvCl=MClMAr

 VAr=1.33×490=651.7 m/s



Q 29 :

The r.m.s. speed of oxygen molecule in a vessel at a particular temperature is (1+5x)1/2v, where v is the average speed of the molecule. The value of x will be: (Take π=227)          [2023]

  • 28

     

  • 27

     

  • 8

     

  • 4

     

(1)

3RTM=(1+5x)128RTπM

 3×227×8=1+5xx=28



Q 30 :

The mean free path of molecules of a certain gas at STP is 1500 d, where d is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K is approximately                    [2023]

  • 1098d

     

  • 2049d

     

  • 750d

     

  • 1500d

     

(2)

Mean free path, λ=RT2πd2NAP

λT

1500dλ=273373  λ=2049d