Q 11 :

A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature 50 °C. The gas is now heated to double its temperature. The modified pressure is ____ Pa.               [2026]



(3730)

As per NTA

V=constant

so PT

Ti=50°C=323 K

Tf=100°C=373 K

PfPi=TfTi

Pf3.23 kPa=373323

Pf=3730 Pa

Other Solution:

As volume is constant

 PT

Since T is doubled (must be in Kelvin),so pressure must be doubled

 Pf=2Pi

Pf=2×3.23=6.46 kPa

Pf=6460 Pa



Q 12 :

The mean free path of a molecule of diameter 5×10-10 m at the temperature 41°C and pressure 1.38×105 Pa is given as ______ m.

(Given kB=1.38×10-23 J/K)    [2026]

  • 102×10-8

     

  • 22×10-8

     

  • 2×10-8

     

  • 22×10-10

     

(2)

λ=kBT2πσ2P

=1.38×10-23×(273+41)×1002×3.14×(5×10-10)2×1.38×105

=22×10-8



Q 13 :

An air bubble of volume 2.9cm3 rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17 °C. The volume of the bubble when it reaches the surface, where the water temperature is 27 °C, is ____ cm3.
(g = 10 m/s2, density of water = 103kg/m3,  and 1 atm pressure is 105 Pa)   [2026]

  • 2.0

     

  • 4.2

     

  • 3.0

     

  • 4.5

     

(4)

For an air bubble rising in water, the number of moles remain constant

P1V1T1=P2V2T2

(Patm+ρgh)2.9 cm3290 K =(Patm)V2300

V2=4.5 cm3



Q 14 :

The r.m.s. speed of oxygen molecules at 47°C is equal to that of the hydrogen molecules kept at ______ °C.

(Mass of oxygen molecule / mass of hydrogen molecule = 32/2)     [2026]

  • −235

     

  • −253

     

  • −20

     

  • −100

     

(2)

Vrms=3RTM

VrmsO2=VrmsH2

TO2=273+47=320K

3RTO2MO2=3RTH2MH2

TO2MO2=TH2MH2

32032=TH22

TH2=20K

TH2=-253°C