A horizontal tube of length l closed at both ends, contains

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 Multiple Choice QuestionsMultiple Choice Questions

51.

22 g of CO2 at 270 C is mixed with 16 g of O2 at 370C. If both gases are considered as ideal, then temperature of mixture is

  • 30.50C

  • 370C

  • 270C

  • 29.30C


52.

At a given temperature the pressure of an ideal of density p is proportional to

  • ρ2

  • ρ

  • 1ρ2

  • 1ρ


53.

A cylinder contains 10kg of gas at pressure of 107 N/m2. The quantity of gas taken out of the cylinder, if final pressure is 2.5×106 N/m2 will be( temperature of gas is constant)

  • 15.2 kg

  • 3.7 kg

  • zero

  • 7.5 kg


54.

The kinetic energy of one molecule of a gas at normal temperature and pressure will be (k = 8.31 J/mole K)

  • 1.7×103 J

  • 10.2×103 J

  • 3.4×103 J

  • 6.8×103 J


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55.

A sphere of diameter 0.2 m and mass 2 kg is rolling on an inclined plane with velocity u = 0.5 mls. The kinetic energy of the sphere is

  • 0.1 J

  • 0.3 J

  • 0.5 J

  • 0.42 J


56.

Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?

  • increase

  • Decrease

  • No change

  • Can't be determined


57.

Pressure of an ideal gas is increased by keeping temperature constant. What is effect on kinetic energy of molecules?

  • Increase

  • Decrease

  • No change

  • Cannot be determined


58.

For a gas RCv= 0.67. This gas is made up of molecules which are

  • monoatomic

  • polyatomic

  • mixture of diatomic and polyatomic molecules

  • diatomic


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59.

A horizontal tube of length l closed at both ends, contains an ideal gas of molecular weight M. The tube is rotated at a constant angular velocity ω about a vertical axis passing through an end. Assuming the temperature to be uniform and constant. If p1 and p2 denote the pressure at free and the fixed end respectively, then choose the correct relation

  • p2p1 = e2 l22RT

  • p1p2 = eM ωl2RT

  • p1p2 = eω l M3RT

  • p2p1 = eM2 ω2 l23RT


A.

p2p1 = e2 l22RT

Consider the diagram

Consider the elementary part of thickness dx at a distance x from the axis of rotation, then force on this part

          Adp = (dm)ω2 x              ......(i)

where, dm = mass of element

           ω = angular velocity

Now, from ideal gas equation

           pV = nRT 

   R - Avagadro constant

   V = Volume

    n = Number of moles of gas

           pA dx = dmMRT

⇒         dm = MpART dx                 ......(ii)

From Eqs. (i) and (ii)

         Adp = M p AR Tω2 x dx

0(minimum) to l (maximum) is the limit

       p1p2dpp = 0lM ω2R T x dx

⇒      ln p = 2RTx220l

⇒     lnp2p1 = 2 l22RT 

⇒       p2p1 = eM ω2 l22RT


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60.

Two rigid boxes containing different ideal gases are placed on table. Box A contains one mole of nitrogen at temperature T0 , while box B contains 1 mole of helium at temperature 7/3 T0. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes) then the final temperature of gases, Tf in terms of T0 is

  • 2T05

  • 3 T05

  • 5 T03

  • 9 T07


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