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Pressure Of An Ideal Gas - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • 18 Questions around this concept.

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For Brownian motion of particle, match columns I and II.

               Factor                                                                    Effect

     (i) Decrease in size of Brownian particle            (P) Increase of Brownian motion

      (ii) Decrease in density of medium                    (Q) Decrease of Brownian motion

       (iii) Increase in temperature of medium             (R) Unaffected

       (iv) Increase in viscosity of the medium  

  An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross-section 8.0 x 10-3 m2. Initially, the gas is at 300K and occupies a volume of 2.4 x 10-3 m3 and the spring is in its relaxed state as shown in the figure. The gas is heated by a small heater until the piston moves out slowly by 0.1 m. The force constant of the spring is 8000 N/m and the atmospheric pressure is 1.0 x 105 N/m2. The cylinder and the piston are thermally insulated.  The piston and the spring are massless and there is no friction between the piston and the cylinder. The final temperature of the gas will be :

(Neglect the heat loss through the lead wires of the heater. The heat capacity of the heater coil is also negligible)

A monoatomic ideal gas is expanded adiabatically to \mathrm{n} times its initial volume, The ratio of the final rate of collision of molecules with a unit area of container walls to the initial rate will be

A gas at pressure \mathrm{P_{0}} is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be

The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressures P1 and P2 are shown in the figure. Here

Gas at pressure \mathrm{P_{0}}  is contained in a vessel. If the masses of all the molecules are doubled and their speed is halved, the resulting pressure \mathrm{P} will be equal to

When an ideal gas at pressure P, temperature T and volume V is isothermally compressed to a V/n, its pressure becomes Pi. If the gas is compressed adiabatically to V/n, its pressure becomes Pa. The ratio Pi/Pa is

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An ideal gas is initially at temperature T and volume V . Its volume is increased by ΔV due to an increase in temperature ΔT, with pressure remaining constant. The quantity δ=ΔV/VΔT varies with temperature as :

An ideal gas expands isothermally from a volume V1 and V2 and then compressed to original volume V1 adiabatically. Initial pressure is P1 and final pressure is P3. The total work done is W Then:

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A cylindrical tube of uniform cross-sectional area A is fitted with two air-tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the pressure of the gas is P0 and temperature is T0, atmospheric pressure is also P0.Now, the temperature of the gas is increased to 2T0, the tension of the wire will be

Concepts Covered - 1

Pressure of an ideal gas

Consider an ideal gas (consisting of N molecules each of mass m) enclosed in a cubical box of side L as shown in the below figure.

1. Instantaneous velocity- 

Any molecule of gas moves with velocity v in any direction
where v=vxi^+vyj^+vzk^
And Due to the random motion of the molecule

vx=vy=vzAsv=vx2+vy2+vz2v=3vx2=3vy2=3vz2

2. The time during a collision-Time between two successive collisions with the wall A1

 I.e Δt= Distance travelled by molecule between two successive collision  Velocity of molecule  or Δt=2Lvx

3. Collision frequency (n) : It means the number of collisions per second.

 I.e n=1Δt=vx2L

4. Change in momentum: This molecule collides with A1 wall (A1) with velocity vx and rebounds with velocity ( vx ) The change in momentum of the molecule is given by

Δp=(mvx)(mvx)=2mvx
 

As the momentum remains conserved in a collision,

 

Δpsystem =0Δpsystem =Δpmolecule +Δpwall =0Δpwall =Δpmolecule 

the change in momentum of wall A1 is Δp=2mvx
5. Force on the wall: Force exerted by a single molecule on the A1 wall is equal to the rate at which the momentum is transferred to the wall by this molecule.
i.e. FSingle molecule =ΔpΔt=2mvx(2L/vx)=mvx2L

The total force on the wall A1 due to N molecules

Fx=mLvx2=mL(vx12+vx22+vx32+)=mNLvx2

where vx2= mean square of x component of the velocity.
6. Pressure-As pressure is defined as force per unit area, hence the pressure on A1 wall

Px=FxA=mNALvx2=mNVvx2 As vx2=vy2=vz2 So v2=vx2+vy2+vz2vx2=vy2=vz2=v23
 

So Total pressure inside the container is given by

P=13mNVv2=13mNVvrms2( where vrms=v2)


Using total mass =M=mN
Pressure due to an ideal gas is given as

P=13ρvrms2=13(MV)vrms2

where
m= mass of one molecule
N= Number of the molecule

vrms2=v12+v22+n

vrms=RMS velocity

 

 

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Pressure of an ideal gas

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Pressure of an ideal gas

Physics Part II Textbook for Class XI

Page No. : 331

Line : 15

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