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Question

The pressure exerted by a perfect gas is equal to


A

23 of mean kinetic energy.

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B

half of mean kinetic energy per unit volume.

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C

mean kinetic energy per unit volume

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D

23 of mean kinetic energy per unit volume.

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Solution

The correct option is D

23 of mean kinetic energy per unit volume.


Step 1: Given parameters

The given gas is an ideal gas.

Step 2: Formulas Used:

  1. P=13ρν2 ; Where P is pressure, ρ is density, and v is r.m.s velocity.
  2. ρ=MV ; Where ρis the density, M is the Mass, and V is the volume.
  3. K.E=12Mν2; Where K.E. is kinetic energy, M is the Mass, and v is r.m.s velocity.

Step 3: Derive the relation between kinetic energy and pressure.

According to the kinetic theory of gasses,

P=13ρν2...1

The density of a gas, ρ=MV...2

Substitute the value of density from equation (2) in equation (1).

P=Mν23V...3

The kinetic energy of gases is expressed as,

K.E=12Mν2...4

Divide equation (3) by (4).

PK.E=Mν23V12Mν2

PK.E=23V

P=23×K.EV

Hence, the pressure exerted by a perfect gas is equal to two-thirds of kinetic energy per unit volume.

Hence, option (D) is correct.


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