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The container shown in figure has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2 = 3.0 litre.

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प्रश्न

The container shown in figure has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2 = 3.0 litre. The chambers contain µ1 = 4.0 and µ2 = 5.0 moles of a gas at pressures p1 = 1.00 atm and p2 = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

V1 V2
µ1, p1 µ2
  p2
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उत्तर

Consider the diagram,

p1V1 p2V2
µ1 µ2

Given, V1 = 2.0 L, V2 = 3.0 L

µ1 = 4.0 mol, µ2 = 5.0 mol

p1 = 1.00 atm, p2 = 2.00 atm

For chamber 1, p1, V1 = µ1RT1

For chamber 2, p2, V2 = µ2RT2

When the partition is removed the gases get mixed without any loss of energy. The mixture Now attains a common equilibrium pressure and the total volume of the system is the sum of the volume of individual chambers V1 and V2.

So, µ = µ1 + µ2, V = V1 + V2

From the kinetic theory of gases,

For `l` mole `pV = 2/3 E`  ......`[(E = "Translational"),("Kinetic energy")]`

For `µ_1` moles, `p_1V_1 = 2/3 µ_1E_1`

For `µ_2` moles, `p_2V_2 = 2/3 µ_2E_2`

The total energy is `(µ_1E_1 + µ_2E_2) = 3/2 (p_1V_1 + p_2V_2)`

From the above relation, `pV = 2/3 E_"total" = 2/3 µE_"per mole"`

`p(V_1 + V_2) = 2/3 xx 3/2 (p_1V_1 + p_2V_2)`

`p = (p_1V_1 + p_2V_2)/(V_1 + V_2)`

= `((1.00 + 2.0 + 2.00 xx 3.0)/(2.0 + 3.0))` atm

= `8.0/5.0` 

= 1.60 atm

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Interpretation of Temperature in Kinetic Theory - Introduction of Kinetic Theory of an Ideal Gas
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अध्याय 13: Kinetic Theory - Exercises [पृष्ठ ९४]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 11
अध्याय 13 Kinetic Theory
Exercises | Q 13.20 | पृष्ठ ९४

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