Question

Calculate the ratio of change in the mass of the molecules of a gas to the initial mass, if its
speed is reduced to half and the ratio of initial and final pressure is 3:4.

Hint : P =1/3 M/V.v²
​

Answer 1

Given

• Speed is reduced to half
• Initial P : Final P = 3:4

To Find

• Ratio of the change in mass of the molecules

Solution

☯ P = ⅓ × (m/volume) × v²

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✭ According to the Question :

➞ Pᵢ/Pբ = ¾

➞ { (⅓ × m₁/volume × v²) ÷ (⅓ × m₂/volume × [v/2]²) } = ¾

➞ (m₁ × v²) ÷ (m₂ × v²/4) = ¾

➞ m₂ = 16m₁/3

So then,

➞ ∆m = m₂ - m₁

➞ ∆m = 16m₁/3 - m₁

➞ ∆m = (16m₁-3m₁)/3

➞ ∆m = 13m₁/3

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✭ Required Radio :

• ∆m/m₁ = (13m₁/3) ÷ m₁
• ∆m/m = 13/3

∴ The ratio of the change in the mass of the molecule is 13 : 3

Answer 2

Answer:

Required Answer is as follows :-

$\sf \dfrac{P _{i}}{ P_{f}} = \dfrac{3}{4}$

3/4 = (⅓ × m₁/Volume × V²)/(⅓ × m₂/Volume) × (V/2)²

¾ = m₁/Volume × (V)²/m₂/Volume × (V/2)²

¾ = m₁ × (V)²/m₂ × (V²/4)

¾ = m₁/m₂ × 4

m₂ = 4 × 4m₁/3 = 16m₁/3

Now,

∆m = m₂ - m₁

∆m = 16m₂/3 - m₁ = 13m₁/3

Ratio = (13m₁/3)/ m₁

Ratio = 13/3

Ratio = 13:3

Know More :-

Mass => It is used to measure it's resistance to its acceleration. SI unit of mass is Kg.