Physics, asked by kondasrinivas059, 9 months ago

Two soap bulles A and B are kept in a closed
chamter where the air is maintained at presua
8 N/m3 The radii of bubbles A and B are 2cm.
and 4cm, respectively surface tension of the
soap water used to make bubbles irooaam
If an and no are the no of moles of air in the
bubbles A and B then the nationale is Neglect
the effect of gravity]​

Answers

Answered by ayushphalke608
0

Explanation:

Excess pressure for soap bubble, ΔP=R4T  where T= surface tension.

Here, PA=P0+RA4T=8+0.024(0.04)=16N/m2

and PB=P0+RB4T=8+0.044(0.04)=12N/m2

As PV=nRt,n∝PV

Thus, nBnA=PBVBPAVA=12×(4/3)π(0.04)316×(4/3)π(0.02)3=1/6

Answered by Fatimakincsem
0

Thus the ratio of nB / nA = 6

Explanation:

Given data:

  • Radius of bubble A = 2 cm = 0.02 m
  • Radius of bubble B = 4 cm = 0.04 m
  • Pressure = 8 N/m^3

ΔP = 4T / R

T = surface tension

P(A) = Po + 4T / R(A) = 8 + 4 ( 0.04) / 0.02 = 16 N/m^2

P(B) = Po + 4T / R(B) = 8 + 4 (0.04) / 0.04 = 12 N/m^2

PV = nRT

n ∝ PV

nA / nB = PAVA / PBVB

nA / nB = 16 x (4/3)π (0.02)^3 / 12 x  (4/3)π (0.04)^3

nA / nB = 1/6

nB / nA = 6

Thus the ratio of nB / nA = 6

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