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Question: Two soap bubbles coalesce to form a single bubble. if \(V\) is the subsequent change in volume of co...

Two soap bubbles coalesce to form a single bubble. if VV is the subsequent change in volume of contained air and SS the change in total surface area, TTis the surface tension and PP atmospheric pressure, which of the following relations is correct ?
(A) 4PV+3ST=04PV + 3ST = 0
(B) 3PV+4ST=03PV + 4ST = 0
(C) 2PV+3ST=02PV + 3ST = 0
(D) 3PV+2ST=03PV + 2ST = 0

Explanation

Solution

Excess pressure formed due to shape of bubble. Use an isothermal equation under which temperature is constantPV = constant{\text{PV = constant}} and conservation of energy.

Complete step by step answer:
Let r1{r_1} is the radius of first bubble before coalesce.
r2{r_2} is the radius of second bubble before coalesce.
r3{r_3} is the radius of sphere after coalesce.
P1{P_1} is excess pressure inside the first bubble.
P1=P+4Tr1{P_1} = P + \dfrac{{4T}}{{{r_1}}} … (i)
Here PPis atmospheric pressure,TT is surface tension.
Similarly
P2{P_2}is excess pressure inside the second bubble
P2=P+4Tr2{P_2} = P + \dfrac{{4T}}{{{r_2}}} … (ii)
and P3{P_3}is excess pressure inside the bubble which formed after coalesce of first and second bubble is
P3=P+4Tr3{P_3} = P + \dfrac{{4T}}{{{r_3}}} … (iii)
and V1{V_1}is volume of first before coalesce
V1=43πr13{V_1} = \dfrac{4}{3}\pi r_1^3 … (iv)
V2{V_2}is volume of second bubble before coalesce
V2=43πr23{V_2} = \dfrac{4}{3}\pi r_2^3 … (v)
V3{V_3}is volume of bubble formed after coalesce
V3=43πr33{V_3} = \dfrac{4}{3}\pi r_3^3 … (vi)
According to energy conservation.
P1V1+P2V2=P3V3{P_1}{V_1} + {P_2}{V_2} = {P_3}{V_3} … (vii)
Use above equation in equation (vii)
(P+4Tr1)43πr13+(P+4Tr2)43πr23=(P+4Tr3)43πr33\left( {P + \dfrac{{4T}}{{{r_1}}}} \right)\dfrac{4}{3}\pi r_1^3 + \left( {P + \dfrac{{4T}}{{{r_2}}}} \right)\dfrac{4}{3}\pi r_2^3 = \left( {P + \dfrac{{4T}}{{{r_3}}}} \right)\dfrac{4}{3}\pi r_3^3
P[43πr13+43πr2343πr33]+4T3[4πr12+4πr224πr32]=0P\left[ {\dfrac{4}{3}\pi r_1^3 + \dfrac{4}{3}\pi r_2^3 - \dfrac{4}{3}\pi r_3^3} \right] + \dfrac{{4T}}{3}\left[ {4\pi r_1^2 + 4\pi r_2^2 - 4\pi r_3^2} \right] = 0 … (viii)
In equation (viii)
V=V = change in volume of bubble
=[initialvolume][findvolume]= \left[ {initial\,\,volume} \right] - \left[ {find\,\,volume} \right]
V=[43πr13+43πr23][43πr33]V = \left[ {\dfrac{4}{3}\pi r_1^3 + \dfrac{4}{3}\pi r_2^3} \right] - \left[ {\dfrac{4}{3}\pi r_3^3} \right] … (ix)
S=S = change in surface area of bubble
=[initialsurfacearea][findsurfacearea]= \,\left[ {initial\,\,surface\,\,area} \right] - \left[ {find\,\,surface\,\,area} \right]
=[4πr12+4πr22][4πr32]= \left[ {4\pi r_1^2 + 4\pi r_2^2} \right] - \left[ {4\pi r_3^2} \right] … (x)
Hence equation (viii) reduces to
PV+4TS3=0PV + \dfrac{{4T\,S}}{3} = 0
3PV+4TS=03PV + 4T\,S = 0

So, the correct answer is “Option B”.

Note:
Let ppand pa{p_a}be the pressure inside the bubble and outside the bubble respectively. The bubble can exist only ifp>pap > {p_a}. The difference in pressure (ppa)\left( {p - {p_a}} \right)is known as excess pressure inside the bubble.