In an isothermal process, two soap bubble of radius a and b combine and form a bubble of radius c. If the external pressure is p, then prove that the surface tension of the soap solution is T-(p(c^3-a^3-b^3))/(4(a^2+b^2-c^2)).

In an isothermal process, two soap bubble of radius a and b combine an

1.	In an isothermal process, two soap bubble of radius a and b combine and form a bubble of radius c. If the external pressure is p, then prove that the surface tension of the soap solution is T-(p(c^3-a^3-b^3))/(4(a^2+b^2-c^2)).
Answer» Solution :We know , the EXCESS pressure inside the soap BUBBLE = internal pressure - external pressure. `THEREFORE`For the bubble of radius a, excess pressure, `(4T)/(a)=p_a-p` `therefore p_a=(p+(4T)/(a))` Similarly, for the bubble of radius b, `p_b=(p+(4T)/(b))` For the bubble of radius c, `p_c=(p+(4T)/(c ))` Boyle.s law is applicable in isothermal process. According to this law. `p_a V_a+p_b V_b=p_c V_c` or, `(p+(4T)/(a))xx(4)/(3)pia^3+(p+(4T)/(b))xx(4)/(3)pib^3` `=(p+(4T)/(c ))xx(4)/(3)pic^3` or, `(p+(4T)/(a))a^3+(p+(4T)/(b))b^3=(p+(4T)/(c ))c^3` or, `4T(a^2+b^2-c^2)=p(c^3-a^3-b^3)` `therefore T=(p(c^3-a^3-b^3))/(4(a^2+b^2-c^2))`.