Under isothermal conditions, two soap bubbles | vedclass

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Question

(B) For isothermal conditions, the total number of moles of air remains constant. The pressure inside a soap bubble of radius $r$ is given by $P_{in}

Vedclass · Accepted Answer

$\frac{P(c^{3}-a^{3}-b^{3})}{4(a^{2}+b^{2}-c^{2})}$

Vedclass · Answer

(B) For isothermal conditions, the total number of moles of air remains constant. The pressure inside a soap bubble of radius $r$ is given by $P_{in} = P + \frac{4T}{r}$, where $P$ is the external pressure and $T$ is the surface tension. Using the ideal gas law $PV = nRT$, since $T$ (temperature) is constant, $PV$ is proportional to the number of moles. Thus, $P_1V_1 + P_2V_2 = P_cV_c$. Substituting the values: $(P + \frac{4T}{a}) \cdot \frac{4}{3}\pi a^3 + (P + \frac{4T}{b}) \cdot \frac{4}{3}\pi b^3 = (P + \frac{4T}{c}) \cdot \frac{4}{3}\pi c^3$. Dividing by $\frac{4}{3}\pi$: $P(a^3 + b^3 - c^3) = 4T(c^2 - a^2 - b^2)$. Rearranging for $T$: $T = \frac{P(a^3 + b^3 - c^3)}{4(c^2 - a^2 - b^2)} = \frac{P(c^3 - a^3 - b^3)}{4(a^2 + b^2 - c^2)}$.

Column-$I$	Column-$II$
$(a)$ Liquid drop in air	$(i)$ $\frac{4T}{R}$
$(b)$ Bubble of liquid in air	$(ii)$ $\frac{2T}{R}$
	$(iii)$ $\frac{2R}{T}$

Under isothermal conditions, two soap bubbles of radii $a$ and $b$ coalesce to form a single bubble of radius $c$. If the external pressure is $P$, then the surface tension of the bubbles is:

Similar Questions

$1000$ droplets of water, each having a diameter of $2 \ mm$, coalesce to form a single drop. Given the surface tension of water is $0.072 \ N/m$. The energy loss in the process is

An air bubble of radius $1.0 \ mm$ is observed at a depth of $20 \ cm$ below the free surface of a liquid having surface tension $0.095 \ J/m^2$ and density $10^3 \ kg/m^3$. The difference between pressure inside the bubble and atmospheric pressure is . . . . . . $N/m^2$. (Take $g = 10 \ m/s^2$)

Formation of bubbles is in Column-$I$ and the pressure difference between them is given in Column-$II$. Match them appropriately.

Column-$I$ Column-$II$

$(a)$ Liquid drop in air $(i)$ $\frac{4T}{R}$

$(b)$ Bubble of liquid in air $(ii)$ $\frac{2T}{R}$

$(iii)$ $\frac{2R}{T}$

The pressure inside a small air bubble of radius $0.1 \, mm$ situated just below the surface of water will be equal to. [Take surface tension of water $T = 70 \times 10^{-3} \, N/m$ and atmospheric pressure $P_0 = 1.013 \times 10^5 \, N/m^2$]

The radius of a soap bubble is increased from $R$ to $2R$. The work done in this process in terms of surface tension $S$ is: (in $\pi R^2 S$)

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