A glass tube of uniform internal radius $(r)$ has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius $r$. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve,

Step 1: Determine the excess pressure inside a soap bubble.

A soap bubble has two liquid-gas interfaces (inner and outer surfaces). The excess pressure $\Delta P$ inside a soap bubble of radius of curvature $R$ is given by:

$$\Delta P = \frac{4T}{R}$$

where $T$ is the surface tension of the soap solution.

Step 2: Find the pressure inside the bubble at End 1.

At End 1, the bubble is hemispherical, so its radius of curvature $R_1$ is equal to the internal radius of the tube $r$.

$$R_1 = r$$

The pressure inside this bubble $P_1$ is:

$$P_1 = P_{\text{atm}} + \frac{4T}{r}$$

Step 3: Find the pressure inside the bubble at End 2.

At End 2, the bubble is sub-hemispherical, which means it is flatter and has a larger radius of curvature $R_2$ than the tube radius $r$.

$$R_2 > r$$

The pressure inside this bubble $P_2$ is:

$$P_2 = P_{\text{atm}} + \frac{4T}{R_2}$$

Step 4: Compare the pressures at both ends.

Since $R_2 > r$, the excess pressure term for End 2 is smaller than that for End 1.

$$\frac{4T}{R_2} < \frac{4T}{r}$$

Comparing the absolute pressures:

$$P_1 > P_2$$

Step 5: Determine the direction of air flow and volume changes.

Air flows spontaneously from a region of higher pressure to a region of lower pressure. Since $P_1 > P_2$, air flows from End 1 to End 2. As a result, the bubble at End 1 loses air and shrinks, while the bubble at End 2 gains air and grows.

Answer: Air flows from End 1 to End 2, and the bubble at End 1 shrinks.