Question

When two soap bubbles touch, the larger is inflated by the smaller until they form a single bubble.

(a) What is the gauge pressure inside a soap bubble with a 1.50-cm radius?
(b) Inside a 4.00-cm-radius soap bubble?
(c) Inside the single bubble they form if no air is lost when they touch?

A. (a) 4.37 kPa, (b) 0.163 kPa, (c) Pressure change but not specified
B. (a) 8.73 kPa, (b) 0.327 kPa, (c) 0 Pa
C. (a) 0.327 kPa, (b) 4.37 kPa, (c) Pressure change but not specified
D. (a) 0.163 kPa, (b) 8.73 kPa, (c) 0 Pa

Answer 1

To find the gauge pressure inside soap bubbles of various radii, a standard formula involving surface tension is used. Correct calculations for specific radii and constants given in reference tables would provide the pressures for cases (a) and (b), while the combined bubble in (c) would change depending on volume and surface area.

Step-by-step explanation:

The question pertains to calculating the gauge pressure inside soap bubbles of different radii and understanding how they change when two bubbles combine. According to the principle of surface tension and pressure in a bubble, the gauge pressure is given by the formula P = 4γ/r, where P is the pressure, γ (gamma) is the surface tension of the soap water, and r is the radius of the bubble.

For a soap bubble with a 1.50-cm radius (a), we can calculate the gauge pressure P using the typical surface tension for soapy water from a reference table. Similarly, for a 4.00-cm-radius soap bubble (b), and we can find the pressure change in the single bubble formed post combination (c).

Correct calculations considering that the surface tension for soapy water is constant and using the given radii will result in specific numerical values for the pressures in (a) and (b). As for (c), when two bubbles combine without air loss, the resulting pressure would depend on the combined volume and surface area, consistent with the law of conservation of air mass and surface tension properties.