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Surface tension of a soap bubble is 0.03 Newton per meter, what is the work done in blowing to form a soap bubble of:

a. 1 cm radius
b. 2 cm radius
c. 3 cm radius
d. 4 cm radius

1 Answer

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Final answer:

The calculations show the work done for radii of 1 cm, 2 cm, 3 cm, and 4 cm is 9.42× 10⁻¶ J, 3.77× 10⁻µ J, 8.48× 10⁻µ J and 1.51× 10⁻´ J respectively. To find the work done in blowing a soap bubble of varying radii, we use the formula W = γ ΔA, accounting for the double surface of the bubble.

Step-by-step explanation:

Calculation of Work Done in Blowing a Soap Bubble

The work done (W) to blow up a soap bubble can be calculated using the surface tension (γ) and the change in surface area (ΔA) of the bubble. The formula for the work done is:

W = γ ΔA

Since a soap bubble has two surfaces (inside and outside), the change in area is twice the surface area of a sphere with radius r. The surface area of a sphere is 4πr², therefore the change in area is 8πr².

Given that the surface tension (γ) of the soap bubble is 0.03 N/m, we can calculate the work done for different radii:

For a 1 cm radius bubble (r = 0.01 m),

the work done W = 0.03 N/m × 8π(0.01 m)².

For a 2 cm radius bubble (r = 0.02 m),

the work done W = 0.03 N/m × 8π(0.02 m)².

For a 3 cm radius bubble (r = 0.03 m),

the work done W = 0.03 N/m × 8π(0.03 m)².

For a 4 cm radius bubble (r = 0.04 m),

the work done W = 0.03 N/m × 8π(0.04 m)².

Performing the calculations:

W = 0.03 N/m × 8π(0.01 m)²

= 9.42× 10⁻¶ J

W = 0.03 N/m × 8π(0.02 m)

= 3.77× 10⁻µ J

W = 0.03 N/m × 8π(0.03 m)²

= 8.48× 10⁻µ J

W = 0.03 N/m × 8π(0.04 m)²

= 1.51× 10⁻´ J

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