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There is a soap bubble of radius 2.4 xx ...

There is a soap bubble of radius `2.4 xx 10^(-4)m` in air cylinder at a pressure of `10^(5) N//m^(2)`. The air in the cylinder is compressed isothermal until the radius of the bubble is halved. Calculate the new pressure of air in the cylinder. Surface tension of soap solution is `0.08 Nm^(-1)`.

Text Solution

Verified by Experts

Similar to question no. `37`,
`p_(2) = 8p_(1) + (245)/(c) = 8 xx 10^(5) + (24 xx .08)/(2.4 xx 10^(-4))`
`= 8.08 xx 10^(5) N//m^(2)`
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Explore conceptually related problems

There is a soap bubble of radius 2.4 xx 10^(-4)m in air cylinder at a pressure of 10^(5) N//m^(2) . The air in the cylinder is compressed isothermal untill the radius of the bubble is halved. Calculate teh new pressure of air in the cylinder. Surface tension of soap solution is 0.08 Nm^(-1) .

There is a soap bubble of radius 2.4xx10^(-4)m in air cylinder which is originally at a pressure of 10^(5)(N)/(m^(2)) . The air in the cylinder is now compressed isothermally untill the radius of the bubble is halved. (the surface tension of the soap film is 0.08Nm^(-1)) . The pressure of air in the cylinder is found to be 8.08xx10^(n)(N)/(m^(2)) . What is the value of n?

Knowledge Check

  • If the excess pressure inside a soap bubble of radius 5 mm is equal to the pressure of a water column of hight 0.8 cm, then the surface tension of the soap solution will be

    A
    980 N/m
    B
    `98 xx10^(-2)`N/m
    C
    `98 xx10^(-3)` N/m
    D
    98 N/m

  • The work done in blowing a soap bubble of radius 0.2 m is (the surface tension of soap solution being 0.06 N/m )

    A
    `192 pi xx10^(-4) J`
    B
    `280pixx10^(-4) j`
    C
    `200 pixx10^(-3)` j
    D
    None of these

  • The work done in blowing a soap bubble of 10 cm radius is (Surface tension of the soap solution is 3/100 N/m)

    A
    `75.36xx10^(-4) ` joule
    B
    `37.68xx10^(-4) ` joule
    C
    `150.72xx10^(-4)` joule
    D
    75.36 joule

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