
What happened to the energy when two small bubbles join to form a bigger one?
A. Energy is released
B. Energy is absorbed
C. Both (A) and (B).
D. None of these
Answer
161.4k+ views
Hint: For liquid bubbles some work is required to be done against the intermolecular force of attraction since molecules on the liquid bubbles surface experience net downward force. when two small bubbles join to form a bigger one, the total surface area of the bigger bubble will be decreased.
Complete answer:
Generally molecules on the liquid surface experience a net downward force, as a result when two small bubbles with higher surface area bring together to form a bigger one, surface area is decreased. As a result energy is liberated in the form of heat by the liquid.
Let the radius of two small bubbles be ${{r}_{1}}$ and the radius of the one big bubble be ${{R}_{1}}$.
As it is given two small bubbles join to form one big bubble, the total initial and final volume will be same.Considering the bubbles to be spherical, then Initial
volume of the two small bubbles is $2\times \dfrac{4}{3}\pi r_{1}^{3}$ …….(1)
Final volume of the bigger bubble is $\dfrac{4}{3}\pi R_{1}^{3}$ ……… (2)
As both the initial and final volume are same,
$\therefore 2\times \dfrac{4}{3}\pi r_{1}^{3}=\dfrac{4}{3}\pi R_{1}^{3}$
Or, ${{R}_{1}}=\sqrt[3]{2}{{r}_{1}}$ ……..(3)
Now Initial surface area of the two bubble$=2\times 4\pi r_{1}^{2}$ ………..(4)
Calculating total surface area of the big bubble,
Final surface area of the bigger bubble$=4\pi R_{1}^{2}$
Putting the value of ${{R}_{1}}$from equation (3), we get:
Then, Final surface area $={{(2)}^{\dfrac{2}{3}}}\times 4\pi r_{1}^{2}$ ……..(5)
Now if we Compare initial and final surface area, we will see,
$2\times4\pi r_{1}^{2}>{{(2)}^{\dfrac{2}{3}}}\times 4\pi r_{1}^{2}$
From the above comparison, when surface area is decreased in case of a bigger bubble, energy is released in the form of heat.
Thus, the correct option is (A).
Note: As we all know bubbles are spherical in shape because there is attractive force called surface tension, here surface tension tends to minimize the surface area and to compensate it the groups of molecules form the best possible shape that is sphere.
Complete answer:
Generally molecules on the liquid surface experience a net downward force, as a result when two small bubbles with higher surface area bring together to form a bigger one, surface area is decreased. As a result energy is liberated in the form of heat by the liquid.
Let the radius of two small bubbles be ${{r}_{1}}$ and the radius of the one big bubble be ${{R}_{1}}$.
As it is given two small bubbles join to form one big bubble, the total initial and final volume will be same.Considering the bubbles to be spherical, then Initial
volume of the two small bubbles is $2\times \dfrac{4}{3}\pi r_{1}^{3}$ …….(1)
Final volume of the bigger bubble is $\dfrac{4}{3}\pi R_{1}^{3}$ ……… (2)
As both the initial and final volume are same,
$\therefore 2\times \dfrac{4}{3}\pi r_{1}^{3}=\dfrac{4}{3}\pi R_{1}^{3}$
Or, ${{R}_{1}}=\sqrt[3]{2}{{r}_{1}}$ ……..(3)
Now Initial surface area of the two bubble$=2\times 4\pi r_{1}^{2}$ ………..(4)
Calculating total surface area of the big bubble,
Final surface area of the bigger bubble$=4\pi R_{1}^{2}$
Putting the value of ${{R}_{1}}$from equation (3), we get:
Then, Final surface area $={{(2)}^{\dfrac{2}{3}}}\times 4\pi r_{1}^{2}$ ……..(5)
Now if we Compare initial and final surface area, we will see,
$2\times4\pi r_{1}^{2}>{{(2)}^{\dfrac{2}{3}}}\times 4\pi r_{1}^{2}$
From the above comparison, when surface area is decreased in case of a bigger bubble, energy is released in the form of heat.
Thus, the correct option is (A).
Note: As we all know bubbles are spherical in shape because there is attractive force called surface tension, here surface tension tends to minimize the surface area and to compensate it the groups of molecules form the best possible shape that is sphere.
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