How much work will be done in increasing the diameter of a soap bubble from $2cm$ to $5cm$ ?
Answer
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Hint
First calculate the change in surface area of the soap bubble due to enlargement. Then assuming the value of surface tension of soap solution to be $30dynes/cm$, use appropriate formula to calculate the work done.
$\Rightarrow dW = S \times dA$ where $S$ is the surface tension of the soap solution and $dA$ is the change in surface area.
Complete step by step answer
In this problem we need to account for the surface tension of the soap bubble. Surface tension is defined as a force per unit length which is exerted in the tangential plane to the surface of a liquid and in a direction perpendicular to any line drawn on the surface.
Let $r$ be the initial radius of the bubble in the process of enlargement and let it increase to a radius of $r + dr$(considering the soap bubbles to be spherical in shape).
So, the increase in surface area is
$\Rightarrow dA = $$4\Pi {\left( {r + dr} \right)^2} - 4\Pi {r^2} = 8\Pi rdr$ (neglecting the term $4\Pi d{r^2}$ since it is very small)
Now work done for this increase is given by $S \times 8\Pi rdr$ where $S$ is the surface tension of the soap solution which has a value of $30dynes/cm$
Since there are two free surfaces, the work done is $dW = 16\Pi Srdr$
Therefore the total work required to enlarge the soap bubble from a radius of $1cm$to $2.5cm$ is
$\Rightarrow W = 16\Pi S\int\limits_1^{2.5} {rdr} $
$\Rightarrow W = \dfrac{{16\Pi S\left( {{{2.5}^2} - 1} \right)}}{2} = 3.96 \times {10^3}erg $
To convert from $erg$ to $J$ we need the multiply the solution with a value of ${10^{ - 7}}$
Doing that we get, $W = 3.96 \times {10^{ - 4}}J$
So, the net work done to increase the diameter of the soap bubble from $2cm$to $5cm$ is $3.96 \times {10^{ - 4}}J$.
Note
It is observed that when this enlargement is done at a faster rate, the amount of work done increases. This is because if the process is made faster, the temperature of the soap bubble falls. Hence, some of the work done is used up to compensate for the fall in temperature. Another interesting fact is that soap bubbles exist and water bubbles don’t because the surface tension of water is very high which makes it behave like it has an elastic membrane at the surface, thus destabilizing the bubble. But when detergent is added to water, it reduces the surface tension making it possible for the bubbles to remain stable.
First calculate the change in surface area of the soap bubble due to enlargement. Then assuming the value of surface tension of soap solution to be $30dynes/cm$, use appropriate formula to calculate the work done.
$\Rightarrow dW = S \times dA$ where $S$ is the surface tension of the soap solution and $dA$ is the change in surface area.
Complete step by step answer
In this problem we need to account for the surface tension of the soap bubble. Surface tension is defined as a force per unit length which is exerted in the tangential plane to the surface of a liquid and in a direction perpendicular to any line drawn on the surface.
Let $r$ be the initial radius of the bubble in the process of enlargement and let it increase to a radius of $r + dr$(considering the soap bubbles to be spherical in shape).
So, the increase in surface area is
$\Rightarrow dA = $$4\Pi {\left( {r + dr} \right)^2} - 4\Pi {r^2} = 8\Pi rdr$ (neglecting the term $4\Pi d{r^2}$ since it is very small)
Now work done for this increase is given by $S \times 8\Pi rdr$ where $S$ is the surface tension of the soap solution which has a value of $30dynes/cm$
Since there are two free surfaces, the work done is $dW = 16\Pi Srdr$
Therefore the total work required to enlarge the soap bubble from a radius of $1cm$to $2.5cm$ is
$\Rightarrow W = 16\Pi S\int\limits_1^{2.5} {rdr} $
$\Rightarrow W = \dfrac{{16\Pi S\left( {{{2.5}^2} - 1} \right)}}{2} = 3.96 \times {10^3}erg $
To convert from $erg$ to $J$ we need the multiply the solution with a value of ${10^{ - 7}}$
Doing that we get, $W = 3.96 \times {10^{ - 4}}J$
So, the net work done to increase the diameter of the soap bubble from $2cm$to $5cm$ is $3.96 \times {10^{ - 4}}J$.
Note
It is observed that when this enlargement is done at a faster rate, the amount of work done increases. This is because if the process is made faster, the temperature of the soap bubble falls. Hence, some of the work done is used up to compensate for the fall in temperature. Another interesting fact is that soap bubbles exist and water bubbles don’t because the surface tension of water is very high which makes it behave like it has an elastic membrane at the surface, thus destabilizing the bubble. But when detergent is added to water, it reduces the surface tension making it possible for the bubbles to remain stable.
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