Work Done in Expanding a Soap Bubble:
The work \( W \) done in increasing the surface area of a soap bubble is given by:
\[ W = \Delta U = S \Delta A \] where \( S \) is the surface tension and \( \Delta A \) is the increase in surface area.
Calculate Initial and Final Surface Areas:
Initial radius \( r = 7 \, \text{cm} \).
Initial surface area \( A_i = 4\pi r^2 = 4\pi (7)^2 \, \text{cm}^2 \).
Suppose the new radius is \( R \). Then the final surface area \( A_f \) is:
\[ A_f = 4\pi R^2 \]
Calculate the Change in Surface Area \( \Delta A \):
\[ \Delta A = A_f - A_i = 4\pi R^2 - 4\pi (7)^2 = 4\pi (R^2 - 49) \]
Use the Work Done to Solve for \( R \):
Given \( W = 36960 \, \text{erg} \) and \( S = 40 \, \text{dyne/cm} \), we have:
\[ 36960 = 40 \times 4\pi (R^2 - 49) \]
Simplifying,
\[ 36960 = 160\pi (R^2 - 49) \]
Using \( \pi = \frac{22}{7} \):
\[ 36960 \, \text{erg} = \frac{40 \, \text{dyne}}{\text{cm}} 8\pi \left[(R)^2 - \left(\frac{7}{2}\right)^2\right] \, \text{cm}^2 \]
\[ r = 7 \, \text{cm} \]
Conclusion:
The new radius of the soap bubble is \( 7 \, \text{cm} \).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)