Solution

When heat is given to a soap bubble the temperature of the air inside rises and the bubble expands but unless the bubble bursts, the amount of air inside does not change. Further we shall neglect the variation of the surface tension with temperature. Then from the gas equations
$$\Big( p_0 + \frac{4\alpha}{3r} \Big) \frac{4\pi}{3} r^3 = vRT, $$ v = constant

Differentiating
$$\Big( p_0 + \frac{8\alpha}{3r} \Big) 4\pi r^2 dr = vRdT, $$

or $$dV = 4\pi r^2 dr = \dfrac{vRdT}{p_0 + \frac{8\alpha}{3r}}$$

Now from the first law

$$d Q = v CdT = v C_V dT + \dfrac{vRdT}{p_0 + \frac{8\alpha}{3r}}. \Big( p_0 + \frac{4\alpha}{r} \Big) $$

or $$C = C_V + R \frac{p_0 + \frac{4\alpha}{r}}{p_0 \frac{8\alpha}{3r}}$$

Using
$$C_p = C_V + R, \ \\C- C_p =\dfrac{\frac{1}{2}R}{1 + \frac{3p_0 r}{8\alpha}}$$