Pressure inside a soap bubble made in a vacuum A question I came across:
A soap bubble is made in vacuum by blowing an ideal diatomic gas in it. Assume the heat capacity of the soap film is much greater than that of the gas in the bubble. What will be the molar heat capacity of the gas in the bubble?
The correct answer is given as $4R$ (R being the ideal gas constant), and I was able to reach this answer using the ideal gas equation to find $dV$, substituting into the first law of thermodynamics and using the fact that $C_v$ for a diatomic gas is $5R/2$.
However, to reach this answer I had to take the pressure inside the bubble as $8\sigma/r$ ($\sigma$ being the surface tension of the soap solution) instead of $4\sigma/r$ which is what I would usually take for the excess pressure inside a soap bubble, and since it is in vacuum I can take as the absolute pressure inside the bubble.
Is there a reason for this which I'm missing, or an error in the question/answer?
 A: Let say you have blown the air bubble to $V$ volume, $P$ Pressuse, $T$ temperature and with $n$ moles of that diatomic gas in it. Also the surface tension of the bubble be $\sigma$
Now, To calculate the molar specific heat i.e $C$, Let's say you gave some heat to the bubble i.e $Q$, due to which there will be some change in pressure, volume and temperature let's say $dP, dV, dT$ respectively and also the number of moles $(n)$ will be constant
As you stated by First Law of Thermodynamics,
$$ Q = dU + W $$
where all the terms have their usual meaning
$$ Q = nCdT$$
$$ dU = nC_v dT$$
Work can be calculated by two ways,
$$ W = PdV$$
$$or$$
$$ W = 2\sigma dA$$
$$W = 2\sigma\times  8\pi r dr$$
where $dA$ is elemental change in surface area
I am going with second you can verify the result for first also
$$nCdT = 2\sigma \times 8\pi rdr + nC_v dT $$
$$C = \mathrm{\frac{2\sigma\ \times 8\pi rdr}{n dT}} + C_v \qquad(1)$$
From ideal gas equation,
$$ P V = nR T$$
differentiating,
$$PdV + VdP = nRdT$$
$$ndT = \mathrm{\frac{PdV+VdP}{R}} $$
So equation (1) becomes,
$$C = \mathrm{\frac{2\sigma\ \times 8\pi rdr\times R}{PdV+VdP}} + C_v$$
$$P = \mathrm{\frac{4\sigma}{r}},V=\mathrm{\frac{4}{3}}\pi r^3, C_v = 5R/2$$
differentiate and substitute you will get the answer.
