= Solution
Use the given spherical-shell <force balance> directly; $R_{\rm out}=30$ nm is the \b[loaded] radius. The radial enlargement is
$$
\Delta R=R_{\rm out}-R_0=\frac{2\pi R_{\rm out}^2p}{k}.
$$
Since $1\ \mathrm{pN\,nm^{-2}}=10^6$ Pa, the <pressure> is $p=40\times101325\ \mathrm{Pa}=4.053\ \mathrm{pN\,nm^{-2}}$. Therefore
$$
\boxed{\Delta R=\frac{2\pi(30\ \mathrm{nm})^2(4.053\ \mathrm{pN\,nm^{-2}})}
{5\times10^5\ \mathrm{pN\,nm^{-1}}}=0.04584\ \mathrm{nm}.}
$$
The diameter increases by $2\Delta R=0.09168$ nm, and $R_0=29.95416$ nm. The relative radial or diametral enlargement is about $1.53\times10^{-3}$, only $0.153\%$. Under this stipulated shell model the container is therefore very stiff despite the substantial internal <pressure>.
The factor of two follows from the particular <elastic energy> convention implicit in the question: differentiating $k(R-R_0)^2$ gives $2k(R-R_0)$. Replacing it by the derivative of $k(R-R_0)^2/2$ would change the answer incorrectly. The calculation treats the wall by an effective spherical radius; local deformation or discrete facets of an icosahedral shell are not resolved by this one-coordinate model.
Back to article page
