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(A) For a spherical vessel of radius $r$ and wall thickness $t$,the volume of metal is $V_m = 4\pi r^2 t$. The hoop stress $\sigma$ for a thin-walled

Vedclass · Accepted Answer

$n = \frac{2}{3} \frac{V_m \sigma}{RT}$

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(A) For a spherical vessel of radius $r$ and wall thickness $t$,the volume of metal is $V_m = 4\pi r^2 t$. The hoop stress $\sigma$ for a thin-walled spherical vessel is given by $\sigma = \frac{Pr}{2t}$,where $P$ is the internal pressure. From the ideal gas law,$PV = nRT$,where $V = \frac{4}{3}\pi r^3$. Substituting $P = \frac{2t\sigma}{r}$ into the ideal gas law: $n = \frac{PV}{RT} = \frac{(2t\sigma/r)(4/3 \pi r^3)}{RT} = \frac{8\pi r^2 t \sigma}{3RT}$. Since $V_m = 4\pi r^2 t$,we have $r^2 t = \frac{V_m}{4\pi}$. Substituting this into the expression for $n$: $n = \frac{8\pi (V_m / 4\pi) \sigma}{3RT} = \frac{2 V_m \sigma}{3RT}$.

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