Answer
$\mathrm{a}.\quad 3.9\times 10^{24}$ atoms
$\mathrm{b}.\quad $increase factor: $1.26$
Work Step by Step
Use $ \quad PV=NkT \quad (17-2)$ to express $N=\displaystyle \frac{PV}{kT}$
($k =1.38\times 10^{-23}\ \mathrm{J}/\mathrm{K}$ is Boltzmann's constant.)
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$V_{sphere}=\displaystyle \frac{4}{3}\pi r^{3},\ \quad \mathrm{T}_{\mathrm{K}}=273.15+\mathrm{T}_{\mathrm{C}}.$
$a.$
$N=\displaystyle \frac{PV}{kT}=\frac{P(\frac{4}{3}\pi r^{3})}{kT}$
$=\displaystyle \frac{(2.4\times 10^{5}\ \mathrm{P}\mathrm{a})[\frac{4}{3}\pi(0.25\ \mathrm{m})^{3}]}{(1.38\times 10^{-23}\ \mathrm{J}/\mathrm{K})(273.15+18\ \mathrm{K})}$
$=3.9\times 10^{24}$ atoms
$\mathrm{b}.$
From $N=\displaystyle \frac{PV}{kT}$, when P and T are fixed, N is proportional to V.
If N increases by factor 2, V increases by factor 2.
$\mathrm{V}_{\mathrm{f}}=2\mathrm{V}_{\mathrm{i}}$
$\displaystyle \frac{4}{3}\mathrm{r}_{\mathrm{f}}^{3}\pi=2(\frac{4}{3}\mathrm{r}_{\mathrm{i}}^{3}\pi)\qquad$ solve for $\mathrm{r}_{\mathrm{f}}$
$\mathrm{r}_{\mathrm{f}}^{3}=2\mathrm{r}_{\mathrm{i}}^{3}$
$\mathrm{r}_{\mathrm{f}}=2^{1/3}\mathrm{r}_{\mathrm{i}}$
$\mathrm{r}_{\mathrm{f}}=1.26\mathrm{r}_{\mathrm{i}}$
