Answer
a) Yes
b) $55\;\rm cm$
c) $9.6$
d) $0.038\%,\;0.114\%$
Work Step by Step
a)
First of all, we need to find the volume of one molecule at STP and then find its radius to compare the distances to find which is greater than the other, the diameter of the molecule, or the separation distance between the molecules.
$$PV=NkT$$
$N=1\;\rm molecule$, $V$ is the volume of space allowed to one molecule from an ideal gas at STP.
$$V=\dfrac{kT}{P}$$
The volume of space is assumed to be spherical, so $V=\frac{4}{3}\pi r^3$
$$\frac{4}{3}\pi r^3=\dfrac{kT}{P}$$
$$r=\sqrt[3]{\dfrac{3kT}{4P\pi }}\tag 1$$
Plugging the known;
$$r=\sqrt[3]{\dfrac{3\cdot 1.38\times 10^{-23}\cdot 273}{4\pi\cdot 1.013\times 10^5 }}$$
$$r= 2.07\times 10^{-9}\;\rm m $$
Thus, the intermolecular distance $D$ for one molecule is twice this radius (diameter), so
$$D= 4.14\times 10^{-9}\;\rm m $$
To compare the intermolecular distance to the diameter of the typical gas, we need to divide $D$ by the diameter of the typical gas $d$.
$$\dfrac{D}{d}=\dfrac{4.14\times 10^{-9}}{0.3\times10^{-9}}=\bf 13.8$$
So, yes, the assumption is reasonable since the distance between molecules is about 14 times larger than the diameter of one molecule.
b)
The next ping-ball will be far about 14 times the diameter of one ping-ball.
$$x=13.8\cdot 4=\boxed{\bf55.2\;\rm cm}$$
c)
using equation (1);
$$r=\sqrt[3]{\dfrac{3\cdot 1.38\times 10^{-23}\cdot 273}{4\pi\cdot 3\cdot 1.013\times 10^5 }}$$
$$r= 1.436\times 10^{-9}\;\rm m $$
Thus, the intermolecular distance $D$ for one molecule is twice this radius (diameter), so
$$D= 2.87\times 10^{-9}\;\rm m $$
Hence,
$$\dfrac{D}{d}=\dfrac{2.87\times 10^{-9}}{0.3\times10^{-9}}=\bf 9.6$$
d)
Estimating the total volume percentage of gas that is taken up by gas's molecules themselves.
$$\dfrac{V_{\rm molecules}}{V_{\rm gas}}\times 100\%$$
Noting that the volume of the gas is the volume of the molecule plus the volume of distance between two molecules. In other words, the volume of molecules is part of the volume of the gas.
$$\dfrac{\frac{4}{3}\pi \left(\frac{d}{2}\right)^3}{\frac{4}{3}\pi r^3}\times 100\%=\dfrac{ d^3 }{ 8r^3}\times 100\%$$
$${\text {At 1 atm} }=\dfrac{ \left(0.3\times10^{-9}\right)^3 }{ 8\left(2.07\times 10^{-9}\right)^3}\times 100\%$$
$${\text {At 1 atm} }=\boxed{\bf 0.038\%}$$
$${\text {At 3 atm} }=\dfrac{ \left(0.3\times10^{-9}\right)^3 }{ 8\left(1.436\times 10^{-9}\right)^3}\times 100\%$$
$${\text {At 3 atm} }=\boxed{\bf 0.114\%}$$
