Answer
See the detailed answer below.
Work Step by Step
a) We know that the number of moles is given by
$$n=\dfrac{m}{M}$$
where $m$ is the whole mass of the sample and $M$ is the atomic mass of the molecule.
Hence, the atomic mass of the Oxygen gas $O_2$ is twice the atomic mass of the Oxygen atom $O$.
$$n=\dfrac{m}{2M_{O}}=\dfrac{50}{2(16)}=\color{red}{\bf 1.56}\;\rm mol$$
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b) We know that the number of molecules is given by
$$N=nN_{\rm A}$$
where $n$ is the number of moles, and $N_{\rm A}$ is Avogadro's number.
$$N=(1.56)(6.02\times 10^{23}) =\color{red}{\bf 9.39\times 10^{23}}\;\rm molecule$$
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c) The number density of oxygen is given by
$$\rho=\dfrac{N}{V}$$
where $V$ is the volume of the oxygen which is the volume of the cylinder.
$$\rho=\dfrac{N}{Ah}=\dfrac{N}{\pi r^2 h}$$
Plugging the known;
$$\rho =\dfrac{9.39\times 10^{23}}{\pi (10\times 10^{-2})^2 (40\times 10^{-2})}$$
$$\rho=\color{red}{\bf7.47 \times 10^{25}}\;\rm molecule/m^3$$
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d) We know, for an ideal gas, that the absolute pressure is given by
$$PV=nRT$$
$$P=\dfrac{nRT}{V}=P_{\rm gauge}+P_a$$
Hence,
$$P_{\rm gauge} =\dfrac{nRT}{V}-P_a$$
$$P_{\rm gauge} =\dfrac{nRT}{\pi r^2 h}-P_a$$
Plugging the known;
$$P_{\rm gauge} =\dfrac{(1.56)(8.31)(20+273)}{\pi (10\times 10^{-2})^2(0.4)}-(1.013\times 10^5)$$
$$P_{\rm gauge} =\color{red}{\bf 2.01\times 10^5}\;\rm P_a$$
