Question

A $20-cm$-diameter cylinder that is $40cm$long contains $50g$of oxygen gas at$20°C$

a. How many moles of oxygen are in the cylinder?

b. How many oxygen molecules are in the cylinder?

c. What is the number density of the oxygen?

d. What is the reading of a pressure gauge attached to the tank

Short answer

a. There are $\mathbf{1}\mathbf{.}\mathbf{56}\mathbf{}\mathit{m}\mathit{o}\mathit{l}$moles of oxygen in the cylinder

b. There are $\mathbf{9}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$oxygen molecules in the cylinder

c. The number density of the oxygen is$\mathbf{7}\mathbf{.}\mathbf{52}\mathbf{\times }{\mathbf{10}}^{\mathbf{25}\mathbf{}}{\mathit{m}}^{\mathbf{3}}$.

d. The reading of a pressure gauge attached to the tank is$\mathbf{2}\mathbf{.}\mathbf{03}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{P}\mathit{a}$

Step-by-step solution

: Given information and Formula used 

Given : Diameter of the cylinder : $20-cm$

Length of the cylinder :$40cm$

Mass of Oxygen :$50g$

Temperature :$20°C$

Theory used :

The ideal gas law is :$PV=nRT$

Finding out how many moles of oxygen are in the cylinder. 

(a) The mass of oxygen gas is $M=50g$. The molecular mass of oxygen is $m=16u$. Oxygen is a diatomic gas data-custom-editor="chemistry" $\mathrm{O}₂$, so its molecular mass is$m=2(16u)=32u$. If we convert this to kg, we get the mass of an oxygen molecule by

$$\begin{gathered} m=32ux\frac{1.66x{10}^{-27}kg}{1u} \\ =5.31x{10}^{-26}kg \end{gathered}$$
Thus, the number of molecules$N$is given by $N=\frac{M}{m}$
Now we substitute the values for $m\mathrm{and}M$ into this equation and obtain$N$:
role="math" localid="1648403857083" $$\begin{gathered} N=\frac{50x{10}^{-3}kg}{5.31x{10}^{-26}kg} \\ =9.4x{10}^{23}\mathrm{molecules}. \end{gathered}$$
Knowing the number of molecules, we obtain the number of moles $n$by using Avogadro's number$N_A$
$$\begin{gathered} n=\frac{N}{N_A} \\ =\frac{9.4x{10}^{23}\mathrm{molecules}}{6.023x{10}^{23}molecules/mole} \\ =\mathbf{1}\mathbf{.}\mathbf{56}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e} \end{gathered}$$

Finding out how many oxygen molecules are in the cylinder.

(b) In part (a), we obtain the number of molecules $N$since it is equal to the mass of the gas $M$ divided by the mass of a molecule $m$, or the molar mass
role="math" localid="1648404286717" $$\begin{gathered} N=\frac{M}{m} \\ =\frac{50\times {10}^{-3}kg}{5.31x{10}^{-26}kg} \\ =\mathbf{}\mathbf{9}\mathbf{.}\mathbf{4}\mathbf{}\mathit{x}\mathbf{}\mathbf{1023}\mathbf{}\mathbf{molecules}\mathbf{} \end{gathered}$$

Finding the number density of the oxygen .

(c)Number density is known as the number of atoms or molecules per cubic meter in a system and determines how densely the atoms in the system are bonded together.
Number density is given by $numberdensity=\frac{N}{V}$
Where $V\mathrm{is}\mathrm{the}\mathrm{volume}\mathrm{and}N$is the number of molecules. The gas is in the form of a cylinder with a radius of

$$\begin{gathered} r=\frac{d}{2} \\ =\frac{20cm}{2} \\ =10cm \\ =0.10m \end{gathered}$$
The volume of the cylinder is given by
$$\begin{gathered} V=\pi r²L \\ =\pi (0.10m)²(0.40m) \\ =0.0125m³ \end{gathered}$$
Now we substitute the values for $N\mathrm{and}V$into number density formula to get the number of oxygen molecules :

$$\begin{gathered} NumberDensity=\frac{9.4x{10}^{23}\mathrm{molecules}}{0.0125m³} \\ =\mathbf{7}\mathbf{.}\mathbf{52}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{25}}\mathbf{}{\mathit{m}}^{\mathbf{-}\mathbf{3}} \end{gathered}$$

Finding the reading of a pressure gauge attached to the tank  

(d) Converting$T_C=20°C$ into Kelvin, we get :
$$\begin{gathered} T_k=T_c+273 \\ =20°C+273 \\ =293K \end{gathered}$$
Now we substitute the values for role="math" localid="1648404800025" $n,R,T\mathrm{and}V$into $p=\frac{nRT}{V}$we get the pressure :

$$\begin{gathered} p=\frac{(1.56mol)(8.314J/mol.K)(293K)}{0.0125m³} \\ =3.04x{10}^{5}Pa \end{gathered}$$
This is the pressure that the gas exerts on the wall of the tank. The pressure gauge shows the pressure difference between the inside of the tank and the outside of the tank. Inside the tank the pressure is $p=3.04x{10}^{5}Pa$ and outside the tank the pressure is equal to atmospheric pressure, i.e. $P_{out}=1.01x10{}^{5}Pa$. Therefore, we obtain the pressure of the manometer by
$$\begin{gathered} ∆p_{gauge}=p-p_{out} \\ =3.04x{10}^{5}Pa-1.01\times {10}^{5}Pa \\ =\mathbf{}\mathbf{2}\mathbf{.}\mathbf{03}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{P}\mathit{a}\mathbf{} \end{gathered}$$