Solved Problem on Gases
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A cylinder containing molecular oxygen O2, a volume equal to 16.4 liters and pressure of 300 atmospheres, is in thermal equilibrium with the environment at 27 °C, given the universal gas constant \( R=0.082\;\frac{\text{atm.}\ell}{\text{mol.K}} \) and considering oxygen as an ideal gas, determine:
a) The number of gas moles in the cylinder;
b) The mass of the gas in the cylinder, the molar mass of the oxygen M = 16 g/mol;
c) The volume of the gas if it were the pressure of 1 atm.
Problem data:
- Volume of oxygen in the cylinder: V = 16.4 \( \ell \) ;
- Pressure of oxygen in the cylinder: p = 300 atm;
- Ambient temperature: t = 27 °C;
- Gas universal constant: \(R=0.082\;\frac{\text{atm.}\ell}{\text{mol.K}} \).
The problem says oxygen in the cylinder "is in thermal equilibrium with the environment at 27 °C", this means that the oxygen temperature is the same as the environment, so its temperature is also 27 °C, we should convert the temperature given in degrees Celsius (°C) to kelvins (K) used in the International System of Units (S.I.)
\[
\begin{gather}
T=t+273=27+273=300\;\text{K}
\end{gather}
\]
a) To calculate the number of moles, we use the Ideal Gas Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{pV=nrT} \tag{I}
\end{gather}
\]
\[
\begin{gather}
n=\frac{pV}{RT}\\[5pt]
n=\frac{300\times 16.4}{0.082\times 300}\\[5pt]
n=\frac{16.4}{0.082}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{n=200\;\text{mols}}
\end{gather}
\]
b) The molar mass of oxygen is given as 16 g/mol, as the O2 molecule has two oxygen atoms, its molecular mass will be
\[
\begin{gather}
M_{O_{2}}=2M\\[5pt]
M_{O_{2}}=2\times 16\\[5pt]
M_{O_{2}}=32\;\text{g/mol}
\end{gather}
\]
The mass of oxygen in the cylinder will be
\[
\begin{gather}
m=nM_{O_{2}}\\[5pt]
m=200\times 32\\[5pt]
m=6400\;\text{g}=6.4\times 10^{3}\;\text{g}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{m=6,4\;\text{kg}}
\end{gather}
\]
c) For a pressure of 1 atm, the volume \( V_{O_{2}} \) occupied by the gas would be using the expression (I)
\[
\begin{gather}
pV_{O_{2}}=nrT\\[5pt]
V_{O_{2}}=\frac{nRT}{p}\\[5pt]
V_{O_{2}}=\frac{200\times 0.082\times 300}{1}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{V_{O_{2}}=4920\;\ell}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .