Question

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of ${{27}^{\circ }}C$. After some oxygen is withdrawn from the cylinder the gauge pressure drops to 11 atm and its temperature drops to ${{17}^{\circ }}C$. Estimate the mass of oxygen taken out of the cylinder.
$\left( R=8.31Jmo{{l}^{-1}}{{K}^{-1}},\text{ }molecular\text{ }mass\text{ }of\text{ }{{O}_{2}}=32u \right)$

Answer 1

Hint: We will use the ideal gas law equation that is PV= nRT which shows that the product of the pressure of a gas times the volume of a gas is constant for a given sample of a gas. We will find the mass of oxygen taken out of the cylinder by calculating the initial mass of oxygen in the cylinder and subtracting it from the final mass of oxygen in the cylinder.

Complete Step by step solution:
We are being provided with the volume of oxygen, let it be ${{V}_{1}}=30\text{ }litres=30\times {{10}^{-3}}{{m}^{3}}$
Gauge pressure, let it be ${{P}_{1}}=15\text{ }atm=15\times 1.013\times {{10}^{5}}Pa$
Temperature, ${{T}_{1}}={{27}^{\circ }}C=300K$
Universal gas constant, ${{R}_{1}}=8.314\text{ }Jmo{{l}^{-1}}{{K}^{-1}}$
Let us take the initial number of moles of oxygen gas in the cylinder as ${{n}_{1}}$,
The gas equation is written as:
\[{{P}_{1}}{{V}_{1}}={{n}_{1}}R{{T}_{1}}\]
So, putting all the values in the given equation we get:
\[\begin{align}
& {{P}_{1}}{{V}_{1}}={{n}_{1}}R{{T}_{1}} \\
& \therefore {{n}_{1}}=\dfrac{{{P}_{1}}{{V}_{1}}}{R{{T}_{1}}} \\
\implies & \dfrac{(15.195\times {{10}^{5}}\times 30\times {{10}^{-3}})}{(8.314\times 300)} \\
\implies & 18.276 \\
\end{align}\]
But, as we know that ${{n}_{1}}=\dfrac{{{m}_{1}}}{M}$
Where,
${{m}_{1}}$= initial mass of oxygen
M= Molecular mass of oxygen=32g
Therefore,
\[\begin{align}
& {{m}_{1}}={{N}_{1}}M=18.276\times 32 \\
\implies & 584.84g \\
\end{align}\]
After sometime when the oxygen is withdrawn from the cylinder, the pressure and the temperature reduces. That is Volume, ${{V}_{2}}=30\text{ }litres=30\times {{10}^{-3}}{{m}^{3}}$
Gauge pressure, ${{P}_{2}}=11\text{ }atm=11\times 1.013\times {{10}^{5}}Pa$
Temperature, ${{T}_{2}}={{17}^{\circ }}C=290K$
Now, let ${{n}_{2}}$ be the number of moles of oxygen left in the cylinder.
The gas equation can be written as:
${{P}_{2}}{{V}_{2}}={{n}_{2}}R{{T}_{2}}$
So, putting all the values in the given equation we get:
$\begin{align}
& {{P}_{2}}{{V}_{2}}={{n}_{2}}R{{T}_{2}} \\
& {{n}_{2}}=\dfrac{{{P}_{2}}{{V}_{2}}}{R{{T}_{2}}} \\
\implies & \dfrac{\left( 11.143\times {{10}^{5}}\times 30\times {{10}^{-30}} \right)}{\left( 8.314\times 290 \right)} \\
 \implies & 13.86 \\
\end{align}$
But, ${{n}_{2}}=\dfrac{{{m}_{2}}}{M}$
Where, ${{m}_{2}}$ is the mass of oxygen remaining in the cylinder
Therefore,
$\begin{align}
& {{m}_{2}}={{n}_{2}}M \\
\implies & 13.86\times 32 \\
\implies & 453.1g \\
\end{align}$Now, we can say that the mass of oxygen taken out of the cylinder is given by the relation:
Initial mass of oxygen in the cylinder - Final mass of oxygen in the cylinder
\[\begin{align}
\implies & {{m}_{1}}-{{m}_{2}} \\
\implies & 584.84g-453.1g \\
\implies & 131.74g \\
\implies & 0.131kg \\
\end{align}\]

Hence, we can conclude that 0.131 kg of oxygen is taken out of the cylinder.

Note: We must not forget to convert the unit of volume, pressure and temperature to ${{m}^{3}},Pa,K$. And should not forget to write the unit in the solution.