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Home » An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to . Estimate the mass of oxygen taken out of the cylinder , molecular mass of .
An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27^{\circ} \mathrm{C} . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17^{\circ} \mathrm{C}. Estimate the mass of oxygen taken out of the cylinder \left(\mathbf{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right., molecular mass of \left.\mathrm{O}_{2}=32 \mathrm{u}\right).
CBSE | Class 11 | Kinetic Theory | NCERT | Physics
An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27^{\circ} \mathrm{C} . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17^{\circ} \mathrm{C}. Estimate the mass of oxygen taken out of the cylinder \left(\mathbf{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right., molecular mass of \left.\mathrm{O}_{2}=32 \mathrm{u}\right).

Volume of gas is given as V_{1}=30 litres =30 \times 10^{-3} \mathrm{~m}^{3}

Gauge pressure is given as \mathrm{P}_{1}=15 \mathrm{~atm}=15 \times 1.013 \times 10^{5} \mathrm{P} a

Temperature is given as T_{1}=27^{0} \mathrm{C}=300 \mathrm{~K}

Universal gas constant is known as \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

Let n_{1} be the initial number of moles of oxygen gas in the cylinder.

We have the gas equation as,

\mathrm{P}_{1} \mathrm{~V}_{1}=\mathrm{n}_{1} \mathrm{RT}_{1}

So,

\begin{array}{l} \mathrm{n}_{1}=\mathrm{P}_{1} \mathrm{~V}_{1} / \mathrm{RT}_{1} \\ =\left(15.195 \times 10^{5} \times 30 \times 10^{-3}\right) /(8.314 \times 300) \\ =18.276 \end{array}

But \mathrm{n}_{1}=\mathrm{m}_{1} / \mathrm{M}

Where,

m_{1} is the Initial mass of oxygen
M is the Molecular mass of oxygen having value 32 \mathrm{~g}

Thus,

m_{1}=N_{1} M=18.276 \times 32=584.84 \mathrm{~g}

The pressure and temperature drop after some oxygen is removed from the cylinder.

Volume is given as V_{2}=30 litres =30 \times 10^{-3} \mathrm{~m}^{3}

Gauge pressure is given as \mathrm{P}_{2}=11 \mathrm{~atm}
=11 \times 1.013 \times 10^{5} \mathrm{P} \mathrm{a}

Temperature is given as \mathrm{T}_{2}=17^{0} \mathrm{C}=290 \mathrm{~K}

Let the number of moles of oxygen left in the cylinder be n_{2}.

The gas equation is given as:

\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{n}_{2} \mathrm{R} \mathrm{T}_{2}

So,

\begin{array}{l} n_{2}=P_{2} V_{2} / R T_{2} \\ =\left(11.143 \times 10^{5} \times 30 \times 10^{-30}\right) /(8.314 \times 290) \\ =13.86 \end{array}

But
\mathrm{n}_{2}=\mathrm{m}_{2} / \mathrm{M}

Where,

\mathrm{m}_{2}= mass of oxygen remaining in the cylinder

Hence,

m_{2}=n_{2} \times M=13.86 \times 32=453.1 \mathrm{~g}

The mass of oxygen taken out of the cylinder can be calculated using the following formula: Mass of oxygen in the cylinder at the start – Mass of oxygen in the cylinder at the end

\begin{array}{l} =\mathrm{m}_{1}-\mathrm{m}_{2} \\ =584.84 \mathrm{~g}-453.1 \mathrm{~g} \end{array}

We get,

\begin{aligned} =& 131.74 \mathrm{~g} \\ =& 0.131 \mathrm{~kg} \end{aligned}

As a result, mass of oxygen is taken out of the cylinder is 0.131 \mathrm{~kg}.

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