NEET · Physics · STD 11 - 12 . kinetic theory of gases
An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature \(27^{\circ} C\). The mass of the oxygen withdrawn from the cylinder is nearly equal to:
[Given, \(R=\frac{100}{12} J mol ^{-1} K^{-1}\), and molecular mass of \(O_2=32,1\) atm pressure \(=1.01 \times 10^5 N / m\)]
- A \(0.125 \text{ kg}\)
- B \(0.144 \text{ kg}\)
- C \(0.116 \text{ kg}\)
- D \(0.156 \text{kg}\)
Answer & Solution
Correct Answer
(C) \(0.116 \text{ kg}\)
Step-by-step Solution
Detailed explanation
\(n _{ f }=\frac{ P _{ f } V }{ RT } \quad\left[ P _{ abs }=12 \text{ atm}\right] \)
\( =\frac{12 \times 1.01 \times 10^5 \times 30 \times 10^{-3}}{\left(\frac{100}{12}\right) \times 300} \)
\( =14.544 \)
\( \therefore \text { moles removed }=18.2-14.544=3.656 \)
\( \therefore \text { mass removed }=3.656 \times \frac{32}{1000}\text{ kg}=0.116\text{ kg}\)
\( =\frac{12 \times 1.01 \times 10^5 \times 30 \times 10^{-3}}{\left(\frac{100}{12}\right) \times 300} \)
\( =14.544 \)
\( \therefore \text { moles removed }=18.2-14.544=3.656 \)
\( \therefore \text { mass removed }=3.656 \times \frac{32}{1000}\text{ kg}=0.116\text{ kg}\)
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