How many moles of dioxygen are present in $8.314 \times 10^{-3} \mathrm{~m}^3$ of it at 318 K having…

How many moles of dioxygen are present in $8.314 \times 10^{-3} \mathrm{~m}^3$ of it at 318 K having pressure $3.18 \times 10^5 \mathrm{Nm}^{-2} ?\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
  1. 0.1 mole
  2. 1.0 mole
  3. 1.5 mole
  4. 2.0 mole

Solution

$\begin{aligned} & \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{3.18 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2} \times 8.314 \times 10^{-3} \mathrm{~m}^3}{8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 318 \mathrm{~K}} \\ & =\frac{26.438 \times 10^2 \mathrm{~N} \mathrm{~m}}{8.314 \mathrm{~J} \mathrm{~mol}^{-1} \times 318}(\because 1 \mathrm{~N} \mathrm{~m}=1 \text { Joule }) \\ & \therefore \quad \mathrm{n}=\frac{26.438 \mathrm{~J} \times 10^2}{2643.85 \mathrm{~J} \mathrm{~mol}^{-1}}=1 \mathrm{~mole} \end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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