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Calculate the pressure of 1.5 mole of gas having volume $3 \mathrm{dm}^3$ at $300 \mathrm{~K}\left(\mathrm{R}=0.0821 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
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The correct answer is:
$12.32 \mathrm{~atm}$
$\begin{aligned} & \mathrm{PV}=\mathrm{nRT} \\ & \mathrm{P} \times 3=1.5 \times 0.0821 \times 300 \\ & \mathrm{P}=12.32 \mathrm{~atm} .\end{aligned}$
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