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At $300 \mathrm{~K}$, the compressibility factor of 1 mole of a gas is 1.1. Its pressure is $2.706 \mathrm{~atm}$. What is its volume in $\mathrm{L}$ ? $\left(\right.$ Given $\left.\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$.
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The correct answer is:
$10$
$\begin{aligned} & Z=\frac{P V}{R T} \Rightarrow V=\frac{Z R T}{P}=\frac{(1.1)(0.082)(300)}{2.706} \\ & =10 L\end{aligned}$
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