AP EAMCET · Chemistry · States of Matter
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 \(\left.\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\).
- A \(1\)
- B \(10\)
- C \(100\)
- D \(0.1\)
Answer & Solution
Correct Answer
(B) \(10\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & Z=\frac{P V}{R T} \Rightarrow V=\frac{Z R T}{P}=\frac{(1.1)(0.082)(300)}{2.706} \\ & =10 L\end{aligned}\)
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