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At 300 K, one mole of a gas present in a 10 L flask exerted a pressure of 2.706 atm. What is its compressibility factor (Z) ?
(Given $R = 0.082 L atm {mol}^{- 1} K^{- 1}$ )

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1.1

1.5

0.91

1.0

answer is D.

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Detailed Solution

$Z = \frac{pV}{nRT} Z = \frac{2.706 \times 10}{1 \times 0.082 \times 300} Z = \frac{27.06}{24.6} = 1.1$

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