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JEE Advanced levelNumerical answerDifficulty 3 of 3rg173

Nitrogen (molar mass $28 g mo l^{- 1}$ ) has density $12.80 g L^{- 1}$ at $10.0 atm$ and $300 K$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , its compressibility factor (three decimals) is

Numerical answer type, tolerance 0.002.

Answer: 0.888

Z = \frac{PM}{ρRT} = \frac{10.0 \times 28}{12.80 \times 0.0821 \times 300} = \frac{280}{315.264} = 0.8881

, i.e.

0.888

.

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